Summary to follow. 15 syllabus statements (10 HL) · 32 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 15 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Kepler's first law
Kepler's first law
Each planet moves in an elliptical orbit with the Sun at one focus of the ellipse (not at its centre). A circle is the special case of an ellipse in which the two foci coincide; in IB calculations orbits are taken as circular.
Kepler's second law
The line joining a planet to the Sun sweeps out equal areas in equal times. In an elliptical orbit the planet therefore moves fastest when nearest the Sun and slowest when furthest from it. In a circular orbit the law means the speed is constant.
Kepler's third law
The square of the time period T of a planet's orbit is proportional to the cube of its mean distance from the Sun: T² ∝ r³, so T²/r³ is the same for every planet of the Sun. For a circular orbit, equating GMm/r² to mv²/r with v = 2πr/T gives T²/r³ = 4π²/GM, so the constant depends on the central mass M: moons of Jupiter share one value, planets of the Sun another.
Time period of an orbit
The time T taken for one complete orbit. For a circular orbit of radius r at speed v, T = 2πr/v. SI unit: second (s), although days or years are often quoted.
Students often think All orbiting bodies travel at the same speed, so the period is simply proportional to the orbital radius (T ∝ r), and a change of orbit does not change the speed. In fact No. Orbital speed decreases with distance from the Sun: from T² ∝ r³ and v = 2πr/T, v ∝ 1/√r. So outer planets are both slower and on longer paths, and T rises as r^(3/2), not in proportion to r.
Students often think T²/r³ is a universal constant, so the same value applies to planets round the Sun, moons round Jupiter and satellites round the Earth. In fact No. For circular orbits T²/r³ = 4π²/GM, where M is the mass of the central body. It is the same for all planets of the Sun, but different for the moons of Jupiter or satellites of the Earth.
Newton's universal law of gravitation
Newton's universal law of gravitation
Every point mass attracts every other point mass with a force along the line joining them, proportional to the product of their masses and inversely proportional to the square of their separation: F = Gm₁m₂/r². The two bodies exert equal and opposite forces on each other (a Newton's third law pair), whatever their masses.
Universal gravitational constant G
The constant in Newton's law of gravitation, G = 6.67 × 10⁻¹¹ N m² kg⁻². It is the same everywhere, which is why the law is called universal. Its small value is why gravitational forces between everyday objects are tiny.
Students often think The more massive body pulls harder, so the Earth pulls the Moon with a larger force than the Moon pulls the Earth. In fact No. The two forces are a Newton's third law pair: equal in magnitude and opposite in direction. F = Gm₁m₂/r² contains the product of both masses, so it is the same number for each body.
Students often think Only large bodies such as planets have gravity; a small body exerts no gravitational pull on a much larger one. In fact Yes. Every mass attracts every other mass. A small body pulls a large one with exactly the same force as the large body pulls it.
Point mass
Point mass
An idealised body whose whole mass is located at a single point, so that its size and shape play no part. 'Point' refers to size, not to how much mass the body has: the Sun can be modelled as a point mass.
Uniform sphere treated as a point mass
At any point outside a sphere of uniform density (or any spherically symmetric body), the gravitational field is exactly the same as if the whole mass were concentrated at the centre. This holds however close to the surface the point is, so r in F = Gm₁m₂/r² is measured centre to centre. Any extended body can also be treated as a point mass when the separation is very large compared with its size.
Students often think The distance r in F = GMm/r² and g = GM/r² (and, at HL, V_g = −GM/r and the orbital and escape speed equations) is the height above the surface, or the gap between the surfaces of two bodies. In fact No. For a spherical body, r is measured from its centre: r = R + h, the radius of the body plus the height above its surface.
Students often think A body can be treated as a point mass because its mass is small compared with the other body's, so the point-mass model is justified by a large mass ratio. In fact No. 'Point' refers to size and shape, not to how much mass there is. A uniform sphere can be treated as a point mass however massive it is, and a small but irregular body close by cannot.
Gravitational field strength g
Gravitational field strength g
The gravitational force per unit mass experienced by a small point mass placed at a point: g = F/m. It is a vector, directed towards the mass producing the field. SI unit: N kg⁻¹, which is equivalent to m s⁻², because g is also the acceleration of free fall at that point. A small test mass is specified so that it does not disturb the field it is measuring.
Field strength of a point or spherical mass
At distance r from the centre of a point mass or uniform sphere of mass M, g = GM/r². The field is radial and follows an inverse-square law: doubling r reduces g to one quarter. It depends on M and r, not on the mass of the object placed in the field.
Radial and uniform fields
Around a point or spherical mass the field is radial: directed towards the centre with a magnitude that falls with distance. Close to the surface of a planet, over a region small compared with its radius, the field is assumed uniform: the same magnitude and direction everywhere.
Resultant field on the line joining two bodies
Between two bodies the two fields point in opposite directions along the line joining them, so the resultant is the difference of their magnitudes. It is zero where GM₁/x² = GM₂/(d − x)², a point nearer the less massive body.
Students often think Gravitational field strength is the gravitational force (weight) on an object, so a heavier object at the same point experiences a greater field strength. In fact No. g is the force per unit mass on a small test mass (N kg⁻¹); the force on an object of mass m is F = mg (N). At a given point g is the same for every object, while the force is proportional to the object's mass.
Students often think The gravitational field strength (or acceleration of free fall) has the value 9.8 m s⁻² at every point, on any planet and at any height. In fact No. 9.8 N kg⁻¹ is the value at the Earth's surface. g = GM/r², so it is different on other planets and decreases with height above the Earth.
Gravitational field lines
Gravitational field lines
Lines showing the direction of the gravitational force on a small mass at each point. For a gravitational field they point towards the mass producing the field; they never cross; and their spacing shows the field strength (closer lines, stronger field). Around an isolated uniform sphere they are radial and point inwards; in a uniform field they are parallel and equally spaced. Field lines are not the paths that moving masses follow.
Students often think Gravitational field lines point outwards from a mass, in the same way as electric field lines point away from a positive charge. In fact No. Field lines show the direction of the force on a small mass, and gravity is always attractive, so gravitational field lines point towards the mass producing the field.
Students often think Gravity is a form of magnetism, so the Earth's gravitational field lines loop out of one pole and back into the other, like its magnetic field. In fact No. Gravity is an attraction between masses; magnetism is a separate interaction between magnets and currents. Gravity acts on all matter, including non-magnetic materials, and its field lines are radial, not loops.
Gravitational potential energy E_p of a system HL
Gravitational potential energy E_p of a system
The work done to assemble the system from infinite separation of its components. Because the components attract, bringing them together from infinity requires negative work from an external agent, so E_p is negative and becomes less negative (increases) as the components move apart. It is a property of the system, not of one body. SI unit: joule (J).
Students often think The minus sign in E_p = −GMm/r (or V_g = −GM/r) shows a direction, towards the planet, as it does for a vector. In fact No. Potential energy is a scalar and has no direction. The minus sign says that the value is below zero, because E_p is defined as zero at infinite separation and the attraction makes it lower at any finite separation.
Students often think Positive work must be done to bring two masses together from infinite separation, and the minus sign in E_p is just a convention. In fact No. The masses attract, so the gravitational force does positive work as they approach. To bring them together without them gaining kinetic energy, an external agent must hold them back and does negative work. This is why E_p is negative.
E_p of a two-body system HL
E_p of a two-body system
E_p = −Gm₁m₂/r, where r is the separation of the centres of mass of the two bodies. E_p = 0 at infinite separation and is negative at any finite separation. The change on moving from r₁ to r₂ is ΔE_p = Gm₁m₂(1/r₁ − 1/r₂), positive when the separation increases. Near a surface, over small heights, this reduces to ΔE_p ≈ mgΔh.
Students often think Moving a mass away from a planet decreases its gravitational potential energy (or potential), because the size of GMm/r (or GM/r) gets smaller. In fact No. E_p = −GMm/r becomes less negative as r increases, so it increases. Moving a mass away from a planet always increases E_p, just as lifting does near the surface.
Students often think ΔE_p = mgΔh, with g equal to its surface value, gives the change in gravitational potential energy for any change of height, even thousands of kilometres. In fact No. ΔE_p = mgΔh assumes g is constant, which holds only for heights much smaller than the planet's radius. For large changes of height ΔE_p = GMm(1/r₁ − 1/r₂) must be used.
Gravitational potential V_g HL
Gravitational potential V_g
The work done per unit mass in bringing a small mass from infinity to a point: V_g = −GM/r for a point or spherical mass M. It is a scalar, zero at infinity and negative everywhere else, rising towards zero with distance. SI unit: J kg⁻¹. The gravitational potential energy of a mass m at that point is mV_g.
Students often think Gravitational field strength and gravitational potential are the same kind of quantity, so g can be defined as potential energy per unit mass and V_g as force per unit mass. In fact No. Field strength g is a vector, the force per unit mass (N kg⁻¹). Potential V_g is a scalar, the work done per unit mass bringing a mass from infinity (J kg⁻¹). g is minus the gradient of V_g.
Students often think Gravitational potential is the potential energy of a mass placed at a point, so V_g and E_p are interchangeable and the mass need not be included. In fact No. V_g is energy per unit mass (J kg⁻¹), a property of a point in the field. E_p = mV_g is the energy of a particular mass at that point (J). Work done moving a mass is mΔV_g, not ΔV_g.
Gravitational potential gradient HL
Gravitational potential gradient
The rate of change of gravitational potential with distance, ΔV_g/Δr. The field strength is minus the potential gradient: g = −ΔV_g/Δr. The minus sign means the field points in the direction in which the potential decreases, towards the mass. On a graph of V_g against r, the magnitude of g is the gradient of the curve.
Students often think The gravitational field points in the direction of increasing potential, so the minus sign in g = −ΔV_g/Δr can be ignored. In fact No. g = −ΔV_g/Δr: the field points in the direction in which the potential decreases, towards the mass, which is towards more negative V_g.
Students often think The field strength or work done can be found from the value of the gravitational potential (or potential energy) at one point, rather than from its change between two points. In fact No. The field strength depends on how fast V_g changes (ΔV_g/Δr), and the work done moving a mass depends on the difference mΔV_g. A single value of V_g at one point gives neither.
Work done moving a mass in a gravitational field HL
Work done moving a mass in a gravitational field
The work done by an external force in moving a mass m between two points, with no change in kinetic energy, is W = mΔV_g, where ΔV_g = V_final − V_initial. It is positive when the mass moves to a higher (less negative) potential and does not depend on the path taken.
Students often think A satellite only needs to be lifted to the height of its orbit; once there it stays up, so the energy needed is just its gain in gravitational potential energy. In fact No. A satellite lifted to a height and left at rest would fall back. It must also be given the orbital speed √(GM/r), so the energy needed is the gain in E_p plus the orbital E_k.
Students often think The kinetic energy of an orbiting satellite is GMm/r, taken directly from mv² = GMm/r without the ½. In fact No. GMm/r² = mv²/r gives mv² = GMm/r, so E_k = ½mv² = GMm/2r. Forgetting the ½ doubles the kinetic energy.
Equipotential surface HL
Equipotential surface
A surface on which every point has the same gravitational potential. Around an isolated uniform sphere the equipotentials are concentric spheres; in a uniform field they are parallel planes. Drawn at equal intervals of potential, they are closer together where the field is stronger, so around a planet they get further apart with distance.
Students often think Equipotential surfaces at equal intervals of potential are equally spaced around a planet, as they are in a uniform field. In fact No. In a radial field the potential changes more slowly with distance further out, so equal steps in V_g need larger and larger steps in r: the surfaces get further apart with distance.
Students often think Equipotentials are another name for field lines, so around a planet they are radial lines pointing towards the centre. In fact No. Field lines show the direction of the force; equipotentials join points of equal potential. They cross at right angles: around a planet the field lines are radial and the equipotentials are concentric spheres.
Field lines and equipotentials HL
Field lines and equipotentials
Gravitational field lines cross equipotential surfaces at right angles. No work is done in moving a mass along an equipotential, because the gravitational force has no component along it; the field is not zero there, it is perpendicular to the surface.
Students often think Because the potential does not change on an equipotential surface, there is no gravitational field there, so no force acts on a mass on it. In fact No. On an equipotential the potential is constant ALONG the surface, so there is no field component along it. The field is perpendicular to the surface and is not zero.
Students often think Whenever a force acts on a moving object it does work on it, whatever the angle between the force and the motion. In fact No. A force does work only if it has a component along the displacement: W = Fs cos θ. A force perpendicular to the motion does no work.
Escape speed v_esc HL
Escape speed v_esc
The minimum speed a body must have at a point to reach infinite separation from a mass M with no further energy input: ½mv_esc² − GMm/r = 0, so v_esc = √(2GM/r). It depends on the distance r from the centre and not on the mass of the escaping body; it is lower at greater distances. SI unit: m s⁻¹.
Students often think Each planet has one escape speed, equal to its surface value, which applies at any height. In fact No. v_esc = √(2GM/r) depends on the distance r from the planet's centre. The familiar value (11.2 km s⁻¹ for the Earth) is for the surface; at greater distances the escape speed is lower.
Students often think The escape speed at a point is twice the orbital speed there, so escaping needs four times the orbital kinetic energy. In fact No. At the same r, v_esc = √(2GM/r) and v_orbital = √(GM/r), so v_esc = √2 × v_orbital ≈ 1.41 v_orbital. The kinetic energy needed to escape is twice the orbital kinetic energy.
Orbital speed v_orbital HL
Orbital speed v_orbital
The speed of a body in a circular orbit of radius r around a large mass M. The gravitational force provides the centripetal force: GMm/r² = mv²/r, so v_orbital = √(GM/r). It is independent of the orbiting body's mass and decreases as r increases. At any r, v_esc = √2 × v_orbital.
Energy of a satellite in a circular orbit
For a satellite of mass m in a circular orbit of radius r: E_k = GMm/2r, E_p = −GMm/r and total energy E = −GMm/2r = −E_k. A larger orbit has less kinetic energy but more (less negative) potential energy and more total energy. To escape from the orbit, the total energy must be raised to zero.
Students often think A satellite in orbit has escaped the planet's gravity, so the speed needed to orbit is the same as the escape speed. In fact No. An orbiting satellite is held in its orbit by gravity, which provides its centripetal force. Its speed √(GM/r) is less than the escape speed √(2GM/r); its total energy is negative, so it is bound.
Students often think A satellite in a higher orbit moves faster, because it has more energy and a rocket must be fired to get it there. In fact No. v_orbital = √(GM/r), so the higher orbit is slower. Its total energy is higher, but that is because its potential energy has increased by more than its kinetic energy has fallen.
Orbital decay due to atmospheric drag HL
Orbital decay due to atmospheric drag
A small viscous drag force from the thin upper atmosphere removes energy from a satellite, so its total energy −GMm/2r becomes more negative and its orbital radius decreases. Because v_orbital = √(GM/r), its speed increases: as it spirals slowly inwards, the loss of E_p is twice the gain in E_k, the other half being dissipated by the drag.
Students often think Drag always slows a moving object, so a satellite affected by atmospheric drag both loses height and slows down. In fact No. Drag removes energy, so the satellite drops to lower orbits, but in each lower orbit its speed is higher: v = √(GM/r). The loss of E_p is twice the gain in E_k.
Students often think Drag reduces the satellite's speed, and since v = √(GM/r) means slower orbits are larger, the satellite moves out to a larger orbit. In fact No. Drag removes energy, and a satellite with less total energy (−GMm/2r more negative) is in a smaller orbit. The satellite spirals inwards and its speed increases.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Kepler's third law describes the orbits of the planets around the Sun. Taking each orbit as a circle of radius r with time period T, which statement expresses the law?
Answer and reasoning
T/r has the same value for every planet that orbits the Sun. — A student who thinks all planets move at the same speed would expect T to be proportional to r and picks this. Outer planets are also slower: from T² ∝ r³ and v = 2πr/T, v ∝ 1/√r, so T grows as r^(3/2): it is T²/r³, not T/r, that is constant.
T has the same value for every planet, whatever its distance. — A student who pictures the planets turning together like points on a wheel picks this. The planets are not linked; each has the period set by gravity at its own distance, from 88 days for Mercury to 165 years for Neptune.
T²/r³ has the same value for every planet that orbits the Sun. — This is Kepler's third law, T² ∝ r³. For circular orbits it follows from GMm/r² = m(2π/T)²r, which gives T²/r³ = 4π²/GM, the same for every planet because M is the Sun's mass.
T²/r³ has one value for all orbiting bodies, whatever they orbit. — A student who treats the constant as universal picks this. T²/r³ = 4π²/GM contains the mass M of the central body, so the moons of Jupiter share a different value from the planets of the Sun.
2 The Earth exerts a gravitational force of 2.0 × 10²⁰ N on the Moon. The mass of the Moon is about 1.2% of the mass of the Earth. Which statement about the gravitational force that the Moon exerts on the Earth is correct?
Answer and reasoning
It is 2.0 × 10²⁰ N, and it pulls the Earth towards the Moon. — The two forces are a Newton's third law pair: equal in size, opposite in direction. F = Gm₁m₂/r² contains the product of both masses, so the same value applies to each body. The force on the Earth points towards the Moon.
It is about 1.2% of 2.0 × 10²⁰ N, as the Moon has far less mass. — A student who thinks the more massive body pulls harder scales the force by the mass ratio. The forces are equal; the Moon's small mass means the same force gives it a large acceleration, while the Earth's acceleration is about 81 times smaller.
It is zero, as a small body cannot attract a much larger one. — A student who thinks only big bodies have gravity picks this. Every mass attracts every other mass, and the Moon's pull on the Earth is strong enough to raise the ocean tides.
It is zero, as the Moon has no air and so has no gravity. — A student who thinks gravity needs an atmosphere picks this. Gravity depends only on mass and distance and acts across a vacuum: the Moon has no atmosphere, yet astronauts on it had weight, and it pulls on the Earth.
Working Newton's third law: F(Moon on Earth) = F(Earth on Moon) = 2.0 × 10²⁰ N, opposite in direction. F = Gm₁m₂/r² is symmetric in m₁ and m₂, so the mass ratio (1.2%) does not enter.
3 Which statement defines the gravitational field strength g at a point?
Answer and reasoning
The gravitational force acting on any object placed at that point, in newtons. — A student who confuses field strength with weight picks this. The force depends on the object's mass; the field strength is the force PER KILOGRAM, so it is the same for every object at that point.
The gravitational force per unit mass on a small point mass placed at that point. — This is the definition in the guide: g = F/m for a small test mass, one small enough not to disturb the field. The unit is N kg⁻¹, equivalent to m s⁻².
The acceleration of free fall at that point, which is 9.8 m s⁻² at every point. — A student who treats g as a universal constant picks this. 9.8 m s⁻² is the value at the Earth's surface; g = GM/r², so it is smaller at height and different on other planets.
The force per unit mass on a small point mass, directed away from the mass producing the field. — A student who carries over the outward direction of electric field lines from a positive charge picks this. Gravity always attracts, so the force on a small mass, and therefore the field, points towards the mass producing the field.
Working Definition: g = F/m, the force per unit mass on a small test mass; unit N kg⁻¹ (= m s⁻²). The value 9.8 m s⁻² applies only at the Earth's surface.
4 A student sketches the gravitational field lines around an isolated planet that is a uniform sphere. Which description of a correct sketch is right?
Answer and reasoning
Straight radial lines pointing inwards to the planet's centre, closer together near its surface. — The force on a small mass points towards the centre of the planet, so the lines are radial and point inwards. They converge on the planet, so they are closest together near the surface, where the field is strongest.
Straight radial lines pointing outwards from the planet's centre, closer together near it. — A student who borrows the pattern for a positive charge picks this. Gravity always attracts, so the arrows on gravitational field lines point towards the mass, not away from it.
Parallel lines all pointing 'down' in one direction, equally spaced at all distances. — A student who pictures gravity as a single fixed 'down' picks this. 'Down' is towards the centre of the planet, so around a whole planet the lines are radial. Parallel, equally spaced lines apply only to a small region near the surface.
Curved lines that leave the planet near one pole and re-enter it near the other. — A student who links gravity with magnetism draws the Earth's magnetic field pattern. Gravitational field lines around a sphere are straight and radial; they never form loops, and gravity acts on all matter, magnetic or not.
5 A satellite of mass 500 kg is moved from a circular orbit of radius 6.80 × 10⁶ m to a circular orbit of radius 2.04 × 10⁷ m around the Earth, of mass 5.97 × 10²⁴ kg. Take G = 6.67 × 10⁻¹¹ N m² kg⁻². The gravitational field strength at the Earth's surface is 9.8 N kg⁻¹. What is the change in the gravitational potential energy of the Earth–satellite system? HL
Answer and reasoning
−1.95 × 10¹⁰ J — A student who compares the sizes of GMm/r sees it get smaller and concludes E_p decreases. E_p is negative: it rises from −2.93 × 10¹⁰ J to −9.76 × 10⁹ J, an increase.
+6.66 × 10¹⁰ J — A student who uses ΔE_p = mgΔh with g = 9.8 N kg⁻¹ over 1.36 × 10⁷ m gets 6.66 × 10¹⁰ J. g falls as 1/r² over this distance, so mgΔh with the surface value overestimates the change several times over.
−2.93 × 10¹⁰ J — A student who takes the potential energy at the starting orbit, E_p = −GMm/r₁, as the change gets −2.93 × 10¹⁰ J. That is the potential energy of the system relative to infinite separation, not the change between the two orbits: ΔE_p = E_p,final − E_p,initial = −9.76 × 10⁹ J − (−2.93 × 10¹⁰ J) = +1.95 × 10¹⁰ J.
6 Which statement about the gravitational potential V_g at a point outside a planet is correct? HL
Answer and reasoning
It is the gravitational potential energy of any mass at the point, measured in joules. — A student who treats potential and potential energy as the same picks this. V_g is energy PER UNIT MASS (J kg⁻¹); the potential energy of a mass m at the point is mV_g, in joules.
It is the work done per unit mass to bring a small mass from infinity to the point, and is negative. — This is the guide's definition, V_g = −GM/r. It is zero at infinity and negative at every finite distance, because the planet attracts the mass as it is brought in. Unit: J kg⁻¹.
It is zero at the planet's surface and increases with height above it, as mgh does. — A student who carries over the ground-level zero from mgh picks this. In this course V_g = 0 at infinity, so at the surface V_g = −GM/R; it does increase with height, but towards zero, never above it.
It is the gravitational force per unit mass on a small mass at the point, directed inwards. — A student who merges potential and field strength picks this. Force per unit mass is the field strength g (a vector, N kg⁻¹); potential is work per unit mass (a scalar, J kg⁻¹).
7 An external force moves a 3.0 kg mass at constant speed from a point where the gravitational potential is −3.2 × 10⁶ J kg⁻¹ to a point where it is −1.1 × 10⁶ J kg⁻¹. What is the work done on the mass by the external force? HL
Answer and reasoning
+6.3 × 10⁶ J — W = mΔV_g = m(V_final − V_initial) = 3.0 × (−1.1 × 10⁶ − (−3.2 × 10⁶)) = 3.0 × 2.1 × 10⁶ = +6.3 × 10⁶ J. The mass moves to a higher (less negative) potential, so positive work is done.
−6.3 × 10⁶ J — A student who sees the size of the potential fall from 3.2 to 1.1 thinks the potential energy decreases and gives a negative answer. −1.1 × 10⁶ is higher than −3.2 × 10⁶, so the potential energy rises and the work done on the mass is positive.
+2.1 × 10⁶ J — A student who treats the potential difference as the work done omits the mass. ΔV_g = 2.1 × 10⁶ J kg⁻¹ is the work per kilogram; for 3.0 kg, W = mΔV_g = 6.3 × 10⁶ J.
−9.6 × 10⁶ J — A student who uses the potential at the starting point alone, W = mV_g = 3.0 × (−3.2 × 10⁶), gets −9.6 × 10⁶ J. mV_g at one point is the work done bringing the mass in from infinity; here the mass moves between two points, so W = mΔV_g, using the difference in potential.
Working ΔV_g = V_final − V_initial = −1.1 × 10⁶ − (−3.2 × 10⁶) = +2.1 × 10⁶ J kg⁻¹. W = mΔV_g = 3.0 kg × 2.1 × 10⁶ J kg⁻¹ = +6.3 × 10⁶ J.
8 Equipotential surfaces are drawn around an isolated planet that is a uniform sphere, at equal intervals of gravitational potential. Which description is correct? HL
Answer and reasoning
Concentric spheres, equally spaced at all distances from the planet's centre. — A student who transfers the uniform-field pattern picks this. In a radial field V_g changes more slowly with r further out, so equal steps of potential are further apart there.
Concentric spheres, further apart the further they are from the planet. — V_g = −GM/r depends only on r, so equipotentials are spheres centred on the planet. Equal steps in V_g need larger steps in r further out, where the field is weaker, so the spacing increases with distance.
Radial straight lines that point inwards towards the planet's centre. — A student who confuses equipotentials with field lines picks this. Radial lines are the field lines; the equipotentials cross them at right angles, forming concentric spheres.
Concentric spheres, on each of which the gravitational field strength is zero. — A student who thinks there is no field where the potential is constant picks this. On each sphere the potential does not change along the surface, but the field is perpendicular to it and equal to GM/r² there.
9 The Moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m. Take G = 6.67 × 10⁻¹¹ N m² kg⁻². What is the escape speed from a point 2.00 × 10⁶ m above the Moon's surface? HL
2.21 × 10³ m s⁻¹ — A student who uses the height, 2.00 × 10⁶ m, as r gets 2.21 × 10³ m s⁻¹. r is measured from the Moon's centre: r = R + h = 3.74 × 10⁶ m.
2.37 × 10³ m s⁻¹ — A student who treats the escape speed as a fixed property of the Moon gives the surface value, √(2GM/R) = 2.37 × 10³ m s⁻¹. Further out less energy is needed to reach infinity, so the escape speed is lower.
1.14 × 10³ m s⁻¹ — A student who thinks an orbiting body has escaped gives the orbital speed, √(GM/r) = 1.14 × 10³ m s⁻¹. A body moving at this speed stays bound in orbit; escape needs √2 times as much.
Working r = R + h = 1.74 × 10⁶ m + 2.00 × 10⁶ m = 3.74 × 10⁶ m. GM = 6.67 × 10⁻¹¹ × 7.35 × 10²² = 4.902 × 10¹² N m² kg⁻¹. v_esc = √(2GM/r) = √(9.805 × 10¹² / 3.74 × 10⁶) = √(2.621 × 10⁶ m² s⁻²) = 1.62 × 10³ m s⁻¹.
10 A navigation satellite orbits the Earth in a circular orbit at a height of 2.02 × 10⁷ m above the surface. The Earth has mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m. Take G = 6.67 × 10⁻¹¹ N m² kg⁻². What is the orbital speed of the satellite? HL
Answer and reasoning
4.44 × 10³ m s⁻¹ — A student who uses the height, 2.02 × 10⁷ m, as r gets 4.44 × 10³ m s⁻¹. The orbit's radius is measured from the Earth's centre: r = R + h = 2.657 × 10⁷ m.
3.87 × 10³ m s⁻¹ — r = 6.37 × 10⁶ + 2.02 × 10⁷ = 2.657 × 10⁷ m. v = √(GM/r) = √(3.982 × 10¹⁴ / 2.657 × 10⁷) = √(1.499 × 10⁷) = 3.87 × 10³ m s⁻¹.
7.91 × 10³ m s⁻¹ — A student who uses the Earth's radius as r finds the orbital speed just above the surface, 7.91 × 10³ m s⁻¹. This satellite is over four Earth radii from the centre, where the orbital speed is much lower.
5.47 × 10³ m s⁻¹ — A student who thinks an orbiting satellite has escaped the Earth's gravity uses the escape speed √(2GM/r). Gravity is what holds the satellite in orbit: GMm/r² = mv²/r gives v = √(GM/r), which is √2 times smaller.
Working r = R + h = 6.37 × 10⁶ m + 2.02 × 10⁷ m = 2.657 × 10⁷ m. GM = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ = 3.982 × 10¹⁴ N m² kg⁻¹. v_orbital = √(GM/r) = √(1.499 × 10⁷ m² s⁻²) = 3.87 × 10³ m s⁻¹.
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22 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A comet moves around the Sun in a highly elliptical orbit. Which statement about the comet agrees with Kepler's laws?
Answer and reasoning
It moves at the same speed at every point of its orbit. — A student who reads 'equal areas in equal times' as 'equal distances in equal times' picks this. The area swept out depends on the comet's distance from the Sun as well as its speed, so near the Sun, where the swept triangle is short, the comet must move faster to sweep the same area.
It moves fastest when it is furthest from the Sun. — A student who pictures the orbit like a turning wheel, sweeping equal angles in equal times, expects the comet to move fastest where it is furthest out and picks this. Kepler's second law is about equal AREAS: far from the Sun a long, thin triangle has a short arc, so the comet is slowest there.
The Sun is at the centre of its elliptical orbit. — A student who places the Sun at the centre, as in most drawings of near-circular orbits, picks this. Kepler's first law puts the Sun at one focus of the ellipse, off-centre; that is why the comet's distance from the Sun varies so much around its orbit.
It moves fastest when it is nearest the Sun. — Kepler's second law: the line from the Sun to the comet sweeps out equal areas in equal times. Near the Sun the swept triangle is short, so the comet must cover a longer arc in the same time: it is fastest at its closest approach and slowest at its furthest point.
2 Io and Europa are moons of Jupiter, and their orbits may be taken as circular. Io has an orbital radius of 4.22 × 10⁸ m and a time period of 1.77 days. Europa has an orbital radius of 6.71 × 10⁸ m. What is the time period of Europa?
Answer and reasoning
2.81 days — A student who assumes the moons move at the same speed takes T ∝ r: 1.77 × 6.71/4.22 = 2.81 days. Europa is further out and also slower, so its period grows faster than its radius.
3.55 days — By Kepler's third law T² ∝ r³ for moons of the same planet, so T_E = T_I × (r_E/r_I)^(3/2) = 1.77 × (6.71/4.22)^(3/2) = 1.77 × 2.005 = 3.55 days.
1.77 days — A student who thinks orbiting bodies share one period, like points on a rotating wheel, gives Io's period again. Each moon's period is set by its own radius: T² ∝ r³.
4.48 days — A student who uses T ∝ r², by analogy with the inverse-square law, gets 1.77 × (6.71/4.22)² = 4.48 days. The law is T² ∝ r³, so the radius ratio is raised to the power 3/2, not 2.
Working Kepler's third law for two moons of the same planet: T_E²/r_E³ = T_I²/r_I³ ⇒ T_E = T_I (r_E/r_I)^(3/2). r_E/r_I = 6.71 × 10⁸ m / 4.22 × 10⁸ m = 1.590; 1.590^(3/2) = 2.005. T_E = 1.77 days × 2.005 = 3.55 days.
3 A satellite of mass 1200 kg orbits the Earth at a height of 600 km above the surface. The Earth has mass 5.97 × 10²⁴ kg and radius 6370 km. Take G = 6.67 × 10⁻¹¹ N m² kg⁻². What is the gravitational force that the Earth exerts on the satellite?
Answer and reasoning
1.33 × 10⁶ N — A student who uses the height, 6.0 × 10⁵ m, as r gets 1.33 × 10⁶ N. The Earth acts as a point mass at its centre, so r = R + h = 6.97 × 10⁶ m.
9.84 × 10⁹ N — A student who correctly adds 6370 km + 600 km but substitutes r = 6970 without converting to metres gets an answer 10⁶ times too large. G is in N m² kg⁻², so r must be in metres: 6.97 × 10⁶ m.
9.84 × 10³ N — r is measured from the Earth's centre: r = 6370 km + 600 km = 6970 km = 6.97 × 10⁶ m. F = GMm/r² = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 1200 / (6.97 × 10⁶)² = 9.84 × 10³ N.
1.18 × 10⁴ N — A student who uses the Earth's radius as r finds the force at the surface, 1.18 × 10⁴ N. The satellite is 600 km higher, so r = 6.97 × 10⁶ m and the force is about 16% smaller.
Working r = R + h = 6370 km + 600 km = 6970 km = 6.97 × 10⁶ m. F = GMm/r² = (6.67 × 10⁻¹¹ N m² kg⁻²)(5.97 × 10²⁴ kg)(1200 kg)/(6.97 × 10⁶ m)² = 4.778 × 10¹⁷ / 4.858 × 10¹³ = 9.84 × 10³ N.
4 A planet is a sphere of uniform density. Why can it be treated as a point mass at its centre when calculating the gravitational force on a satellite orbiting it?
Answer and reasoning
A satellite is usually far away compared with the planet's radius, so the planet looks like a point. — A student who thinks the point-mass model is only an approximation for distant bodies picks this. For a uniform sphere the result is exact at every point outside it, however close: a satellite in low orbit is only a little more than one planet radius from the centre, yet the model applies exactly.
The satellite's mass is tiny compared with the planet's, so the planet's size is irrelevant. — A student who hears 'point mass' as 'small mass' picks this. The point-mass model depends on the shape and density of the planet, not on the mass ratio; a very massive uniform star is also treated as a point mass.
All of the planet's mass is concentrated at its centre, where its gravity comes from. — A student who takes 'as if concentrated at the centre' literally picks this. The mass is spread through the whole planet and every part attracts the satellite; only the combined effect equals that of a point mass at the centre.
At any point outside a uniform sphere, its field is that of a point mass at its centre. — This is the condition in the guide: at any point outside a sphere of uniform density, however close to its surface, the combined attraction of all its parts equals that of a point mass at the centre. So r in F = GMm/r² is measured from the planet's centre.
5 At the surface of the Earth, g = 9.8 N kg⁻¹. Planet X has the same mass as the Earth but half its radius. Both may be treated as uniform spheres. What is the gravitational field strength at the surface of planet X?
Answer and reasoning
39.2 N kg⁻¹ — g = GM/R². With the same M and half the R, g increases by a factor 1/(0.5)² = 4: 4 × 9.8 = 39.2 N kg⁻¹.
19.6 N kg⁻¹ — A student who treats g as inversely proportional to R (forgetting the square) doubles the value. g = GM/R², so halving R multiplies g by four.
9.80 N kg⁻¹ — A student who assumes g has the same value everywhere gives the Earth's value. g = GM/R² depends on the planet's radius as well as its mass.
4.90 N kg⁻¹ — A student who thinks a smaller planet must have weaker gravity halves the value. Planet X has the same mass, but its surface is closer to its centre, so the field there is stronger.
Working g = GM/R². g_X/g_E = (M_X/M_E)(R_E/R_X)² = 1 × 2² = 4. g_X = 4 × 9.8 N kg⁻¹ = 39.2 N kg⁻¹.
6 The Earth (mass 5.97 × 10²⁴ kg) and the Moon (mass 7.35 × 10²² kg) are 3.84 × 10⁸ m apart, centre to centre. At one point on the line joining their centres, the resultant gravitational field strength is zero. How far is this point from the centre of the Earth?
Answer and reasoning
3.79 × 10⁸ m — A student who forgets that the fields fall as 1/r² sets M_E/x = M_M/(d − x) and gets x = d × 81.2/82.2 = 3.79 × 10⁸ m. The inverse-square law puts the square root of the mass ratio into the answer.
3.46 × 10⁸ m — Set GM_E/x² = GM_M/(d − x)²: x/(d − x) = √(M_E/M_M) = √81.2 = 9.01, so x = 9.01d/10.01 = 3.46 × 10⁸ m, about 90% of the way to the Moon.
1.92 × 10⁸ m — A student who assumes the fields cancel halfway picks the midpoint. The Earth is 81 times more massive, so its field is much stronger at the midpoint; the zero point is much closer to the Moon.
4.67 × 10⁶ m — A student who confuses the zero-field point with the centre of mass gets x = d × M_M/(M_E + M_M) = 4.67 × 10⁶ m, which is inside the Earth. The centre of mass balances masses × distances; the zero-field point balances masses ÷ distances².
Working Fields are antiparallel on the line joining the centres, so the resultant is zero where GM_E/x² = GM_M/(d − x)². √(M_E/M_M) = √(5.97 × 10²⁴ / 7.35 × 10²²) = √81.2 = 9.01. x = 9.01(d − x) ⇒ x = 9.01 × 3.84 × 10⁸ m / 10.01 = 3.46 × 10⁸ m from the Earth's centre.
7 Close to the surface of a planet, over a region a few kilometres across, the gravitational field is drawn as parallel, equally spaced field lines pointing vertically downwards. What does this pattern show about the field in that region?
Answer and reasoning
Any object released or thrown in the region moves along one of the field lines. — A student who reads field lines as paths picks this. The lines show the direction of the force, not of the motion. A ball thrown sideways follows a parabola across the lines; only an object released from rest falls along one.
The field acts only along the lines drawn, with no field in the gaps between them. — A student who takes the lines literally picks this. The lines are a sample; the field exists at every point, including between the lines, and more lines could be drawn anywhere.
The field strength has the same size and direction at every point in the region. — Parallel lines show the same direction everywhere, and equal spacing shows the same strength everywhere: a uniform field. This is the assumed uniform field close to the surface of a planet.
A heavier object placed in the region experiences a larger field strength. — A student who confuses field strength with force picks this. A heavier object feels a larger FORCE, mg, but the field strength g, the force per kilogram, is the same for every object at a point.
8 The gravitational potential energy E_p of a system is the work done to assemble the system from infinite separation of its components. Why is E_p of an Earth–satellite system negative? HL
Answer and reasoning
The minus sign shows that the potential energy is directed towards the centre of the Earth. — A student who reads every minus sign as a direction picks this. Energy is a scalar and has no direction; the minus sign says E_p is less than its value at infinite separation, which is zero.
Positive work is needed to bring them together, so the minus sign is only a convention. — A student who equates assembling with putting energy in picks this. The attraction does the pulling; the agent holds the bodies back and does negative work. The sign follows from the definition, not from an arbitrary choice.
Energy has been used up in bringing them together, leaving the system with an energy deficit. — A student who thinks energy is consumed reads the negative value as a debt. No energy is destroyed: the E_p lost as the bodies come together goes into other stores. The value is negative only because zero is set at infinite separation.
The bodies attract, so the work done to bring them together from infinity is negative. — As the bodies approach, gravity pulls them together; to assemble the system without them speeding up, an external agent must hold them back, so it does negative work. E_p is zero at infinity and negative at any finite separation.
9 A probe of mass m is lifted from the surface of a planet of radius R to a height R above the surface. The field strength at the surface is g. A student calculates the gain in gravitational potential energy using ΔE_p = mgΔh with Δh = R. How does the student's answer compare with the true gain, found from E_p = −GMm/r? HL
Answer and reasoning
1.0 times the true gain — A student who thinks mgΔh holds at any height expects the two to agree. g at height R is only g/4, so the work done per metre falls as the probe rises and mgR overestimates the gain.
4.0 times the true gain — A student who takes the true gain as m(g/4)R, using the field strength at the final height for the whole journey, gets a ratio of 4. The field is stronger (up to g) over the lower part of the path, so the true gain is ½mgR, not ¼mgR.
1.6 times the true gain — A student who averages the end values of g, (g + g/4)/2 = 5g/8, gets a ratio of 1.6. That average is valid only if g falls linearly; g ∝ 1/r², so the exact result from E_p = −GMm/r must be used.
2.0 times the true gain — True gain: GMm(1/R − 1/2R) = GMm/2R. Since g = GM/R², GMm/2R = ½mgR. The student's mgR is therefore 2.0 times the true gain, because g falls as the probe rises and less work is needed for each metre higher up.
Working True ΔE_p = GMm(1/R − 1/2R) = GMm/2R. With GM = gR²: ΔE_p = mgR/2. Student: mgΔh = mgR. Ratio = mgR ÷ (mgR/2) = 2.0.
10 The Moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m. Take G = 6.67 × 10⁻¹¹ N m² kg⁻². What is the gravitational potential due to the Moon at a point 1.00 × 10⁶ m above its surface? HL
−4.90 × 10⁶ J kg⁻¹ — A student who uses the height, 1.00 × 10⁶ m, as r gets −4.90 × 10⁶ J kg⁻¹. r is measured from the Moon's centre: r = R + h = 2.74 × 10⁶ m.
−2.82 × 10⁶ J kg⁻¹ — A student who uses the Moon's radius as r finds the potential at the surface, −2.82 × 10⁶ J kg⁻¹. The point is 1.00 × 10⁶ m higher, where the potential is less negative.
+1.03 × 10⁶ J kg⁻¹ — A student who takes V_g = 0 at the surface finds the potential difference from the surface, +1.03 × 10⁶ J kg⁻¹. In this course V_g = 0 at infinity, so the potential at any finite distance is negative.
Working r = R + h = 1.74 × 10⁶ m + 1.00 × 10⁶ m = 2.74 × 10⁶ m. V_g = −GM/r = −(6.67 × 10⁻¹¹ N m² kg⁻²)(7.35 × 10²² kg)/(2.74 × 10⁶ m) = −4.902 × 10¹² / 2.74 × 10⁶ = −1.79 × 10⁶ J kg⁻¹.
11 Two equipotential surfaces around the Earth are 1.00 × 10⁵ m apart along a radial line. The inner surface has V_g = −6.251 × 10⁷ J kg⁻¹ and the outer surface has V_g = −6.155 × 10⁷ J kg⁻¹. Taking r as increasing outwards from the Earth's centre, what is the average gravitational field strength g between the two surfaces? HL
Answer and reasoning
+9.6 N kg⁻¹ (directed away from the Earth) — A student who drops the minus sign in g = −ΔV_g/Δr gets +9.6 N kg⁻¹. The potential increases outwards, so the field points the other way, towards lower potential and towards the Earth.
−9.8 N kg⁻¹ (directed towards the Earth’s centre) — A student who takes g as 9.8 N kg⁻¹ everywhere picks this. The inner surface is close to the Earth's surface, where g is about 9.8 N kg⁻¹, but g decreases with height (g = GM/r²), so it is smaller across the gap. The average field between the surfaces must come from the data: g = −ΔV_g/Δr = −(9.6 × 10⁵ J kg⁻¹)/(1.00 × 10⁵ m) = −9.6 N kg⁻¹.
0 N kg⁻¹ (each surface is an equipotential) — A student who thinks there is no field on an equipotential picks this. The potential is constant ALONG each surface but changes BETWEEN them, and it is this change across the gap that gives g.
−9.6 N kg⁻¹ (directed towards the Earth) — g = −ΔV_g/Δr = −(−6.155 × 10⁷ − (−6.251 × 10⁷))/1.00 × 10⁵ = −(9.6 × 10⁵)/(1.00 × 10⁵) = −9.6 N kg⁻¹. The minus sign means the field points towards decreasing r, towards the Earth.
Working ΔV_g = V_outer − V_inner = −6.155 × 10⁷ − (−6.251 × 10⁷) = +9.6 × 10⁵ J kg⁻¹. Δr = +1.00 × 10⁵ m. g = −ΔV_g/Δr = −9.6 N kg⁻¹, i.e. 9.6 N kg⁻¹ directed towards the Earth.
12 An 800 kg satellite is launched from rest on the surface of a non-rotating planet of mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m, and placed in a circular orbit of radius 1.00 × 10⁷ m. Take G = 6.67 × 10⁻¹¹ N m² kg⁻². Ignoring air resistance, what is the minimum energy that must be supplied to the satellite? HL
Answer and reasoning
1.82 × 10¹⁰ J — A student who thinks the satellite only has to be lifted to the orbit's height calculates the gain in E_p alone, GMm(1/R − 1/r) = 1.82 × 10¹⁰ J. Left at rest there, it would fall back; it also needs E_k = GMm/2r = 1.59 × 10¹⁰ J.
5.00 × 10¹⁰ J — A student who takes the orbital kinetic energy as GMm/r, forgetting the ½, adds 3.19 × 10¹⁰ J to the gain in E_p and gets 5.00 × 10¹⁰ J (which is the energy to escape from the surface). E_k = ½mv² = GMm/2r = 1.59 × 10¹⁰ J.
1.59 × 10¹⁰ J — A student who thinks reaching orbit is only a matter of reaching orbital speed gives the orbital kinetic energy alone. Rising from the surface to r = 1.00 × 10⁷ m also increases E_p by 1.82 × 10¹⁰ J.
Working GMm = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(800) = 3.186 × 10¹⁷ J m. Initial energy (rest, non-rotating planet): E = −GMm/R = −5.001 × 10¹⁰ J. Final energy in circular orbit: E = −GMm/2r = −1.593 × 10¹⁰ J. Energy supplied = −1.593 × 10¹⁰ − (−5.001 × 10¹⁰) = 3.41 × 10¹⁰ J (ΔE_p = 1.82 × 10¹⁰ J plus E_k = 1.59 × 10¹⁰ J).
13 A small mass is moved along an equipotential surface in a gravitational field. Which statement is correct? HL
Answer and reasoning
There is no gravitational field on the surface, so no force acts on the mass. — A student who thinks constant potential means no field picks this. The field is not zero; it is perpendicular to the surface, so it does no work along it. A satellite in a circular orbit moves along an equipotential under a large gravitational force.
Gravity does work on the mass, as its weight acts on it all the time it moves. — A student who thinks any force on a moving object does work picks this. Work needs a force component along the displacement, W = Fs cos θ; here θ = 90°, so the work done is zero.
The field acts along the surface, in the same direction as the mass moves. — A student who thinks field lines run along equipotentials picks this. If the field had a component along the surface, the potential would change along it, contradicting the definition of an equipotential.
The field is perpendicular to its path, so gravity does no work on the mass. — Field lines cross equipotential surfaces at right angles, so the gravitational force has no component along the path. With W = mΔV_g and ΔV_g = 0, no work is done by gravity.
14 A satellite is in a circular orbit around a planet and has kinetic energy E_k. What is the minimum extra energy that must be given to the satellite for it to escape the gravitational influence of the planet? HL
Answer and reasoning
2E_k, the magnitude of its gravitational potential energy — A student who uses the escape condition for a body starting from rest supplies the full |E_p| = 2E_k. The satellite already has kinetic energy E_k, so only E_k more is needed to bring the total to zero.
3E_k, since escape speed is twice the orbital speed — A student who takes v_esc = 2v_orbital needs four times the kinetic energy, so 3E_k extra. In fact v_esc = √(2GM/r) = √2 × v_orbital: the kinetic energy only doubles.
E_k, which raises its total energy from −E_k to zero — In a circular orbit E_k = GMm/2r and E_p = −GMm/r = −2E_k, so the total energy is −E_k. Escape needs a total energy of at least zero, so E_k more must be supplied: its kinetic energy doubles, and its speed rises by √2.
No finite amount, as the field reaches to infinity — A student who pictures escape as reaching a place beyond gravity picks this. The field weakens as 1/r², so the total work to reach infinity is finite. A satellite with zero total energy slows forever but never returns.
Working Circular orbit: E_k = GMm/2r, E_p = −GMm/r = −2E_k, E_total = −E_k. Escape requires E_total ≥ 0, so the minimum extra energy is E_k (v rises from v_orbital to √2 v_orbital = v_esc).
15 A satellite is moved from one circular orbit to another circular orbit of larger radius around the same planet. How does its total energy in the new orbit compare with its total energy in the old orbit? HL
Answer and reasoning
Greater, because it moves faster in the larger orbit and so has more kinetic energy. — A student who links more energy with more speed picks this. The larger orbit needs less centripetal force, which gravity supplies at a lower speed: v = √(GM/r) and E_k = GMm/2r both fall. The total energy is greater, but the extra energy is in E_p, not E_k.
Lower, because it moves more slowly in the larger orbit and so has less kinetic energy. — A student who counts only kinetic energy picks this. E_k = GMm/2r does fall, but E_p = −GMm/r rises by twice as much, so the total energy −GMm/2r increases. Energy must be supplied to raise an orbit.
Lower, because the magnitude of its potential energy, GMm/r, is smaller in the larger orbit. — A student who compares magnitudes and ignores the sign picks this. GMm/r is smaller in the larger orbit, but E_p = −GMm/r, so E_p becomes less negative: it increases. It rises by twice as much as E_k falls, so the total energy −GMm/2r increases.
Greater, because its potential energy rises by twice as much as its kinetic energy falls. — Moving from r₁ to r₂ > r₁, E_k = GMm/2r falls by (GMm/2)(1/r₁ − 1/r₂) while E_p = −GMm/r rises by GMm(1/r₁ − 1/r₂), twice as much. The total energy −GMm/2r therefore becomes less negative: greater. The satellite ends up slower but with more energy.
16 A satellite in a low circular orbit experiences a small viscous drag force from the thin upper atmosphere. Over many orbits its path remains approximately circular. How do its orbital radius and its speed change? HL
Answer and reasoning
The radius gradually decreases and the orbital speed also decreases. — A student who expects any resistive force to slow an object picks this. The lost energy comes from E_p: as the satellite drops, E_p falls by twice the gain in E_k, so it speeds up while it loses height.
The radius gradually decreases while the orbital speed gradually increases. — Drag removes energy, so the total energy −GMm/2r becomes more negative and r decreases. In each smaller orbit v = √(GM/r) is larger: gravity does more positive work as the satellite spirals in than the drag does negative work.
The radius gradually increases because the orbital speed decreases. — A student who reasons 'drag slows it, and slower orbits are larger' picks this. Start from energy: drag lowers the total energy −GMm/2r, which means a smaller r, and a smaller orbit has a higher speed.
The radius gradually decreases while the orbital speed stays the same. — A student who thinks the orbital speed does not depend on the orbit picks this. For a circular orbit there is only one speed at each radius, v = √(GM/r), so the smaller orbit the satellite drops into must be travelled faster.
17 A space probe approaches an asteroid that is irregular in shape and not of uniform density. While the probe is far away, the gravitational force on it is given accurately by F = GMm/r², with r measured from the asteroid's centre of mass. Close to the asteroid's surface the same equation is no longer accurate. Which statement explains these observations?
Answer and reasoning
No body acts as a point mass when it is close: even a sphere of uniform density can be treated as a point mass at its centre only when the distance to it is large compared with its radius, as here. — A student who thinks the point-mass model is only a far-away approximation picks this. For a sphere of uniform density the result is exact at every point outside it, however close. The failure here is due to the asteroid's irregular shape and non-uniform density, not to closeness alone.
The point-mass model needs the probe's mass to be negligible compared with the asteroid's; an asteroid is so small that this condition only holds while the probe is far away, and it breaks down as the probe closes in. — A student who hears 'point mass' as 'small mass' picks this. The mass ratio of probe to asteroid is the same at every distance and is irrelevant to whether a body's field is that of a point mass; what matters is how the asteroid's mass is distributed in space.
Far away, the asteroid's size is negligible compared with r, so it acts as a point mass; close up, its mass is not distributed spherically, so its field is not that of a point mass at its centre. — Any body acts as a point mass when r is large compared with its dimensions. The exact result at every distance holds only for a sphere of uniform density (or spherical shells); an irregular, non-uniform asteroid gives a field close to its surface that differs from GM/r² about its centre of mass.
Close to the asteroid, r should be measured from the asteroid's surface to the probe rather than from its centre of mass, because the surface is where the attraction on the probe begins. — A student who takes r as the gap between the surfaces picks this. In Newton's law r is the separation of the centres of mass; changing where r is measured from does not turn an irregular body into a point mass. Every part of the asteroid attracts the probe, and close up these attractions do not add to the field of a point at the centre.
18 Two identical stars, each of mass M, are at rest with their centres a distance r apart. A student argues that each star 'has' a gravitational potential energy of −GM²/r in the field of the other, so that the total gravitational potential energy of the two-star system is −2GM²/r. Which statement correctly assesses this claim? HL
Answer and reasoning
The claim is right: potential energy belongs to a body placed in a field, and each star sits in the gravitational field of the other, so each star has −GM²/r of its own. — A student who attaches potential energy to an individual body picks this. E_p is a property of the interacting pair, defined by the work done to assemble the system. Counting −GM²/r once for each star counts the same assembly work twice.
The claim has the sign wrong: work must be done against gravity to bring the stars together from infinity, so each star has +GM²/r and the total for the two is +2GM²/r. — A student who thinks assembling the system needs positive work picks this. The stars attract, so the external agent holds them back and does negative work; E_p at any finite separation is negative. The doubling is also wrong, since E_p belongs to the pair.
The claim ignores direction: each star's potential energy is directed towards the other star, so the two values cancel and the system's total is zero. — A student who reads the minus sign as a direction picks this. Energy is a scalar; the sign of E_p records that the system has less energy than at infinite separation, and scalars cannot point in opposite directions and cancel.
The claim double-counts: −GM²/r is the whole work done in bringing the stars together from infinite separation, and belongs to the pair, not to either star. — E_p of a system is the work done to assemble it from infinite separation. Whether star A is brought up to star B or both are moved, the total work done by the external agent is −GM²/r. That single quantity is the potential energy of the pair; there is no separate E_p for each star to add.
19 A graph shows the gravitational potential V_g along a radial line from the centre of a planet, plotted against the distance r from the centre. Which statement about the gravitational field strength g at a point outside the planet on this line is correct? HL
Answer and reasoning
g is the gradient of the graph at that point, so the field is directed towards higher V_g, outwards. — A student who drops the minus sign picks this. The gradient of V_g against r is positive outside a planet, but the field points inwards, towards lower potential; g = −ΔV_g/Δr makes this explicit.
g is fixed by the value of V_g at that point alone, and does not depend on how V_g changes with r near the point. — A student who reads a field from a single potential value picks this. The potential at one point can be shifted by choosing a different zero without changing the field; g depends on the rate of change of V_g with distance, the gradient of the graph.
g is minus the gradient of the graph at that point, so the field is directed towards lower V_g. — g = −ΔV_g/Δr: the field strength is the negative of the potential gradient. Outside a planet V_g increases (becomes less negative) with r, so the gradient is positive and g is negative, meaning the field points towards decreasing r, where the potential is lower.
g is zero at that point whenever the point lies on an equipotential surface, however steep the graph is there. — A student who thinks constant potential means no field picks this. Every point lies on some equipotential surface. The potential is constant along the surface but changes across it, and it is that change with r, the gradient of the graph, that gives g.
20 The Earth has mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m, and the gravitational field strength at its surface is 9.8 N kg⁻¹. Take G = 6.67 × 10⁻¹¹ N m² kg⁻². An equipotential surface around the Earth has gravitational potential V_g = −2.50 × 10⁷ J kg⁻¹. What is the radius of this surface, measured from the centre of the Earth? HL
Answer and reasoning
2.23 × 10⁷ m — A student who thinks the r in V_g = −GM/r is the height above the surface computes GM/|V_g| = 1.59 × 10⁷ m, calls it a height, and adds the Earth's radius. The r in the equation is already the distance from the centre, so nothing should be added.
3.99 × 10³ m — A student who confuses potential with field strength uses |V_g| = GM/r² and gets r = √(GM/|V_g|) = 3.99 × 10³ m, a distance far inside the Earth. Potential falls as 1/r, not 1/r²: r = GM/|V_g|.
1.59 × 10⁷ m — An equipotential around a spherical mass is a sphere on which V_g = −GM/r has one value, so r = GM/|V_g| = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴)/(2.50 × 10⁷) = 1.59 × 10⁷ m, measured from the centre.
8.92 × 10⁶ m — A student who takes potential as zero at the surface and equal to gh above it sets h = 2.50 × 10⁷/9.8 = 2.55 × 10⁶ m and adds the Earth's radius. Potential is zero at infinity and V_g = −GM/r; the field is far from uniform over such heights, so gh does not apply.
Working V_g = −GM/r, so r = GM/|V_g| = (6.67 × 10⁻¹¹ N m² kg⁻²)(5.97 × 10²⁴ kg)/(2.50 × 10⁷ J kg⁻¹) = 3.982 × 10¹⁴ / 2.50 × 10⁷ = 1.593 × 10⁷ m ≈ 1.59 × 10⁷ m. This is about 2.5 Earth radii from the centre, i.e. a height of 9.56 × 10⁶ m above the surface.
21 A small mass is at a point P on an equipotential surface in the gravitational field of a planet. It is to be moved a short distance from P. In which direction should it be moved so that the gravitational force does the greatest positive work on it? HL
Answer and reasoning
Perpendicular to the equipotential surface and towards the planet, so that it moves along the field line through P towards lower potential. — Field lines cross equipotential surfaces at right angles and point towards lower potential. Work done by gravity is W = −mΔV_g, so the greatest positive work over a short distance comes from the largest fall in V_g, which is along the field line, perpendicular to the surface and towards the planet.
Perpendicular to the equipotential surface and away from the planet, so that it moves towards higher potential more quickly. — A student who thinks the field points towards increasing potential picks this. Moving to higher V_g means ΔV_g > 0, so gravity does negative work, W = −mΔV_g < 0; positive work by gravity requires a fall in potential, which is towards the planet.
Along the equipotential surface, because the gravitational field acts along the surface rather than across it. — A student who thinks field lines run along equipotentials picks this. The field is perpendicular to an equipotential surface; along the surface ΔV_g = 0, so gravity does no work at all.
In any direction, because the gravitational force does the same work over the same distance whichever way the mass moves. — A student who thinks a force does work whatever the angle to the motion picks this. W = Fs cos θ: the work depends on the direction of the displacement relative to the force, and is greatest when the mass moves along the field, towards the planet.
22 A satellite in a low circular orbit around a planet experiences a small viscous drag force from the thin upper atmosphere. Over many orbits it spirals slowly inwards, its orbit remaining approximately circular. Which statement about the energy changes is correct? HL
Answer and reasoning
The kinetic energy of the satellite decreases, because the drag force acts against its motion and so does negative work on it throughout. — A student who expects any resistive force to slow an object picks this. Drag does do negative work, but as the satellite spirals inwards gravity does more positive work on it than drag does negative work, so its kinetic energy rises and it moves faster in the smaller orbit.
The gravitational potential energy of the system increases as the satellite moves inwards, because GMm/r becomes larger. — A student who looks at the size of GMm/r and forgets the sign picks this. E_p = −GMm/r is negative and becomes more negative as r decreases, so the potential energy falls as the satellite moves inwards.
The kinetic energy of the satellite increases, because the fall in potential energy exceeds the energy transferred to the atmosphere by drag. — For a circular orbit E_k = GMm/2r and E_p = −GMm/r, so as r falls E_p drops by twice the rise in E_k. Half the lost E_p goes into E_k and half is transferred to the atmosphere by drag; the satellite therefore speeds up while it loses height.
The total energy of the satellite–planet system is unchanged, because the loss of gravitational potential energy equals the gain in kinetic energy. — A student who applies energy conservation without including the atmosphere picks this. Drag transfers energy out of the satellite–planet system, so its total energy −GMm/2r falls; E_k rises by only half of the fall in E_p, the other half going to the atmosphere.
That was your twenty minutes. Real practice on D.1 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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