D.1 Gravitational fields – IB Physics exam questions with mark scheme

1Core3 marks

Mars has mass 6.42 × 10²³ kg and radius 3.39 × 10⁶ m.

(a) Define gravitational field strength at a point. [1]

(b) Calculate the gravitational field strength at the surface of Mars. [2]

(a) Force per unit mass acting on a (small) test mass placed at that point(1)
(b) Uses g = GM/R² with correct substitution (6.67 × 10⁻¹¹ × 6.42 × 10²³ ÷ (3.39 × 10⁶)²)(1)
(b) g = 3.73 N kg⁻¹ (allow m s⁻²)(1)
Answer: (a) force per unit mass on a test mass; (b) 3.73 N kg⁻¹
Examiner tip: Define it as force per unit mass – 'force on an object' alone does not earn the mark.

2Standard4 marks

A satellite orbits the Earth in a circular orbit at a height of 600 km above the surface.

Mass of Earth = 5.97 × 10²⁴ kg; radius of Earth = 6.37 × 10⁶ m.

(a) Show that the orbital speed is about 7.6 km s⁻¹. [3]

(b) Calculate the orbital period in seconds. [1]

(a) Gravitational force provides the centripetal force: GMm/r² = mv²/r(1)
(a) Orbital radius r = 6.37 × 10⁶ + 6.0 × 10⁵ = 6.97 × 10⁶ m, and v = √(GM/r)(1)
(a) v = 7.56 × 10³ m s⁻¹ ≈ 7.6 km s⁻¹ (must show an unrounded value)(1)
(b) T = 2πr/v = 5.79 × 10³ s (5794 s; allow 5.8 × 10³ s)(1)
Answer: v = 7.56 × 10³ m s⁻¹; T = 5.79 × 10³ s
Examiner tip: Orbital radius is measured from Earth's centre: add the height to the radius of the Earth.

3StandardHL only5 marks

The Moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m.

(a) Calculate the gravitational potential at the Moon's surface. [2]

(b) Calculate the escape speed from the Moon's surface. [2]

(a) V = −GM/R with correct substitution(1)
(a) V = −2.82 × 10⁶ J kg⁻¹ (negative sign required)(1)
(b) Kinetic energy at surface equals the energy needed to reach V = 0: ½mv² = GMm/R (so v = √(2GM/R))(1)
(b) v = 2.37 × 10³ m s⁻¹(1)
(c) Mass m cancels from ½mv² = GMm/R (both KE and the change in potential energy are proportional to m)(1)
Answer: (a) −2.82 × 10⁶ J kg⁻¹; (b) 2.37 × 10³ m s⁻¹; (c) m cancels
Examiner tip: Gravitational potential is always negative (zero at infinity) – a missing minus sign loses the mark.

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4Core5 marks

Io and Europa are moons of Jupiter. Both move in near-circular orbits.

Moon	Orbital radius / m	Orbital period / days
Io	$4.22\times10^{8}$	1.769
Europa	$6.71\times10^{8}$	?

(a) State Kepler’s third law. [1]

(b) Calculate the orbital period of Europa, in days. [2]

(c) Kepler’s second law states that a line joining a planet to the Sun sweeps out equal areas in equal times. Outline what this law implies about the speed of a planet in an elliptical orbit. [2]

The square of the orbital period is proportional to the cube of the mean orbital radius / semi-major axis ($T^2\propto r^3$) (for bodies orbiting the same central mass) (allow $T^2/r^3$ = constant; accept $T\propto r^{3/2}$)(1)
$T_E^2 = T_I^2\left(\dfrac{r_E}{r_I}\right)^3$ or $\dfrac{T_E^2}{(6.71\times10^8)^3}=\dfrac{1.769^2}{(4.22\times10^8)^3}$ seen(1)
$T_E = 3.55$ days (accept 3.5 to 3.6; award [2] for a bald correct answer)(1)
The planet moves faster when it is closer to the Sun / slower when further away(1)
Because a shorter radius line must sweep the same area in the same time, it must cover a longer arc (OR speed is greatest at perihelion and least at aphelion) (allow reference to conservation of angular momentum)(1)
Answer: (b) 3.55 days (c) faster near the Sun, slowest at the furthest point
Examiner tip: Use the ratio form of $T^2\propto r^3$ so the units of $T$ and $r$ cancel; there is no need to convert days to seconds.

5Core5 marks

In a laboratory measurement of the gravitational constant, a small lead sphere of mass 0.73 kg is placed so that its centre is 0.23 m from the centre of a large lead sphere of mass 158 kg.

(a) Calculate the magnitude of the gravitational force between the spheres. [2]

(b) Explain why the spheres can be treated as point masses in this calculation. [1]

(c) Describe the pattern of gravitational field lines around the large sphere when it is isolated, and state what the spacing of the lines represents. [2]

$F=\dfrac{GMm}{r^2}=\dfrac{6.67\times10^{-11}\times158\times0.73}{0.23^2}$ seen(1)
$F=1.5\times10^{-7}$ N (accept $1.45\times10^{-7}$ N; award [2] for bald correct answer)(1)
The mass of each (uniform) sphere acts as if concentrated at its centre (for points outside the sphere) / the distance used is measured between the centres (do not accept “the spheres are small” alone)(1)
Radial lines, pointing inwards towards the centre of the sphere (direction arrows inward required)(1)
Spacing shows field strength: lines closer together means a stronger field (so the field weakens with distance as the lines spread out)(1)
Answer: (a) $1.5\times10^{-7}$ N
Examiner tip: Field lines must carry arrows pointing towards the mass; outward arrows lose the mark.

6Standard4 marks

A spacecraft travels in a straight line from the Earth towards the Moon.

Mass of Earth $=5.97\times10^{24}$ kg; mass of Moon $=7.35\times10^{22}$ kg; distance between their centres $=3.84\times10^{8}$ m.

(a) Explain why there is a point between the Earth and the Moon at which the resultant gravitational field strength is zero. [1]

(b) Calculate the distance of this point from the centre of the Earth. [3]

The fields of the Earth and the Moon point in opposite directions between them, and at one point they are equal in magnitude (so they cancel)(1)
$\dfrac{GM_E}{x^2}=\dfrac{GM_M}{(d-x)^2}$ seen (allow with $G$ cancelled)(1)
Correct rearrangement, e.g. $\dfrac{d-x}{x}=\sqrt{\dfrac{M_M}{M_E}}$ OR $x=\dfrac{d}{1+\sqrt{M_M/M_E}}$(1)
$x=3.46\times10^{8}$ m (accept $3.4$ to $3.5\times10^{8}$ m; ECF for a solution of a quadratic giving the root between the bodies only)(1)
Answer: (b) $3.46\times10^{8}$ m (about 90% of the way to the Moon)
Examiner tip: Take the square root of both sides early; expanding to a quadratic invites algebra slips and a second root that lies beyond the Moon.

7StandardHL only6 marks

The graph shows how the gravitational potential $V$ varies with distance $r$ from the centre of a planet, for points from the planet’s surface outwards.

(a) Use the graph to determine the mass of the planet. [2]

(b) By drawing a tangent, determine the gravitational field strength at $r=2.0\times10^{7}$ m. [2]

(c) A probe of mass 500 kg is moved from $r=2.0\times10^{7}$ m to $r=4.0\times10^{7}$ m. Calculate the minimum work done on the probe. [2]

Pair of values read from the graph and used in $V=-\dfrac{GM}{r}$, e.g. $-4.0\times10^{7}=-\dfrac{6.67\times10^{-11}M}{1.0\times10^{7}}$(1)
$M=6.0\times10^{24}$ kg (accept 5.8 to $6.2\times10^{24}$ kg)(1)
Tangent drawn at $r=2.0\times10^{7}$ m and gradient calculated from a large triangle, $g=-\dfrac{\Delta V}{\Delta r}$(1)
$g=1.0$ N kg$^{-1}$ (accept 0.9 to 1.1 N kg$^{-1}$; allow J kg$^{-1}$ m$^{-1}$ or m s$^{-2}$; accept value calculated as $GM/r^2$ from (a) only if a tangent is also shown — else [1] max)(1)
$\Delta V = -1.0\times10^{7}-(-2.0\times10^{7}) = 1.0\times10^{7}$ J kg$^{-1}$ read from graph(1)
$W=m\Delta V=5.0\times10^{9}$ J (accept 4.8 to $5.2\times10^{9}$ J)(1)
Answer: (a) $6.0\times10^{24}$ kg (b) 1.0 N kg$^{-1}$ (c) $5.0\times10^{9}$ J
Examiner tip: Field strength is the negative gradient of $V$ against $r$: the slope here is positive, so state $g$ as a magnitude directed towards the planet.

8HardHL only6 marks

A satellite of mass $m$ moves in a circular orbit of radius $r$ around the Earth, of mass $M$.

(a) Show that the kinetic energy of the satellite is $E_k=\dfrac{GMm}{2r}$. [2]

(b) Hence state an expression for the total energy of the satellite. [1]

(c) A 1200 kg satellite is transferred from a circular orbit of radius $6.77\times10^{6}$ m to a circular orbit of radius $4.22\times10^{7}$ m. Mass of Earth $=5.97\times10^{24}$ kg.

Calculate the minimum energy that must be supplied to the satellite. [3]

Gravitational force provides the centripetal force: $\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}$(1)
$mv^2=\dfrac{GMm}{r}$ so $E_k=\tfrac12mv^2=\dfrac{GMm}{2r}$ (answer given, so the step $mv^2 = GMm/r$ must be explicit)(1)
$E_T=E_k+E_p=-\dfrac{GMm}{2r}$ (sign required)(1)
$\Delta E=\dfrac{GMm}{2}\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$ OR both total energies calculated: $-3.53\times10^{10}$ J and $-5.66\times10^{9}$ J(1)
Correct substitution: $\dfrac{6.67\times10^{-11}\times5.97\times10^{24}\times1200}{2}\left(\dfrac{1}{6.77\times10^{6}}-\dfrac{1}{4.22\times10^{7}}\right)$(1)
$\Delta E=2.96\times10^{10}$ J (accept $3.0\times10^{10}$ J; award [1] max for $5.9\times10^{10}$ J from using the change in $E_p$ only)(1)
Answer: (b) $E_T=-\dfrac{GMm}{2r}$ (c) $3.0\times10^{10}$ J
Examiner tip: Energy supplied equals the change in total energy, not potential energy: the satellite ends up slower, so half of the potential energy gain is offset by lost kinetic energy.

9HardHL only6 marks

A satellite is in a low circular orbit around the Earth, where it experiences a small resistive force from the thin upper atmosphere.

Discuss how atmospheric drag affects the orbit of the satellite. In your answer, refer to the orbital radius, the speed and the kinetic, potential and total energies of the satellite. [6]

Drag does negative work on the satellite, so its total energy decreases (becomes more negative)(1)
Total energy $E_T=-\dfrac{GMm}{2r}$, so a decrease in $E_T$ means the orbital radius decreases(1)
The satellite spirals inwards gradually (orbit stays approximately circular as drag is small)(1)
Orbital speed $v=\sqrt{GM/r}$ so the speed increases as $r$ decreases (counter-intuitive result)(1)
Kinetic energy $\dfrac{GMm}{2r}$ increases(1)
Potential energy $-\dfrac{GMm}{r}$ decreases, by twice the gain in kinetic energy(1)
Loss of $E_p$ = gain in $E_k$ + energy dissipated as thermal energy by drag(1)
As $r$ falls the atmosphere is denser so drag increases and the decay accelerates; the orbital period decreases ($T^2\propto r^3$)(1)
Satellites need periodic boosts (thruster firings) to maintain their orbit, otherwise they re-enter(1)

Level	Marks	What it looks like
Level 3	5–6	A logically ordered discussion that links the loss of total energy to a smaller radius and correctly concludes, with justification from an orbital relationship, that speed and kinetic energy increase while potential energy falls by more. Energy accounting (including dissipation) is clear and terminology is precise.
Level 2	3–4	States that energy is lost and the radius decreases, and makes a correct claim about speed or kinetic energy with partial justification. Some links between quantities are missing or the energy accounting is incomplete.
Level 1	1–2	Isolated relevant statements, e.g. the satellite loses energy or falls lower. May wrongly state that the satellite slows down. Little or no reasoning.
Level 0	0	No relevant content.

Answer: Total energy falls, radius falls, speed and $E_k$ rise, $E_p$ falls by twice the rise in $E_k$; the difference is dissipated as heat.
Examiner tip: Many answers say drag makes the satellite slow down: in orbit the opposite happens, and you must justify it with $v=\sqrt{GM/r}$.

10StretchHL only8 marks

A small asteroid is a uniform sphere of mass $9.4\times10^{20}$ kg and radius $4.7\times10^{5}$ m. It has no atmosphere.

(a) Calculate the escape speed from the surface of the asteroid. [2]

(b) A rock is projected vertically upwards from the surface at $400$ m s$^{-1}$. Determine the maximum height reached above the surface. [3]

(c) A student calculates the height in (b) using $v^2=2gh$ with the surface field strength. Explain why this gives an answer that is too small. [1]

(d) Equipotential surfaces are drawn around the asteroid at equal intervals of potential. Describe their shape and how their separation changes with distance from the asteroid. [2]

$v_{esc}=\sqrt{\dfrac{2GM}{R}}=\sqrt{\dfrac{2\times6.67\times10^{-11}\times9.4\times10^{20}}{4.7\times10^{5}}}$(1)
$v_{esc}=520$ m s$^{-1}$ (accept 516 m s$^{-1}$)(1)
Energy conservation with non-uniform field: $\tfrac12v^2-\dfrac{GM}{R}=-\dfrac{GM}{r}$ (allow with $m$ included)(1)
$r=1.17\times10^{6}$ m (from the centre)(1)
Height $=r-R=7.0\times10^{5}$ m (accept $7.0$ to $7.1\times10^{5}$ m; ECF for their $r$ minus $R$)(1)
$g$ decreases with height, so the rock decelerates less than the surface value suggests (the uniform-field result is $2.8\times10^{5}$ m)(1)
Concentric spheres centred on the asteroid’s centre (perpendicular to the radial field lines)(1)
Separation increases with distance because $g$ decreases, and $\Delta r=\dfrac{\Delta V}{g}$ (allow $V\propto 1/r$ argument)(1)
Answer: (a) 520 m s$^{-1}$ (b) $7.0\times10^{5}$ m (d) concentric spheres, spacing increasing outwards
Examiner tip: When the height is comparable to the radius, $mg\Delta h$ is not valid; use $E_p=-GMm/r$ at both points.

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