Gravitational fields and orbits: practice worksheet

1. Question 1warm-up · calculator

   A planet of radius $R$ has a gravitational field strength $g_0$ at its surface. The planet may be treated as a uniform sphere.

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

   Markscheme

   force per unit mass on a «small/point» test mass placed at the point

   1

   Accept $g = \frac{F}{m}$ with symbols defined. OWTTE.

   Answer: The force per unit mass acting on a small test mass placed at the point.

   (b)Find, in terms of $g_0$, the gravitational field strength at a height $2R$ above the surface of the planet.[2]

   Markscheme

   distance from centre of planet is $3R$ «so $g \propto \frac{1}{(3R)^2}$»	1
   $g = \frac{g_0}{9}$	1

   Using $2R$ as the distance from the centre scores zero.

   Answer: $\frac{g_0}{9}$

2. Question 2warm-up · calculator

   Two small moons, A and B, move in circular orbits about the same planet. Moon A has orbital radius $r$ and period $T$. Moon B has orbital radius $4r$.

   (a)State Kepler's third law of planetary motion.[1]

   Markscheme

   «square of» period is proportional to «cube of» orbital radius / $T^2 \propto r^3$

   1

   Accept 'mean distance from the central body' or 'semi-major axis' in place of radius.

   Answer: $T^2 \propto r^3$ for bodies orbiting the same central mass.

   (b)Find the orbital period of moon B in terms of $T$.[2]

   Markscheme

   $\left(\frac{T_B}{T}\right)^2 = 4^3 = 64$	1
   $T_B = 8T$	1

   Answer: $8T$

   (c)Write down the ratio $\frac{\text{orbital speed of B}}{\text{orbital speed of A}}$.[1]

   Markscheme

   $\frac{v_B}{v_A} = \frac{1}{2}$ «since $v \propto \frac{1}{\sqrt{r}}$»

   1

   Accept 0.5.

   Answer: $\frac{1}{2}$

3. Question 3exam-style · calculator

   A satellite moves in a circular orbit about a planet. Data for the orbit are given below.
   
   Orbital radius $r = 8.0 \times 10^{6}$ m
   Orbital period $T = 2.0$ hours
   
   The planet may be treated as a uniform sphere of mass $M$.

   (a)Show that the speed $v$ of a satellite in a circular orbit of radius $r$ is given by $v = \sqrt{\frac{GM}{r}}$.[2]

   Markscheme

   gravitational force provides the centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$	1
   cancels $m$ and one $r$ to give $v^2 = \frac{GM}{r}$ «hence result»	1

   Answer: $v = \sqrt{\frac{GM}{r}}$

   (b)Determine the mass of the planet.[3]

   Markscheme

   uses $v = \frac{2\pi r}{T}$ OR $T^2 = \frac{4\pi^2 r^3}{GM}$	1
   $M = \frac{4\pi^2 (8.0\times10^{6})^3}{6.67\times10^{-11}\times(7200)^2}$	1
   $M = 5.8 \times 10^{24}$ kg	1

   Accept $5.8$–$5.9 \times 10^{24}$ kg. ECF for a period left in hours at the substitution line only.

   Answer: $M = \frac{4\pi^2 r^3}{GT^2} \approx 5.85 \times 10^{24}$ kg

   (c)Calculate the orbital speed of the satellite.[2]

   Markscheme

   $v = \frac{2\pi \times 8.0\times10^{6}}{7200}$ OR $v = \sqrt{\frac{6.67\times10^{-11}\times5.8\times10^{24}}{8.0\times10^{6}}}$	1
   $v = 7.0 \times 10^{3}$ m s$^{-1}$	1

   ECF from (b).

   Answer: $v = \frac{2\pi r}{T} \approx 6.98 \times 10^{3}$ m s$^{-1}$

4. Question 4exam-style · calculator

   A planet and its moon may be treated as point masses. Their centres are separated by a fixed distance $d$.
   
   Mass of planet $M_P = 5.0 \times 10^{24}$ kg
   Mass of moon $M_m = 2.0 \times 10^{23}$ kg
   Separation of centres $d = 3.0 \times 10^{8}$ m
   
   Point X lies on the line joining the two centres. At X the resultant gravitational field strength is zero.

   (a)Explain why there must be a point on the line joining the centres at which the resultant gravitational field strength is zero.[2]

   Markscheme

   between the bodies the field of the planet and the field of the moon act in opposite directions «along the line joining the centres»	1
   field strength of planet decreases and that of the moon increases with distance from the planet, so at some point the magnitudes are equal «and cancel»	1

   OWTTE.

   Answer: The two fields point in opposite directions along the line; the planet's field falls and the moon's field rises as one moves from planet to moon, so their magnitudes are equal somewhere between them.

   (b)Determine the distance of X from the centre of the planet.[3]

   Markscheme

   $\frac{GM_P}{x^2} = \frac{GM_m}{(d-x)^2}$	1
   $\frac{d-x}{x} = \sqrt{\frac{2.0\times10^{23}}{5.0\times10^{24}}} = \frac{1}{5}$	1
   $x = 2.5 \times 10^{8}$ m	1

   Accept equivalent forms, e.g. $x = \frac{5}{6}d$.

   Answer: $x = \frac{5d}{6} = 2.5 \times 10^{8}$ m

   (c)Calculate the gravitational potential at X.[2]

   Markscheme

   $V = -6.67\times10^{-11}\left(\frac{5.0\times10^{24}}{2.5\times10^{8}} + \frac{2.0\times10^{23}}{5.0\times10^{7}}\right)$	1
   $V = -1.6 \times 10^{6}$ J kg$^{-1}$	1

   Sign of potential is implied at the substitution line but required in the final answer. ECF from (b).

   Answer: $V = -G\left(\frac{M_P}{x} + \frac{M_m}{d-x}\right) = -1.60 \times 10^{6}$ J kg$^{-1}$

   (d)Explain why the gravitational potential at X is not zero even though the gravitational field strength at X is zero.[1]

   Markscheme

   potential is a scalar, so the two «negative» contributions add rather than cancel / field is a vector so opposite fields can cancel

   1

   OWTTE.

   Answer: Potential is a scalar; both contributions are negative and add, whereas the fields are vectors in opposite directions and cancel.

5. Question 5exam-style · calculator

   A satellite of mass $m = 450$ kg is in a circular orbit about a planet of mass $M = 6.0 \times 10^{24}$ kg.
   
   Initial orbital radius $r_1 = 7.0 \times 10^{6}$ m
   Final orbital radius $r_2 = 1.4 \times 10^{7}$ m

   (a)Show that the total energy of a satellite of mass $m$ in a circular orbit of radius $r$ is $E = -\frac{GMm}{2r}$.[2]

   Markscheme

   kinetic energy $= \frac{1}{2}mv^2 = \frac{GMm}{2r}$ «using $v^2 = \frac{GM}{r}$»	1
   $E = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$	1

   Potential energy $E_p = -\frac{GMm}{r}$ must appear explicitly for the second mark.

   Answer: $E = -\frac{GMm}{2r}$

   (b)Calculate the energy that must be supplied to move the satellite from the orbit of radius $r_1$ to the orbit of radius $r_2$.[3]

   Markscheme

   $\Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$	1
   $\Delta E = \frac{6.67\times10^{-11}\times6.0\times10^{24}\times450}{2}\left(\frac{1}{7.0\times10^{6}} - \frac{1}{1.4\times10^{7}}\right)$	1
   $\Delta E = 6.4 \times 10^{9}$ J	1

   Accept $6.4$–$6.5 \times 10^{9}$ J. Award [max 2] if the change in potential energy alone ($1.3 \times 10^{10}$ J) is given.

   Answer: $\Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right) \approx 6.43 \times 10^{9}$ J

   (c)When in the lower orbit the satellite experiences a small resistive force from the upper atmosphere of the planet. Explain why the speed of the satellite increases as a result of this force.[3]

   Markscheme

   resistive force does negative work / total energy of the satellite decreases	1
   total energy $-\frac{GMm}{2r}$ becomes more negative so the orbital radius decreases / satellite spirals inwards	1
   kinetic energy $\frac{GMm}{2r}$ increases as $r$ decreases, so speed increases	1
   decrease in potential energy is larger than the work done against the resistive force / potential energy falls by twice the rise in kinetic energy	1

   Award [max 3]. OWTTE.

   Answer: Drag removes total energy; the orbit radius decreases; since $E_k = \frac{GMm}{2r}$ the kinetic energy and hence speed rise, the loss of potential energy being greater than the work done against drag.

6. Question 6challenge · calculator

   A small moon of a distant planet may be treated as a uniform sphere.
   
   Mass of moon $M = 1.2 \times 10^{23}$ kg
   Radius of moon $R = 1.5 \times 10^{6}$ m
   
   A probe of mass $120$ kg rests on the surface of the moon. The moon does not rotate.

   (a)Define gravitational potential at a point.[1]

   Markscheme

   work done per unit mass in bringing a «small/point» test mass from infinity to the point

   1

   OWTTE. Reference to infinity is required.

   Answer: The work done per unit mass in bringing a small test mass from infinity to the point.

   (b)Show that the escape speed from the surface of the moon is $v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$.[2]

   Markscheme

   total energy at surface $= \frac{1}{2}mv^2 - \frac{GMm}{R}$ and equals zero «at infinity with zero speed»	1
   $\frac{1}{2}mv^2 = \frac{GMm}{R}$ leading to $v^2 = \frac{2GM}{R}$	1

   Accept an argument in terms of $\frac{1}{2}v^2 = -V_g$.

   Answer: $v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$

   (c)Calculate the escape speed from the surface of the moon.[2]

   Markscheme

   $v_{\text{esc}} = \sqrt{\frac{2\times6.67\times10^{-11}\times1.2\times10^{23}}{1.5\times10^{6}}}$	1
   $v_{\text{esc}} = 3.3 \times 10^{3}$ m s$^{-1}$	1

   Answer: $v_{\text{esc}} \approx 3.27 \times 10^{3}$ m s$^{-1}$

   (d)Determine the minimum energy that must be supplied to place the probe, starting from rest on the surface, into a circular orbit of radius $2R$ about the moon.[3]

   Markscheme

   energy in orbit $= -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}$	1
   $\Delta E = -\frac{GMm}{4R} - \left(-\frac{GMm}{R}\right) = \frac{3GMm}{4R}$	1
   $\Delta E = \frac{3\times6.67\times10^{-11}\times1.2\times10^{23}\times120}{4\times1.5\times10^{6}} = 4.8 \times 10^{8}$ J	1

   Award [max 2] if only the change in potential energy ($3.2 \times 10^{8}$ J) is calculated, i.e. the orbital kinetic energy is omitted.

   Answer: $\Delta E = \frac{3GMm}{4R} \approx 4.80 \times 10^{8}$ J

   (e)The moon has a very thin atmosphere of temperature $150$ K containing hydrogen molecules of mass $3.3 \times 10^{-27}$ kg. Estimate the root mean square speed of the hydrogen molecules. Take $k_{\mathrm{B}} = 1.38 \times 10^{-23}$ J K$^{-1}$.[2]

   Markscheme

   $\frac{1}{2}m\langle v^2\rangle = \frac{3}{2}k_{\mathrm{B}}T$ OR $v_{\text{rms}} = \sqrt{\frac{3\times1.38\times10^{-23}\times150}{3.3\times10^{-27}}}$	1
   $v_{\text{rms}} = 1.4 \times 10^{3}$ m s$^{-1}$	1

   Answer: $v_{\text{rms}} = \sqrt{\frac{3k_{\mathrm{B}}T}{m}} \approx 1.37 \times 10^{3}$ m s$^{-1}$

   (f)Using your answers to (c) and (e), discuss whether the moon can retain hydrogen in its atmosphere over a long period of time.[2]

   Markscheme

   compares $v_{\text{rms}}$ «$1.4 \times 10^{3}$ m s$^{-1}$» with the escape speed «$3.3 \times 10^{3}$ m s$^{-1}$»: $v_{\text{rms}}$ is smaller, by a factor of only about 2.4	1
   molecular speeds are distributed / some molecules have speeds greater than the escape speed, so these escape and the hydrogen is gradually lost	1

   Award [max 2]. ECF from (c) and (e). A conclusion consistent with the candidate's own values is required for the second mark. OWTTE.

   Answer: The rms speed is below the escape speed, but molecular speeds are distributed and a fraction of molecules always exceed $3.3 \times 10^{3}$ m s$^{-1}$; these escape and, over a long time, the hydrogen is lost.

Original questions and notes written by Pietro Meloni, PhD, from the IB syllabus. Not endorsed by or affiliated with the International Baccalaureate Organization. Found a mistake? Please tell me privately at pietro@pietromeloni.com. More questions on this topic: D. Fields practice.