### Question

A moon \(\mathrm{M}\) orbits a planet \(\mathrm{P}\). The gravitational field strength at the surface of \(\mathrm{P}\) due to \(\mathrm{P}\) is \(g_{\mathrm{P}}\). The gravitational field strength at the surface of \(M\) due to \(M\) is \(g_M\).

For \(\mathrm{M}\) and \(\mathrm{P}\) :

$

\frac{\text { radius of } M}{\text { radius of } P}=0.27 \text { and } \frac{\text { mass of } M}{\text { mass of } P}=0.055

$

(a) Determine \(\frac{g_M}{g_p}\).

(b) Point O lies on the line joining the centre of M to the centre of P.

The graph shows the variation of gravitational potential V with distance x from the

surface of P to O.

The gradient of the graph is zero at point O.

(i) State and explain the magnitude of the resultant gravitational field strength at \(\mathrm{O}\).[2]

(ii) Outline why the graph between \(\mathrm{P}\) and \(\mathrm{O}\) is negative.[2]

(iii) Show that the gravitational potential \(V_P\) at the surface of \(P\) due to the mass of \(P\) is given by \(V_{\mathrm{p}}=-g_{\mathrm{p}} R_{\mathrm{p}}\) where \(R_{\mathrm{p}}\) is the radius of the planet.[2]

(iv) The gravitational potential due to the mass of \(\mathrm{M}\) at the surface of \(\mathrm{P}\) can be assumed to be negligible.

Estimate, using the graph, the gravitational potential at the surface of \(\mathrm{M}\) due to the mass of \(\mathrm{M}\). [2]

(v) Draw on the axes the variation of gravitational potential between O and M. [1]

**▶️Answer/Explanation**

Ans:

Work using \(g \propto \frac{m}{r^2} \checkmark\)

$

\frac{g_M}{g_p}=\frac{m_M}{m_p}\left(\frac{r_p}{r_M}\right)^2=0.75

$

b i $

g=0

$

As \(g \&=-\frac{\Delta V_g}{\Delta r}\) which \(»\) is the gradient of the graph

OR

As the force of attraction/field strength of \(\mathrm{P}\) and \(\mathrm{M}\) are equal

b ii The gravitational field is attractive so that energy is required «to move away from \(P_n \checkmark\) the gravitational potential is defined as 0 at \(\infty\), (the potential must be negative) \(\checkmark\)

iii \(V_{\mathrm{P}}=-\frac{G M}{R_{\mathrm{p}}}\) AND \(g_{\mathrm{P}}=\frac{G M}{R_{\mathrm{p}}^2}\) (at surface)

Suitable working and cancellation of \(G\) and \(M\) seen \(\checkmark\) \(V_p=-g_p R_p\)

iv \(\begin{aligned} & \alpha \frac{V_M}{V_p}=\frac{g_M R_M}{g_p R_p}=0.75 \times 0.27 w=0.20^{\checkmark} \\ & V_M=\alpha-6.4 \times 10^7 \times 0.2=» \approx-» 1.3 \times 10^7 \ll \mathrm{J} \mathrm{kg}^{-1} \rightsquigarrow \\ & \end{aligned}\)

v.Line always negative, of suitable shape and end point below -8 and above -20 unless awarding \(E C F\) from \(b\) (iv)

This question is about a probe in orbit.

A probe of mass *m* is in a circular orbit of radius *r* around a spherical planet of mass *M*.

a.

State why the work done by the gravitational force during one full revolution of the probe is zero.

Deduce for the probe in orbit that its

(i) speed is \(v = \sqrt {\frac{{GM}}{r}} \).

(ii) total energy is \(E = – \frac{{GMm}}{{2r}}\).

It is now required to place the probe in another circular orbit further away from the planet. To do this, the probe’s engines will be fired for a very short time.

State and explain whether the work done on the probe by the engines is positive, negative **or** zero.

**Answer/Explanation**

## Markscheme

a.

because the force is always at right angles to the velocity / motion/orbit is an equipotential surface; *Do not accept answers based on the displacement being zero for a full revolution.*

(i) equating gravitational force \(\frac{{GMm}}{{{r^2}}}\);

to centripetal force \(\frac{{m{v^2}}}{r}\) to get result;

(ii) kinetic energy is \(\frac{{GMm}}{{2r}}\);

addition to potential energy −\(\frac{{GMm}}{{r}}\) to get result;

the total energy (at the new orbit) will be greater than before/is less negative;

hence probe engines must be fired to produce force in the direction of motion / positive work must be done (on the probe); *Award [1] for mention of only potential energy increasing.*