# IB DP Physics D.1 Gravitational fields IB Style Question Bank HL Paper 2

### 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)

Question

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.

[1]
b.

Deduce for the probe in orbit that its

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

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

[4]
c.

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.

[2]

## 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.

b.

(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;

c.

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.

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