Outline why Phobos moves with uniform circular motion.

Show that the orbital speed of Phobos is about \({\text{2 km}}\,{{\text{s}}^{ - 1}}\).

Deduce the mass of Mars.

gravitational provides centripetal force / gravitational provides force towards centre; (because radius is implied constant) (centripetal) force is constant;

at 90° to velocity (vector)/orbit/direction / OWTTE / \(\frac{{GmM}}{{{r^2}}} = \frac{{m{v^2}}}{r}\) (or re-arranged) and therefore speed is constant (and motion is uniform); } (do not allow “inwards/centripetal” for this mark. The right angle must be explicit)

\(v = \omega r\) and \(\omega  = \frac{{2\pi }}{T}\) combined; } (allow approach from speed \( = \frac{s}{T}\), do not allow approach from \({v^2} = ar\) or \(f = \frac{1}{T}\))

\(v = \left( {\frac{{2\pi r}}{T} = } \right)\frac{{2\pi  \times 9.4 \times {{10}^6}}}{{7.7 \times 3600}}\) or \(2.1(3) \times {10^3}{\text{ (m}}\,{{\text{s}}^{ - 1}})\);

\(m\frac{{{v^2}}}{r} = G\frac{{mM}}{{{r^2}}}\) or \({F_{\text{c}}} = {F_{\text{G}}}\);

\(M = \frac{{{v^2}r}}{G}\) or \(\frac{{{{(2.13 \times {{10}^3})}^2} \times 9.4 \times {{10}^6}}}{{6.67 \times {{10}^{ - 11}}}}\); (allow power of ten error in this mark)

\(M = 6.4 \times {10^{23}}{\text{ (kg)}}\) from 2.13 or \(5.6 \times {10^{23}}{\text{ (kg)}}\) from 2;

Award [3] for a bald correct answer.

Candidates were asked to outline the real meaning of “uniform circular motion”. They were required to link the gravitational force acting on Phobos due to Mars (and the constancy of this force) to the dynamics of the force direction associated with the orbit and its consequences for the change in velocity (and lack of change in speed). Few managed to score all points with the majority managing to score 2 out of the 3 available.

This was a particularly simple “show that” question. Once again, examiners saw considerable numbers of answers that gave little information about the origin of the solution. As in past examinations, examiners saw much pure substitution without any explanation of its origin. This does not score well. It is best practice for candidates to present a full argument in calculations, and in “show that” and “deduce” questions it is essential.

Candidates were on surer ground with the deduction of the mass of Mars. An algebraic starting point was allowed and many scored all 3 marks. However, a very large number failed to arrive at the correct numerical answer due to errors in powers of ten from the data provided.