Date | May 2011 | Marks available | 2 | Reference code | 11M.2.HL.TZ1.2 |
Level | Higher level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Deduce | Question number | 2 | Adapted from | N/A |
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.
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.
Markscheme
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.