Date | November 2013 | Marks available | 6 | Reference code | 13N.2.HL.TZ0.8 |
Level | Higher level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Determine | Question number | 8 | Adapted from | N/A |
Question
Part 2 Gravitational potential
Define gravitational potential at a point in a gravitational field.
The graph shows how the gravitational potential V of Earth varies with distance R from the centre of Earth in the range R=2.0×108 m to R=5.0×108 m.
The Moon is at a distance of 4.0×108 m from the centre of Earth. At some time in the past it was at a distance of 2.7×108 m from the centre of Earth.
Use the graph opposite to determine
(i) the present day magnitude of the acceleration of the Moon.
(ii) by how much the potential energy of the Moon has changed as a result of moving from R=2.7×108 m to R=4.0×108 m. The mass of the Moon is 7.4×1022 kg.
State why the change of potential energy in (f)(ii) is an increase.
Markscheme
work done per unit mass;
in bringing (test) mass from infinity to point;
reference to small/point (test) mass;
(i) tangent construction attempted at R=4.0×108 m;
triangle/pair of coordinates used in calculation;
attempt to calculate gradient;
2.5×10–3 ms−2; (accept answers in the range of 2.2 to 2.7)
Award [1 max] for \(\frac{V}{R}\) to give (–)2.1×10–3.
(ii) change in V=0.45×106-0.50×106 Jkg−1;
change in PE = (0.5×106×7.4×1022=)3.3-3.7×1028 J;
work is done against the gravitational field of Earth / Moon is now closer to infinity/further from Earth / \(\frac{{ - GMm}}{R}\) means that as R increases potential increases/becomes less negative;