| Date | May 2011 | Marks available | 3 | Reference code | 11M.2.HL.TZ2.8 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Deduce | Question number | 8 | Adapted from | N/A |
Question
Part 2 Orbital motion
A satellite, of mass m, is in orbit about Earth at a distance r from the centre of Earth. Deduce that the kinetic energy EK of the satellite is equal to half the magnitude of the potential energy EP of the satellite.
The graph shows the variation with distance r of the Earth’s gravitational potential V. Values of V for r<R, where R is the radius of Earth, are not shown.
The satellite in (a) has a mass of 8.2×102kg and it is in orbit at a distance of 1.0×107m from the centre of Earth. Using data from the graph and your answer to (a), calculate for the satellite
(i) its total energy.
(ii) its orbital speed.
(iii) the energy it must gain to move to an orbit a distance 2.0×107 m from the centre of the Earth.
Markscheme
\(\frac{{m{v^2}}}{r} = \frac{{GMm}}{{{r^2}}}\);
EK = \(\frac{1}{2}\)mv2=\(\frac{{GMm}}{{2r}}\);
\({E_P} = - \frac{{GMm}}{r}\) (hence magnitude of Ek=½ magnitude of EP);
(i) total energy=(KE+PE=)−\(\frac{{Vm}}{2}\);
\( = \left( { - \frac{{4.0 \times {{10}^7} \times 8.2 \times {{10}^2}}}{2} = } \right) - 1.6 \times {10^{10}}{\rm{J}}\);
(ii) v=\(\sqrt V \); (or use of Ek=½mv2)
=6.3×103ms-1;
(iii) total energy in new orbit=\(\left( { - \frac{{2.0 \times {{10}^7} \times 8.2 \times {{10}^2}}}{2} = } \right) - 0.82 \times {10^{10}}\left( {\rm{J}} \right)\);
energy required=(1.6×1010−0.82×1010=)7.8×109J;
or
total energy is proportional to EP;
so energy required =−(b)(i)÷2=8 or 8.2×109J; (allow ECF from (b)(i))
