| Date | May 2015 | Marks available | 2 | Reference code | 15M.2.HL.TZ2.6 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Outline | Question number | 6 | Adapted from | N/A |
Question
Part 2 Motion of a rocket
A rocket is moving away from a planet within the gravitational field of the planet. When the rocket is at position P a distance of 1.30×107m from the centre of the planet, the engine is switched off. At P, the speed of the rocket is 4.38×103ms–1.
At a time of 60.0 s later, the rocket has reached position Q. The speed of the rocket at Q is 4.25×103ms–1. Air resistance is negligible.
Outline, with reference to the energy of the rocket, why the speed of the rocket is changing between P and Q.
Estimate the average gravitational field strength of the planet between P and Q.
(i) An object is a distance r from the centre of a planet. Show that the minimum speed required to escape the gravitational field is equal to
\[\sqrt {2g'r} \]
where g′ is the gravitational field strength at distance r from the centre of a planet.
(ii) Discuss, using a calculation, whether the rocket at P can completely escape the gravitational field of the planet without further use of the engine.
A space station is in orbit at a distance r from the centre of the planet in (e)(i). A satellite is launched from the space station so as just to escape from the gravitational field of the planet. The launch takes place in the same direction as the velocity of the space station. Outline why the launch velocity relative to the space station can be less than your answer to (e)(i).
Markscheme
gravitational potential energy is being gained;
this is at the expense of kinetic energy (and speed falls);
\(\left( {{\rm{acceleration}} = \frac{{\left( {v - u} \right)}}{t} = \frac{{4.25 \times {{10}^3} - 4.38 \times {{10}^3}}}{{60}} = } \right)\left( - \right)2.17\left( {{\rm{m}}{{\rm{s}}^{ - 2}}} \right)\);
gravitational field strength = acceleration of rocket (=2.17 N kg–1); } (allow g = a in symbols)
or
computes potential difference from KE per unit mass change (5.61\( \times \)105),
computes distance travelled (0.259 Mm), uses \(g = \frac{{\left( - \right)\Delta V}}{{\Delta r}}\);
g=(\( - \))2.17(ms-2);
(i) \(\left( {{\rm{gravitational force}}} \right){\rm{ = }}mg'{\rm{ = }}\frac{{{\rm{GMm}}}}{{{r^2}}}\);
\(\left( {{\rm{kinetic energy}}} \right){\rm{ = }}\frac{{{\rm{m}}{{\rm{v}}^{\rm{2}}}}}{2} = \frac{{GMm}}{r}\);
\({v^2} = 2\frac{{GM}}{{{r^2}}}r\left( { = 2g'r} \right)\);
Award [0] for centripetal force argument as rocket is not in orbit.
(ii) calculation of speed at a relevant distance eg: \(\sqrt {2 \times g' \times 13 \times {{10}^6}} = 7500\left( {{\rm{m}}{{\rm{s}}^{ - 1}}} \right)\);
speed is less than this so will not escape; } (allow ECF from (d) which could lead to rocket able to escape)
Award [1 max] for use of g=9.81 and r which gives 16000 ms–1 .
the satellite has velocity/kinetic energy as it is orbiting with the space station;
