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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.

 

[1]
a.

Deduce for the probe in orbit that its

(i) speed is \(v = \sqrt {\frac{{GM}}{r}} \).

(ii) total energy is \(E =  - \frac{{GMm}}{{2r}}\).

[4]
b.

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.

 

[2]
c.

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.

a.

(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;

b.

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.

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Additional higher level (AHL) » Topic 10: Fields » 10.2 – Fields at work
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