Date | November 2014 | Marks available | 3 | Reference code | 14N.2.HL.TZ0.4 |
Level | Higher level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Show that | Question number | 4 | Adapted from | N/A |
Question
This question is about the energy of an orbiting satellite.
A space shuttle of mass m is launched in the direction of the Earth’s South Pole.
The shuttle enters a circular orbit of radius \(R\) around the Earth.
The kinetic energy \({E_{\text{K}}}\) given to the shuttle at its launch is given by the expression
\[{E_{\text{K}}} = \frac{{7GMm}}{{8{R_{\text{E}}}}}\]
where \(G\) is the gravitational constant, \(M\) is mass of the Earth and \({R_{\text{E}}}\) is the radius of the Earth. Deduce that the shuttle cannot escape the gravitational field of the Earth.
Show that the total energy of the shuttle in its orbit is given by \( - \frac{{GMm}}{{2R}}\). Air resistance is negligible.
Using the expression for \({E_{\text{K}}}\) in (a) and your answer to (b)(i), determine \(R\) in terms of \({R_{\text{E}}}\).
In practice, the total energy of the shuttle decreases as it collides with air molecules in the upper atmosphere. Outline what happens to the speed of the shuttle when this occurs.
Markscheme
KE needs to be \( \ge \) (magnitude of) GPE at surface \(\left( { - \frac{{GMm}}{{{R_E}}}} \right)\);
But KE is \(\frac{{7GMm}}{{8{R_{\text{E}}}}} < \frac{{GMm}}{{{R_{\text{E}}}}}\) / OWTTE;
or
shows that total energy at launch \( = - \frac{{GMm}}{{8{R_{\text{E}}}}}\); (appropriate working required)
this is \( < 0\), so escape impossible;
or
states that escape velocity needed is \(\sqrt {\frac{{2GM}}{{{R_{\text{E}}}}}} \);
shows launch velocity is only \(\sqrt {\frac{{7GM}}{{4{R_{\text{E}}}}}} \); (appropriate working required)
\({E_{{\text{tot}}}} = {\text{PE}} + {\text{KE}}\);
shows that kinetic energy \( = (\frac{1}{2}m{v^2} = ){\text{ }}\frac{{GMm}}{{2R}}\); (appropriate working required)
adds PE \(\left( { - \frac{{GMm}}{R}} \right)\) and KE to get given answer; (appropriate working required)
\( - \frac{{GMm}}{{{R_{\text{E}}}}} + \frac{{7GMm}}{{8{R_{\text{E}}}}} = - \frac{{GMm}}{{2R}}\); (equating total energy at launch and in orbit)
\(\frac{1}{{8{R_{\text{E}}}}} = \frac{1}{{2R}}\);
\(R = 4{R_{\text{E}}}\);
Award [0] for an answer such as \(R = \frac{{4{R_{\text{E}}}}}{7}\).
total energy decreases/becomes a greater negative value, so R decreases;
as \(R\) decreases kinetic energy increases;
speed increases;
Allow third marking point even if reasoning is incorrect.
Examiners report
Candidates were asked to show that the shuttle could not escape the Earth’s field. There are many ways of approaching this, but in general answers were good, with only the weakest candidates failing to know where to start.
The determination of total energy of a mass in Earth orbit is a standard classroom derivation. Most were able to reproduce it, but not everyone explained how the formula for orbital KE was derived.
Determining the radius of the orbit proved difficult for most candidates with many obtaining negative values or orbits with a radius less than the radius of the Earth. This was due to carelessness with the symbols used for the two different radii.
Few candidates equated a decrease in total energy with an increasingly negative value. The consequent fact that the radius of the orbit decreases and velocity increases was counter-intuitive for most candidates. Most incorrectly opted for a decrease in KE due to resistive forces.