State Newton’s universal law of gravitation.

A satellite of mass m orbits a planet of mass M. Derive the following relationship between the period of the satellite T and the radius of its orbit R (Kepler’s third law).

\({T^2} = \frac{{4{\pi ^2}{R^3}}}{{GM}}\)

A polar orbiting satellite has an orbit which passes above both of the Earth’s poles. One polar orbiting satellite used for Earth observation has an orbital period of 6.00 × 10³s.
                                                                   Mass of Earth                = 5.97 × 10²⁴ kg
                                                                   Average radius of Earth = 6.37 × 10⁶ m

(i) Using the relationship in (b), show that the average height above the surface of the Earth for this satellite is about 800 km.

(ii) The satellite moves from an orbit of radius 1200 km above the Earth to one of radius 2500 km. The mass of the satellite is 45 kg.

Calculate the change in the gravitational potential energy of the satellite.

(iii) Explain whether the gravitational potential energy has increased, decreased or stayed the same when the orbit changes, as in (c)(ii).

the (attractive) force between two (point) masses is directly proportional to the product of the masses;
and inversely proportional to the square of the distance (between their centres of mass); 
Use of equation is acceptable:
Award [2] if all five quantities defined. Award [1] if four quantities defined.

\(G\frac{{Mm}}{{{R^2}}} = \frac{{m{v^2}}}{R}\) so \({v^2} = \frac{{Gm}}{R}\);
\(v = \frac{{2\pi R}}{T}\);
\({v^2} = \frac{{4{\pi ^2}{R^2}}}{{{T^2}}} = \frac{{Gm}}{R}\);

or

\(G\frac{{Mm}}{{{R^2}}} = m{\omega ^2}R\);
\({\omega ^2} = \frac{{4{\pi ^2}}}{{{T^2}}}\);
\(\frac{{4{\pi ^2}}}{{{T^2}}} = \frac{{GM}}{{{R^3}}}\);
Award [3] to a clear response with a missing step.

(i) \({R^3} = \frac{{6.67 \times {{10}^{ - 11}} \times 5.97 \times {{10}^{24}} \times {{6000}^2}}}{{4 \times {\pi ^2}}}\);
R=7.13x10⁶(m);
h=(7.13x10⁶-6.37x10⁶)=760(km);
Award [3] for an answer of 740 with π taken as 3.14.

(ii) clear use of \(\Delta V = \frac{{\Delta E}}{m}\) and \(V =  - \frac{{Gm}}{r}\) or \(\Delta E = GMm\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)\);
one value of potential energy calculated (2.37×10⁹ or 2.02×10⁹ );
3.5×10⁸ (J);
Award [3] for a bald correct answer.
Award [2] for 7.7x10⁹. Award [1] for 7.7x10¹².
Award [0] for answers using mgΔh.

(iii) increased;
further from Earth / closer to infinity / smaller negative value;
Award [0] for a bald correct answer.

(a) Many candidates stated that the Newton’s law force is proportional to the masses of the objects in question, rather than the product of the masses.

This part was generally well done with most candidates coming to the correct outcome; too often steps were missed out in the derivations and this cost candidates mark. It is essential that they realise that a derivation must include every step. The presentation of this part left much to be desired in quite a large minority with mathematically incorrect statements being given.

(i) Most candidates were able to substitute values into the equation and rearrange it to find a value for R. Many then fail to subtract the radius of the Earth.

(ii) Very few candidates completed this part correctly. Confusion between potential and potential energy was common as were adding the height in kilometres to the radius of the Earth in metres. A sizeable minority of candidates attempted to use the mgh equation.

(iii) This part was quite well answered with most candidates realising that the increase in height meant an increase in potential energy. Several argued that the magnitude decreased but being a negative quantity this meant an increase.