M3 January 2012 Q5
5. Above the Earth’s surface, the magnitude of the gravitational force on a particle due to the Earth is inversely proportional to the square of the distance of the particle from the centre of the Earth. The Earth is modelled as a sphere of radius \(R\) and the acceleration due to gravity at the Earth’s surface is \(g\). A particle \(P\) of mass \(m\) is at a height \(x\) above the surface of the Earth.
A rocket is fired vertically upwards from the surface of the Earth. When the rocket is at height \(2R\) above the surface of the Earth its speed is \(\sqrt{\left(\dfrac{gR}{2}\right)}\). You may assume that air resistance can be ignored and that the engine of the rocket is switched off before the rocket reaches height \(R\).
Modelling the rocket as a particle,
| Scheme | Marks |
|---|---|
| Distance of P from the centre of the Earth \(= R + x\) \(F = \dfrac{k}{(R + x)^2}\) | |
| \(x = 0,\ F = mg,\quad k = mg(R)^2\) | M1 A1 |
| \(F = \dfrac{mgR^2}{(R + x)^2}\) * | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(F = ma,\ \ -\dfrac{gR^2}{(R + x)^2} = v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | M1 A1 |
| \(\displaystyle\int_V^{\sqrt{\frac{gR}{2}}} v\,\mathrm{d}v = \int_R^{2R} -\dfrac{gR^2}{(R + x)^2}\,\mathrm{d}x\) | M1 A1 |
| \(\left[\dfrac{1}{2}v^2\right]_V^{\sqrt{\frac{gR}{2}}} = \left[\dfrac{gR^2}{R + x}\right]_R^{2R}\) | M1 A1 |
| \(\dfrac{1}{2} \times \dfrac{gR}{2} - \dfrac{1}{2}V^2 = \dfrac{gR^2}{3R} - \dfrac{gR^2}{2R} = -\dfrac{gR}{6}\) | M1 |
| \(\dfrac{V^2}{2} = \dfrac{gR}{4} + \dfrac{gR}{6} = \dfrac{5gR}{12} \quad V^2 = \dfrac{5gR}{6},\quad V = \sqrt{\dfrac{5gR}{6}}\) | A1, A1 |
| (9) | |
| (12 marks) |