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1999 Paper 2 Q9
D: 1600.0 B: 1500.0
variable force differential equation energy conservation inverse power law radial motion velocity-displacement escape velocity optimisation

In the \(Z\)--universe, a star of mass \(M\) suddenly blows up, and the fragments, with various initial speeds, start to move away from the centre of mass \(G\) which may be regarded as a fixed point. In the subsequent motion the acceleration of each fragment is directed towards \(G\). Moreover, in accordance with the laws of physics of the \(Z\)--universe, there are positive constants \(k_1\), \(k_2\) and \(R\) such that when a fragment is at a distance \(x\) from \(G\), the magnitude of its acceleration is \(k_1x^3\) if \(x < R\) and is \(k_2x^{-4}\) if \(x \ge R\). The initial speed of a fragment is denoted by \(u\).

  1. For \(x < R\), write down a differential equation for the speed \(v\), and hence determine \(v\) in terms of \(u\), \(k_1\) and \(x\) for \( x < R\).
  2. Show that if \(u < a\), where \(2a^2=k_1 R^4\), then the fragment does not reach a distance \(R\) from \(G\).
  3. Show that if \(u \ge b\), where $ 6b^2= 3k_1R^4 + 4k_2 /R^3, $ then from the moment of the explosion the fragment is always moving away from \(G\).
  4. If \(a < u < b\), determine in terms of \(k_2\), \(b\) and \(u\) the maximum distance from \(G\) attained by the fragment.

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1992 Paper 1 Q13
D: 1500.0 B: 1484.0
variable force gravitational field inverse square law energy conservation escape velocity integration differential equation radial motion comet

A comet, which may be regarded as a particle of mass \(m\), moving in the sun's gravitational field, at a distance \(x\) from the sun, experiences a force \(Gm/x^{2}\) (where \(G\) is a constant) directly towards the sun. Show that if, at some time, \(x=h\) and the comet is travelling directly away from the sun with speed \(V\), then \(x\) cannot become arbitrarily large unless \(V^{2}\geqslant2G/h\). A comet is initially motionless at a great distance from the sun. If, at some later time, it is at a distance \(h\) from the sun, how long after that will it take to fall into the sun?

Solution: Consider \(E = \frac12 m \dot{x}^2 - \frac{Gm}{x}\), notice that \begin{align*} && \dot{E} &= m \dot{x} \ddot{x} + \frac{Gm}{x^2} \dot{x} \\ &&&= \dot{x} \underbrace{\left (m\ddot{x} + \frac{Gm}{x^2} \right)}_{=0 \text{ by N2}} \end{align*} Therefore \(E\) is conserved. Therefore if \(x \to \infty\) \(\frac12 m V^2 - \frac{Gm}{h} = \frac12 m u^2 - 0 \geq 0\) so \(V^2 \geqslant 2G/h\) Since \(E \approx 0\) we want to solve \begin{align*} && \dot{x} &= -\sqrt{\frac{2G}{x}} \\ \Rightarrow && -\int_h^0 \sqrt{x} \d x &= \int_0^T \sqrt{2G} \d t \\ \Rightarrow && \frac{2h^{3/2}}{3} &= \sqrt{2G}T \\ \Rightarrow && T &= \frac{\sqrt{2}h^{3/2}}{3\sqrt{G}} = \frac13 \sqrt{\frac{2h^3}{G}} \end{align*}

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