1 problem found
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*}