A body of mass $m$ moves in the gravitational field of a much larger spherical object of mass $M$ located at the origin. Starting from the equations of motion

$$\begin{aligned} \ddot{r}-r \dot{\theta}^{2} &=-\frac{G M}{r^{2}}, \\ r \ddot{\theta}+2 \dot{r} \dot{\theta} &=0, \end{aligned}$$

show that:

(i) the body moves in an orbit of the form

$$\frac{h^{2} u}{G M}=1+e \cos \left(\theta-\theta_{0}\right)$$

where $u=1 / r, h$ is the constant angular momentum per unit mass, and $e$ and $\theta_{0}$ are constants;

(ii) the total energy of the body is

$$E=\frac{m G^{2} M^{2}}{2 h^{2}}\left(e^{2}-1\right)$$

A meteorite is moving very far from the Earth with speed $V$, and in the absence of the effect of the Earth's gravitational field would miss the Earth by a shortest distance $b$ (measured from the Earth's centre). Show that in the subsequent motion

$$h=b V$$

and

$$e=\left[1+\frac{b^{2} V^{4}}{G^{2} M^{2}}\right]^{\frac{1}{2}}$$

Use equation $(*)$ to find the distance of closest approach, and show that the meteorite will collide with the Earth if

$$b<\left[R^{2}+\frac{2 G M R}{V^{2}}\right]^{\frac{1}{2}}$$

where $R$ is the radius of the Earth.