A particle is projected vertically upwards at speed $V$ from the surface of the Earth, which may be treated as a perfect sphere. The variation of gravity with height should not be ignored, but the rotation of the Earth should be. Show that the height $z(t)$ of the particle obeys

$$\ddot{z}=-\frac{g R^{2}}{(R+z)^{2}},$$

where $R$ is the radius of the Earth and $g$ is the acceleration due to gravity measured at the Earth's surface.

Using dimensional analysis, show that the maximum height $H$ of the particle and the time $T$ taken to reach that height are given by

$$H=R F(\lambda) \quad \text { and } \quad T=\frac{V}{g} G(\lambda)$$

where $F$ and $G$ are functions of $\lambda=V^{2} / g R$.

Write down the equation of conservation of energy and deduce that

$$T=\int_{0}^{H} \sqrt{\frac{R+z}{V^{2} R-\left(2 g R-V^{2}\right) z}} d z$$

Hence or otherwise show that

$$F(\lambda)=\frac{\lambda}{2-\lambda} \quad \text { and } \quad G(\lambda)=\int_{0}^{1} \sqrt{\frac{2-\lambda+\lambda x}{(2-\lambda)^{3}(1-x)}} d x$$