Paper 4, Section II, 9C\mathbf{9 C}

Dynamics and Relativity | Part IA, 2015

A particle is projected vertically upwards at speed VV 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)z(t) of the particle obeys

z¨=gR2(R+z)2,\ddot{z}=-\frac{g R^{2}}{(R+z)^{2}},

where RR is the radius of the Earth and gg is the acceleration due to gravity measured at the Earth's surface.

Using dimensional analysis, show that the maximum height HH of the particle and the time TT taken to reach that height are given by

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

where FF and GG are functions of λ=V2/gR\lambda=V^{2} / g R.

Write down the equation of conservation of energy and deduce that

T=0HR+zV2R(2gRV2)zdzT=\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(λ)=λ2λ and G(λ)=012λ+λx(2λ)3(1x)dxF(\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

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