Question

A particle is projected from point ${}^{\prime }{A}^{\prime }$ , that is at a distance $4R$ from the centre of the Earth, with speed $V_0$ as shown. If the particle passes grazing the earth's surface, then it is found that $V_0=4000\times \sqrt{N}$. Find the value of N.
[Take $\frac{GM}{R}=6.4\times {10}^7\frac{{\text{m}}^2}{{\text{s}}^2}$]

Answer 1

Applying conservation of angular momentum with respect to the centre of earth,
$4R\times m{v}_0\sin {30}^0=mvR$
$v=2v_0$
applying energy conservation
$\frac{1}{2}m(2v_0{)}^2-\frac{GMm}{R}=-\frac{GMm}{4R}+\frac{1}{2}m{v}_0^2$
$\therefore v_0=\sqrt{\frac{GM}{2R}}=\sqrt{\frac{64\times {10}^6}{2}}=4\sqrt{2}\times 1000\frac{m}{s}$
On comparing
$\therefore N=2$