Question

Minimum and maximum distance of a satellite from the center of earth are $2R$ and $4R$ respectively where $R$ is radius of earth. Radius of curvature at the point when satellite is at a distance $2R$ is $nR$, here $n$ is(Answer upto two decimal places)

Answer 1

$\frac{GMm}{(2R{)}^2}=\frac{m{V}_1^2}{r_1}\Rightarrow r_1=V_1^2\frac{4R^2}{GM}$

By conservation of angular momentum,
$m{V}_1\times 2R=m{V}_2\times 4R$
$V_1=2V_2$ ………..(i)

By Consevation of energy
$-\frac{GMm}{2R}+\frac{1}{2}m{V}_1^2=-\frac{GMm}{4R}+\frac{1}{2}m{V}_2^2$ ……(ii)

From (i) and (ii)
$r_1=\frac{8}{3}R$