The minimum and maximum distances of a satellite from the centre of the earth are $2R$ and $4R$ respectively, where $R$ is the radius of earth and $M$ is the mass of the earth. The radius of curvature at the point of minimum distance is  $\frac{nR}{3}$. Then $n$ = ______.

Applying conservation of angular momentum
$m{v}_1\left(2R\right)=m{v}_2\left(4R\right)\Rightarrow v_1=2v_2\mathrm{..........}\left(1\right)$
From conservation of energy  
$\frac{1}{2}m{v}_1^2-\frac{GMm}{2R}=\frac{1}{2}m{v}_2^2-\frac{GMm}{4R}$………. (2)
Solving equations (1) and (2), we get  
$v_2=\sqrt{\frac{GM}{6R}},v_1=\sqrt{\frac{2GM}{3R}}$
If r is the radius of curvature at point A
$\frac{m{v}_1^2}{r}=\frac{GMm}{{\left(2R\right)}^2}\Rightarrow r=\frac{4v_1^2{R}^2}{GM}=\frac{8R}{3}\left(puttingvalueof{v}_1\right)$