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Question

For energy in radius $r_{i}=\left|\frac{-G M m}{2 r}\right|=n k$

where $n$ is integer $k$ is constant energy, now

$\frac{G M m}{2 R}=n k$

$\f

Vedclass · Accepted Answer

$3R$

Vedclass · Answer

For energy in radius $r_{i}=\left|\frac{-G M m}{2 r}\right|=n k$

where $n$ is integer $k$ is constant energy, now

$\frac{G M m}{2 R}=n k$

$\frac{G M m}{2\left(\frac{3 R}{2}\right)}=(n-1) k$

so solving $\mathrm{k}=\frac{G M m}{6 R}$

now this implies that for $R_{\max }$

$\frac{G M m}{2 R_{\max }}=(1) \frac{G M m}{6 R} \Rightarrow R_{\max }=3 R$

On a hypothetical planet satellite can only revolve in quantized energy level i.e. magnitude of energy of a satellite is integer multiple of a fixed energy. If two successive orbit have radius $R$ and $\frac{3R}{2}$ what could be maximum radius of satellite

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