The orbital angular momentum of a satellite is \(L\), when it is revolving in a circular orbit at height \(h\) from earth surface. If the distance of satellite from the earth centre is increased by eight times to its initial value, then the new angular momentum will be \(............\,L\)

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Answer & Solution

Correct Answer

(D) \(3\)

Detailed explanation

\(L = mvr\)

\(v =\sqrt{\frac{ GM _{\odot}}{ r }}\)

\(L=m \sqrt{\frac{ GM _e}{ r }} \cdot r\)

\(L \propto r ^{\frac{1}{2}}\)

Now distance from centre is increased by \(8\) times.

So new distance from centre \(=r+8 r=9 r\)

Now angular momentum \(L^{\prime} \propto(9 r )^{1 / 2}\)

So new distance from centre \(=r+8 r=9 r\)

Now angular momentum \(L^{\prime} \propto(9 r)^{1 / 2}\)

\(\frac{L}{L^{\prime}}=\frac{ r ^{1 / 2}}{(9 r )^{1 / 2}}=\frac{1}{3}\)

\(L^{\prime}=3 L\)

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