Gravitation

JEE Main level, single correct, medium

208 of 223

Question 208 of 223

The orbital angular momentum of a satellite is LL, when it is revolving in a circular orbit at height hh 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

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Answer: (D) 3L3L

For a satellite revolving around the Earth in a circular orbit, it can be written that
mv2r=GMmr2...(1)\dfrac{m{v}^{2}}{r}=\dfrac{GMm}{{r}^{2}}...\left(1\right) (r=R+hr=R+h is the distance from the centre of the Earth)
Simplify equation (1) to obtain the linear speed of the satellite.
v=GMrv=\sqrt{\dfrac{GM}{r}}
Thus, the angular momentum of the satellite is given by
L=mvr=mrGMr=mGMr...(2)\begin{matrix}L & = & mvr \\ & = & mr\sqrt{\dfrac{GM}{r}} \\ & = & m\sqrt{GMr}...\left(2\right)\end{matrix}
Similarly, for distance r′r' from the centre of the Earth, the angular momentum is given by
L′=mGMr′...(3)L'=m\sqrt{GMr'}...\left(3\right)
Divide equation (3) by equation (2) to obtain the required angular momentum.
L′L=mGMr′mGMr=r′r...(4)\begin{matrix}\dfrac{L'}{L} & = & \dfrac{m\sqrt{GMr'}}{m\sqrt{GMr}} \\ & = & \sqrt{\dfrac{r'}{r}}...\left(4\right)\end{matrix}
Now, r′=r+8r=9r{r}^{'}=r+8r=9r
Substitute the known values of the parameters into equation (4) to obtain the required angular momentum.
L′L=9⇒L′=3L\begin{matrix}\dfrac{L'}{L} & = & \sqrt{9} \\ & \Rightarrow & L'=3L\end{matrix}

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