The orbital angular momentum of a satellite i | vedclass

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(D) The orbital angular momentum $L$ of a satellite of mass $m$ revolving in a circular orbit of radius $r$ is given by $L = mvr$.The orbital velocity

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$3$

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(D) The orbital angular momentum $L$ of a satellite of mass $m$ revolving in a circular orbit of radius $r$ is given by $L = mvr$.The orbital velocity $v$ is given by $v = \sqrt{\frac{GM_e}{r}}$,where $M_e$ is the mass of the earth.Substituting $v$ in the expression for $L$:$L = m \sqrt{\frac{GM_e}{r}} \cdot r = m \sqrt{GM_e} \cdot r^{1/2}$.This shows that $L \propto r^{1/2}$.Let the initial distance from the earth's centre be $r_1 = r$. The new distance from the earth's centre is increased by eight times its initial value,meaning the new distance $r_2 = r + 8r = 9r$.Now,the ratio of the new angular momentum $L'$ to the initial angular momentum $L$ is:$\frac{L'}{L} = \left( \frac{r_2}{r_1} \right)^{1/2} = \left( \frac{9r}{r} \right)^{1/2} = (9)^{1/2} = 3$.Therefore,the new angular momentum $L' = 3L$.

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

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