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(D) The angular momentum $L$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ is given by $L = mvr$.Since the orbital veloci

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$4 L$

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(D) The angular momentum $L$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ is given by $L = mvr$.Since the orbital velocity $v = \sqrt{\frac{GM}{r}}$,we have $L = m \sqrt{\frac{GM}{r}} \cdot r = \sqrt{m^2 GMr}$.This shows that $L \propto \sqrt{r}$.Given the initial distance $r_1 = r$ and final distance $r_2 = 16r$,we can write the ratio of angular momenta as:$\frac{L_2}{L_1} = \sqrt{\frac{r_2}{r_1}} = \sqrt{\frac{16r}{r}} = \sqrt{16} = 4$.Therefore,the new angular momentum $L_2 = 4L_1 = 4L$.

$A$ satellite is orbiting at a distance $r$ from the center of a planet with an angular momentum $L$. If the distance is increased to $16r$,what will be the new angular momentum?

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