The orbital angular momentum of a satellite revolving at a distance r from the centre is L. If the distance is increased to 16r , then new angular momentum will be

$16 L$

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$64 L$

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$\frac{L}{4}$

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$4 L$

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Solution

The correct option is D
$4 L$

Angular momentum $L = \sqrt{m^{2} G M r} ∴ L \propto \sqrt{r}$

$\frac{L_{2}}{L_{1}} = \sqrt{\frac{r_{2}}{r_{1}}} = \sqrt{\frac{16 r}{r}} = 4 L_{2} = 4 L_{1} = 4 L$

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The orbital angular momentum of a satellite revolving at a distance r from the centre is L. If the distance is increased to 16r , then new angular momentum will be

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The orbital angular momentum of a satellite revolving at a distance r from the centre is L. If the distance is increased to 16r , then new angular momentum will be

The correct option is D 4 L Angular momentum L=√m2GMr ∴ L∝ √r L2L1=√r2r1=√16rr=4L2=4L1=4L

The correct option is D
$4 L$

Angular momentum $L = \sqrt{m^{2} G M r} ∴ L \propto \sqrt{r}$

$\frac{L_{2}}{L_{1}} = \sqrt{\frac{r_{2}}{r_{1}}} = \sqrt{\frac{16 r}{r}} = 4 L_{2} = 4 L_{1} = 4 L$