First slide

Keplers 3 Laws

Question

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 the new angular momentum will be

Easy

A

16 L
B

65 L
C

$\frac{L}{4}$
D

4 L

Solution

$\begin{array}{l} L = m v r v \propto \frac{1}{\sqrt{r}} \\ \frac{L_{1}}{L_{2}} = \frac{V_{1}}{V_{2}} \frac{r_{1}}{r_{2}} = \sqrt{\frac{r_{2}}{r_{1}}} \frac{r_{1}}{r_{2}} = \sqrt{\frac{r_{1}}{r_{2}}} = \sqrt{\frac{r}{16 r}} = \frac{1}{4} \\ L_{2} = 4 L_{1} \end{array}$

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