First slide

Gravitational force and laws

Question

An artificial satellite revolves around earth in circular orbit of radius r with time period T. The satellite is made to stop in the orbit which makes it fall onto earth. Time of fall of the satellite on to earth is given by

Easy

A

$\sqrt{3} \frac{T}{6}$
B

$\frac{\sqrt{2}}{8} T$
C

$\frac{T}{\sqrt{3}}$
D

$\sqrt{\frac{2}{3}} \frac{T}{π}$

Solution

On stopping the satellite will fall along the radius r of the orbit which can be regarded as a limiting case of an ellipse with semi major axis $\frac{r}{2}$
From keplers’s 3rd law $T^{2} \propto r^{3} \Rightarrow T \propto r^{\frac{3}{2}} \Rightarrow \frac{T_{1}}{T_{2}} = {(\frac{r_{1}}{r_{2}})}^{\frac{3}{2}} = \Rightarrow \frac{T_{1}}{T_{2}} = {(\frac{r_{1}}{\frac{r_{1}}{2}})}^{\frac{3}{2}} = {(2)}^{\frac{3}{2}} = \sqrt{2^{3}} = \sqrt{8} \Rightarrow T_{2} = \frac{T}{\sqrt{8}}$
Time of fall from semi major axes to centre of ellipse = $\frac{T_{2}}{2} = \frac{T}{2 \sqrt{8}} = \frac{T}{2 \sqrt{8}} \times \frac{\sqrt{8}}{\sqrt{8}} = \frac{2 \sqrt{2} T}{2 \times 8} = \frac{\sqrt{2} T}{8}$

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