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Q.

A satellite is geostationary in a particular orbit. It is allowed to go to another orbit having orbital radius 2 times that of the earlier orbit from the centre of the earth. The time period in the second orbit is

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a

48 hrs

b

24 hrs

c

$48 \sqrt{2} hrs$

d

$24 \sqrt{2} hrs$

answer is C.

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Detailed Solution

The time period of geostationary satellite $T_{1} = 24 h$
$T^{2} \propto R^{3}$
$\begin{array}{l} \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{R_{1}^{3}}{R_{2}^{3}} \\ T_{1} = 24 h R_{1} = R \\ T_{2} = ? R_{2} = 2 R \\ \frac{(24)^{2}}{T_{2}^{2}} = \frac{R^{3}}{8 R^{3}} \\ \frac{(24)^{2}}{T_{2}^{2}} = \frac{1}{8} \\ T_{2}^{2} = 8 \times (24)^{2} \\ T_{2} = 2 \sqrt{2} \times 24 \\ = 48 \sqrt{2} h \end{array}$

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