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Question

A satellite is moving in a circular orbit at a certain height above the earth's surface. It takes $$5.26 \times 10^3s$$ to complete one revolution with a centripetal acceleration equal to 9.32 m/s$$^2$$. The height of satellite orbiting above the earth is
(Earth's radius $$=6.37 \times 10^6 m$$)


A
220 km
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B
160 km
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C
70 km
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D
120 km
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Solution

The correct option is C 160 km
Given : $$T = 5.26 \times 10^3 s$$,  $$ a = 9.32 m/s^2$$
Centripetal acceleration $$a = \dfrac{v^2}{r} = 9.32$$
or $$v^2 = 9.32 r$$
or $$v = \sqrt{9.32 (R_e + h)}$$
Time period of revolution is given by  $$T = \dfrac{2\pi (R_e + h)}{v} = \dfrac{2\pi (R_e + h)}{\sqrt{9.32 (R_e + h)}}$$
or $$T = \dfrac{2\pi \sqrt{R_e + h}}{\sqrt{9.32}}$$
$$\therefore \  5.26 \times 10^3 = \dfrac{2 \times 3.14 \sqrt{R_e + h}}{3.05}$$
$$\sqrt{R_e + h} = 2.55 \times 10^3$$
$$R_e + h = 6.53 \times 10^6$$
$$h = 6.53 \times 10^6 - 6.37 \times 10^6$$           $$(\because R_e = 6.37\times 10^6 \ m)$$
$$h= 0.16 \times 10^6 m$$
$$ h= 160 \times 10^3 m = 160 km$$

Physics

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