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
B
160 km
C
70 km
D
120 km

Solution

## The correct option is C 160 kmGiven : $$T = 5.26 \times 10^3 s$$,  $$a = 9.32 m/s^2$$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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