Question

A satellite orbits the earth at an altitude of 600 km with an orbital speed of 27 000km/hr. What is its period of revlution. Take the radius of the earth as 6400 km.

97.7 min

36.4 min

120.3 min

73.7 min

Answer 1

Answer on Question #41463, Physics, Mechanics

A satellite orbits the earth at an altitude of 600 km with an orbital speed of 27 000 km/hr. What is its period of revolution? Take the radius of the earth as 6400 km.

97.7 min 36.4 min 120.3 min 73.7 min

Solution

The angular velocity of a satellite is

$$\omega = \frac {v}{r} = \frac {v}{R + h},$$

where $v = 27000\frac{\mathrm{km}}{\mathrm{hr}}$ is an orbital speed of a satellite, $R = 6400\mathrm{km}$ is the radius of the earth,

$h = 600\mathrm{km}$ is the distance from the earth surface.

A period of revolution of a satellite is

$$T = \frac {2 \pi}{\omega} = \frac {2 \pi}{v} (R + h) = \frac {2 \pi}{27000} (6400 + 600) = 1.62\ \mathrm{hr} = 97.7\ \mathrm{min}.$$

Answer: 97.7 min.

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