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Chapter 13: Q. 30 (page 354)

An earth satellite moves in a circular orbit at a speed of 5500m/s. What is its orbital period?

Short Answer

Expert verified

The orbital period is4.18hours.

Step by step solution

01

Given information

Speed of orbiting =4.18hours.

02

Calculation

The centripetal acceleration required for circular motion of satellite is provided by the gravitational force between the satellite and the earth, therefore : GMems/rs2=msvs2/rs

Here, Meis the mass of earth, msis the mass of satellite, G is the gravitational constant, r is the radius of the orbit.

Now solving we get,

GMems/rs2=msvs2/rsrs=GMe/vs2rs=(6.67×10-11Nm2/kg2)(5.98×1024kg)(5500m/s)2r=13.18×106m

Orbital period T is given by according to Kepler's Law :

T=2πrs3GMeT=2π13.18×106m3(6.67×10-11Nm2/kg2)(5.98×1024kg)T=2π5740096.7s2T=2π×2395.84sT=15053secT=4.18hours

Therefore, time period = 4.18hours.

03

Final answer

The orbital period is4.18hours.

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