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An earth satellite moves in a circular orbit with orbital speed of 6km/s. Find the time period of one revolution and radial acceleration of the satellite.


Let r be the radius of the circular orbit.

The orbital velocity of satellite is given by, 

                    straight v subscript straight o space equals space square root of GM over straight r end root

i.e.,                  straight r equals GM over straight v subscript straight o squared 

The time period of one revolution is, 

                  straight T equals square root of fraction numerator 4 straight pi squared over denominator GM end fraction straight r cubed end root 

rightwards double arrowor           straight T equals square root of fraction numerator 4 straight pi squared over denominator GM end fraction open parentheses GM over straight v subscript straight o squared close parentheses cubed end root

                    equals 2 straight pi GM over straight v subscript straight o cubed equals 2 straight pi gR squared over straight v subscript straight o cubed 

Here, we have

Acceleration due to gravity, straight g equals 9.8 space straight m divided by straight s squared

Radius of the Earth, straight R equals 6.4 cross times 10 to the power of 6 straight m

Orbital velocity,space space space straight v subscript straight o equals 6000 straight m divided by straight s

Therefore,

Time period, straight T equals 2 straight pi fraction numerator 9.8 cross times left parenthesis 6.4 cross times 10 to the power of 6 right parenthesis squared over denominator left parenthesis 6000 right parenthesis cubed end fraction equals 11676 straight s 

The radial acceleration of satellite is,

                   straight a equals rω squared equals vω equals fraction numerator straight v 2 straight pi over denominator straight T end fraction 

i.e.,               straight a equals fraction numerator 6000 cross times 2 cross times 22 over denominator 7 cross times 11676 end fraction equals 3.23 straight m divided by straight s squared 

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