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](/application/zrc/images/qvar/PHEN11038874.png)
i.e.,
The time period of one revolution is,
or ![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](/application/zrc/images/qvar/PHEN11038874-3.png)
Here, we have
Acceleration due to gravity, ![straight g equals 9.8 space straight m divided by straight s squared](/application/zrc/images/qvar/PHEN11038874-5.png)
Radius of the Earth, ![straight R equals 6.4 cross times 10 to the power of 6 straight m](/application/zrc/images/qvar/PHEN11038874-6.png)
Orbital velocity,![space space space straight v subscript straight o equals 6000 straight m divided by straight s](/application/zrc/images/qvar/PHEN11038874-7.png)
Therefore,
Time period,
The radial acceleration of satellite is,
i.e.,
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