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A gestationary satellite is orbiting the...

A gestationary satellite is orbiting the earth at a height 6 R above the surface of earth, where R is the radius of the earth. The time period of another statellight at a height of 2.5 R from the surface of earth in hours is

A

`6sqrt(2)`

B

`(6)/(sqrt(2))`

C

`5`

D

`10`

Text Solution

Verified by Experts

The correct Answer is:
A
According to Kepler's third law, `T^(2) prop r^(3)`
`(T_(2))/(T_(1))=((r_(2))/(r_(1)))^(1//2)`
Give `T_(1)=24h`
`r_(1)+R+h_(1)=R+5R=6R`
`r_(2)=R+h_(2)=R+2R=3R`
On solving we get
`T_(2)=6sqrt(2)h`
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