If the orbital radius of the moon is `3.84xx10^(8)m` and period, is 27 dyas, then the orbital radius of communication satellite placed in orbit above the equator will be
A. `4.26xx10^(7)m`
B. `5.25xx10^(7)m`
C. `3.26xx10^(7)m`
D. `2.26xx10^(7)m`

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1 Answers

Correct Answer - a
`(T_(2))/(T_(1))=((r_(2))/(r_(1)))^(3//2)`
`((1)/(27))^(2//3)=(r_(2))/(r_(1))`
`(1)/(9)=(r_(2))/(r_(1))`
`r_(2)=(r_(1))/(9)=(3.84xx10^(8))/(9)=4.26xx10^(7)m`.

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