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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 beA. `4.26xx10^(7)m`B. `5.25xx10^(7)m`C. `3.26xx10^(7)m`D. `2.26xx10^(7)m` |
Answer» 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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