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The time period `T` of the moon of planet mars (mass `M_(m)`) is related to its orbital radius `R` as (`G`=gravitational constant)A. `T^(2)=(4pi^(2)R^(3))/(GM_(m))`B. `T^(2)=(4pi^(2)GR^(3))/(M_(m))`C. `T^(2)=(4piR^(3)G)/(M_(m))`D. `T^(2)=4piM_(m)GR^(3)` |
Answer» Correct Answer - A (a) Time period, `T=(2pi R)/(sqrt((GM_(m))/( R)))=(2piR^(3//2))/(sqrt(GM_(m))` where the symbols have their meanings as given in the question. Squaring both sides, we get, `T^(2)=(4pi^(2)R^(3))/(GM_(m))` |
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