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The time period T of the moon of planet ...

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)=(2piGR^(3))/(M_(m))`

D

`T^(2)=(4piM_(m)GR^(3))`

Text Solution

Verified by Experts

The correct Answer is:
A
Time period, `T=(2piR)/((sqrt(GM_(m))/R))=(2piR^(3//2))/(sqrt(GM_(m)))`
Where the symbols have their meanings as given. Squaring boty sides, we get
`T^(2)=(4pi^(2)R^(3))/(GM_(m))`
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Knowledge Check

  • The acceleration due to gravity'g' for objects on or near the surface of earth is related to the universal gravitational constant 'G' as ('M' is the mass of the earth and 'R' is its radius):

    A
    `G=g(M)/(R^(2))`
    B
    `g=G(M)/(R^(2))`
    C
    `M=(gG)/(R^(2))`
    D
    `R=(gG)/(M^(2))`

  • What is the mass of the planet that has a satellite whose time period is T and orbital radius is r ?

    A
    `(4pi^(3)r^(3))/(GT^(2))`
    B
    `(4pi^(2)r^(3))/(GT^(2))`
    C
    `(4pi^(2)r^(3))/(GT^(3))`
    D
    `(4pi^(2)T)/(GT^(2))`

  • If M is the mass of the earth and R its radius, then ratio of the gravitational acceleration and the gravitational constant is

    A
    `(R^(2))/(M)`
    B
    `(M)/(R^(2))`
    C
    `MR^(2)`
    D
    `(M)/(R)`

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