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if the gravitational force were to vary inversely as `m^(th)` power of the distance then the time period of a planet in circular orbit of radius `r` around the sun will be proportional to

A

`r^(3m//2)`

B

`r^(3m//2)`

C

`r^(m+1//2)`

D

`r^(m+1)//2`

Text Solution

Verified by Experts

The correct Answer is:
D

`Fprop(1)/(r^(m)),F=(C)/(r^(m))`
this force will provide the required centripetal force
therefore `momega^(2)r=(C)/(r^(m)),omega^(2)=(C)/(mr^(m+1))`
`T=(2pi)/(omega)impliesTpropr^((m+1)//2)`
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