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In a satellite if the time of revolution is T, then kinetic energy is proportional to

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a
1T
b
1T2
c
1T3
d
T−2/3

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detailed solution

Correct option is B

Time period  of a sattelite is  T= 2πR3GMKinetic energy = 12m v02=m 2π2R2T2


Similar Questions

In the following four periods     
(i) Time of revolution of a satellite just above the earth’s surface  (Tst)
(ii) Period of oscillation of mass inside the tunnel bored along the diameter of the earth  (Tma)
(iii) Period of simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 N/kg  (Tsp)
(iv) Period of an infinite length simple pendulum in the earth’s real gravitational field  (Tis)
 


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