Questions
In a satellite if the time of revolution is T, then K.E. is proportional to
detailed solution
Correct option is D
v=GMr ∴K.E.∝v2∝1r---(1) and T2∝r3⇒T23∝r⇒1r∝T-23---(2) From (1) and (2) we get ∴ K.E. ∝T−2/3Talk to our academic expert!
Similar Questions
In the following four periods
(i) Time of revolution of a satellite just above the earth’s surface
(ii) Period of oscillation of mass inside the tunnel bored along the diameter of the earth
(iii) Period of simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 N/kg
(iv) Period of an infinite length simple pendulum in the earth’s real gravitational field
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