For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is :
4
\(\frac{1}{2}\)
\(\frac{1}{\sqrt 2}\)
\(\frac{1}{4}\)
The Correct Option is B
Solution and Explanation
The satellite's potential energy is equal to \(\frac{-GMm}{r}\).
Satellite's Kinetic Energy equals \(\frac{GMm}{2r}\), where G is the gravitational constant.
The planet's mass is M, and the satellite's mass is M.
R stands for radius.
The necessary ratio is \(\frac{1}{2}\) = \(\frac{(\frac{Gm}{2r})}{(\frac{GMm}{r})}\)
Kinetic energy to potential energy is split in half.
Therefore, the correct option is (B): \(\frac{1}{2}\)
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- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].