A body of mass m is placed on earth surface which is taken from earth surface to a height of h = 3R then change in gravitational potential energy is :
A body of mass m is placed on earth surface which is taken from earth surface to a height of h = 3R then change in gravitational potential energy is :
\(\frac{mgR}{4}\)
\(\frac{2}{3}mgR\)
\(\frac{3}{4}mgR\)
\(\frac{mgR}{2}\)
The Correct Option is C
Solution and Explanation
The correct option is (C) : \(\frac{3}{4}mgR\)
Change in G.P.E. = final energy – initial energy
= \(-\frac{GMm}{4R}+\frac{GMm}{R}\) = \(\frac{GMm}{R}[\frac{1}{\frac{1}{4}}]\)
\(\frac{3}{4}\frac{GMm}{R}=\frac{3}{4}\frac{GM}{R^2}mR=\frac{3}{4}gmR\)
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- 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].