Question:
Abody of mass m is raised through a distance equal to the radius of the earth, (R), from the earth surface. Increase in potential energy is
Abody of mass m is raised through a distance equal to the radius of the earth, (R), from the earth surface. Increase in potential energy is
Updated On: Jul 31, 2023
- $mgR$
- $2 mgR$
- $\frac{mgR}{2}$
- $3 mgR$
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The Correct Option is C
Solution and Explanation
Potential energy of mass $m$ on the surface of the earth, $U_1 = -mgR$.
Potential energy of mass $m$ at a distance $R$ above the surface of the earth, $U_1= -mg'(2R)$
Here, $g' = \frac{GM}{\left(R +R\right)^{2}} = \frac{g}{4}$
$\therefore\:\: U_{f} =-\frac{mgR}{2} $
Increase in potential energy of mass, m
$=U_{f} - U_{i} = - \frac{mgR}{2}+ mgR = mgR 2$
Potential energy of mass $m$ at a distance $R$ above the surface of the earth, $U_1= -mg'(2R)$
Here, $g' = \frac{GM}{\left(R +R\right)^{2}} = \frac{g}{4}$
$\therefore\:\: U_{f} =-\frac{mgR}{2} $
Increase in potential energy of mass, m
$=U_{f} - U_{i} = - \frac{mgR}{2}+ mgR = mgR 2$
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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].