Question:
From a solid sphere of mass $M$ and radius $R$, a spherical portion of radius $\frac{R}{2}$ is removed, as shown in the figure. Taking gravitational potential $V = 0$ at $r = \infty$, the potential at the centre of the cavity thus formed is ($G =$ gravitational constant)
From a solid sphere of mass $M$ and radius $R$, a spherical portion of radius $\frac{R}{2}$ is removed, as shown in the figure. Taking gravitational potential $V = 0$ at $r = \infty$, the potential at the centre of the cavity thus formed is ($G =$ gravitational constant)
Updated On: Oct 10, 2024
- $-\frac{2GM}{3R}$
- $\frac{-2GM}{R}$
- $\frac{-GM}{2R}$
- $\frac{-GM}{R}$
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The Correct Option is D
Solution and Explanation
Solid sphere is of mass $M$, radius $R$.
Spherical portion removed have radius $R / 2$, therefore its mass is $M / 8$.
Potential at the centre of cavity $= V _{\text {solid sphere }}+ V _{\text {removed part }}$
Spherical portion removed have radius $R / 2$, therefore its mass is $M / 8$.
Potential at the centre of cavity $= V _{\text {solid sphere }}+ V _{\text {removed part }}$
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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].