The gravitational field, due to the 'left over part' of a uniform sphere (from which a part as shown, has been 'removed out'), at a very far off point, \(P\), located as shown, would be (nearly):
- $\frac{5}{6} \frac{GM}{x^{2}}$
- $\frac{8}{9} \frac{GM}{x^{2}}$
- $\frac{7}{8} \frac{GM}{x^{2}}$
- $\frac{6}{7} \frac{GM}{x^{2}}$
The Correct Option is C
Solution and Explanation
Radius $=\frac{R}{2}$ (from figure)
$\frac{M}{\frac{4}{3}\pi R^{3}}= \frac{m}{\frac{4}{3}\pi\left(\frac{R}{2}\right)^{3}}$
$\Rightarrow m = \frac{M}{8}$
Mass of the left over part of the sphere
$M' =M - \frac{M}{8} = \frac{7}{8}M$
Therefore gravitational field due to the left over part of the sphere
$= \frac{GM'}{X^{2}} = \frac{7}{8} \frac{GM}{x^{2}}$
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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].