Question:
If the Earth shrinks in its radius by 4%, mass remaining the same, the value of 'g' on its surface
If the Earth shrinks in its radius by 4%, mass remaining the same, the value of 'g' on its surface
Updated On: Oct 13, 2024
- decreases by about 4%
- increases by about 8%
- decreases by about 8%
- remains constant
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The Correct Option is B
Solution and Explanation
We know, $g = \frac{GM}{R^2} $
Since G and M are constants, so g increases as R decreases
$\frac{\Delta g}{g} = \frac{\Delta R}{R}$
As $ \frac{\Delta R}{R} =4\% $
So, $\frac{\Delta g}{g} = 8\%$
Since G and M are constants, so g increases as R decreases
$\frac{\Delta g}{g} = \frac{\Delta R}{R}$
As $ \frac{\Delta R}{R} =4\% $
So, $\frac{\Delta g}{g} = 8\%$
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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].