Question:
If radius of earth shrinks by $1 %$ when mass remains same, the acceleration due to gravity on the surface of earth would
If radius of earth shrinks by $1 %$ when mass remains same, the acceleration due to gravity on the surface of earth would
Updated On: Jul 6, 2022
- remain unchanged
- decrease by 2%
- become zero
- increase by 2%.
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The Correct Option is B
Solution and Explanation
$g = \frac{GM}{R^2}$
Here $g \propto \frac{1}{R^2}$
and $g' \propto \frac{1}{(0.99R)^2} \propto \frac{0.98}{R^2}$ i.e. $\frac{g'}{g} $ = 0.98 i.e. g' = 0.98 g
Clearly g' decreases by 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].