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If $g$ is the acceleration due to gravity on the surface of earth, its value at a height equal to double the radius of earth is
If $g$ is the acceleration due to gravity on the surface of earth, its value at a height equal to double the radius of earth is
Updated On: May 15, 2024
- $ g $
- $ \frac{g}{2} $
- $ \frac{g}{3} $
- $ \frac{g}{9} $
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The Correct Option is D
Solution and Explanation
Acceleration due to gravity at a height $h$ from the surface of the earth
$g'=g \frac{1}{\left(1+\frac{h}{R}\right)^{2}}$
Given $h=2 R$
$\therefore g'=g \frac{1}{(1+2)^{2}}$
or $g'=\frac{g}{9}$
$g'=g \frac{1}{\left(1+\frac{h}{R}\right)^{2}}$
Given $h=2 R$
$\therefore g'=g \frac{1}{(1+2)^{2}}$
or $g'=\frac{g}{9}$
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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].
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