Question:
At what height over the earth's pole, the free fall acceleration decreases by $1%$. (Assume the radius of earth to be $6400 km$)
At what height over the earth's pole, the free fall acceleration decreases by $1%$. (Assume the radius of earth to be $6400 km$)
Updated On: May 30, 2022
- 1.253 km
- 64 km
- 32 km
- 80 km
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The Correct Option is C
Solution and Explanation
$g' = g( 1- \frac{2h}{R}) $ i.e. $\frac{g'}{g} = 1 - \frac{2h}{R}$ i.e. $\frac{99}{100}= 1 - \frac{2h}{6400} $ i.e. $h = \frac{3200}{100} $ = 32 km
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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].