Question:

At what height over the earth's pole, the free fall acceleration decreases by $1%$ . (Assume the radius of earth to be $6400 km$ )

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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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].

At what height over the earth's pole, the free fall acceleration decreases by 11%1. (Assume the radius of earth to be 6400km6400 km6400km)

The Correct Option is C

Solution and Explanation

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At what height over the earth's pole, the free fall acceleration decreases by $1%$ . (Assume the radius of earth to be $6400 km$ )