Question:

Height at which the value of $g$ becomes $1/4th$ to that on earth is

Updated On: Jul 6, 2022
  • R
  • 4R
  • 2R
  • $\frac {3}{2}R$
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The Correct Option is A

Solution and Explanation

$g' = \frac{r\, R^2}{(R + h)^2} $ i.e. $\frac{1}{4} = \frac{R^2}{R+h^2}$ i.e. $\frac{1}{2} = \frac{R}{R+h}$ i.e. (R + h) = 2 R i.e. h = R
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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].