Question:

The weight of a body on the surface of the earth is 100 N. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:

Updated On: Mar 21, 2025
  • 25 N
  • 64 N
  • 100 N
  • 50 N
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The Correct Option is B

Solution and Explanation

Using Newton's formula: \[ F = \frac{GMm}{r^2} \] At surface of the earth, \[ F = \frac{GMm}{R_e^2} \quad \text{(eq. 1)} \] At \(r = R_e + \frac{R_e}{4} = \frac{5R_e}{4}\), \[ F' = \frac{GMm}{\left(\frac{5R_e}{4}\right)^2} = \frac{16GMm}{25R_e^2} \] Thus, \[ F' = \frac{16}{25} F = \frac{16}{25} \times 100 = 64 \, N \]
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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].