Question:

A spherical planet has a mass $M_p$ and diameter $D_p$. A particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity, equal to

Updated On: May 3, 2024
  • $\frac {4GM_p}{D_p^2} $
  • $\frac {GM_pm}{D_p^2} $
  • $\frac {GM_p}{D_p} $
  • $\frac {4GM_pm}{D_p^2} $
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The Correct Option is A

Solution and Explanation

$g =\frac{ GM _{ p }}{ R _{ p }{ }^{2}}$ where $R _{ p }= D _{ p } / 2$
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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].