Question:

If the radius of earth shrinks by one percent and its mass remaining the same, then acceleration due to gravity on the earth?s surface will

Updated On: Apr 23, 2024
  • remain constant
  • decrease
  • increase
  • either (b) or (c)
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The Correct Option is C

Solution and Explanation

Acceleration due to gravity on the earth's surface is given by $g=\frac{G M_{e}}{R_{e}^{2}}$ Now, $R_{e}$ shrinks by $1 \%$, so the new value for radius of the earth is $0.09\, R_{e}$. $\therefore\,\,\,\, g'=\frac{G M_{e}}{(0.09)^{2} R_{e}^{2}}=\frac{g}{0.0081} > g$
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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].