What will happen to the value of acceleration due to gravity if the radius of Earth is shrink to half?

Option 4 : \(\frac{9}{4}\)g

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10 Questions 10 Marks 10 Mins

Concept:

• Acceleration due to gravity: The acceleration achieved by any object due to the gravitational force of attraction by the earth is called acceleration due to gravity by the earth.
  • As each planet has a different mass and radius so the acceleration due to gravity will be different for a different planet.

Acceleration due to gravity of the earth having mass (M) and radius (R) on earth surface is given by:

\(g = \frac{{GM}}{{{R^2}}}\)

Here G is the universal gravitational constant.

Acceleration due to gravity at any depth (h) of the earth’s surface whose distance from the centre of the earth is r is given by:

\(Acceleration\;due\;to\;gravity\;at\;depth\;\left( {g'} \right) = \frac{{g\;r}}{R}\)

Acceleration due to gravity at height (h’) whose distance from the centre of the earth is r is given by:

\(Acceleration\;due\;to\;gravity\;at\;height\;\left( {g''} \right) = \frac{{g\;{R^2}}}{{r{'^2}}}\)

Where G is Universal gravitational constant, r = (R - h) and r’ = (R + h).

Calculation:

Given: radius of earth shrink by 1/3.

\(R' = R - \frac{R}{3} = \frac{{2R}}{3}\)

If new gravitational acceleration is g’ then,

\(g = \frac{{GM}}{{{R^2}}}\)

\(g ∝\frac{{1}}{{{R^2}}}\)

\(\frac{{g'}}{g} = \frac{{{R^2}}}{{{{\left( {2R/3} \right)}^2}}} = \frac{9}{4}\)

\(g' = \frac{9}{4}g\)

Let's discuss the concepts related to Gravitation and Acceleration due to gravity below and above the surface of earth. Explore more from Physics here. Learn now!

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