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 The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by

(R=Earth’s radius) :


  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)


Answers (1)

As we learnt in

Value of 'g' at ∞ -

g'\alpha\: \frac{1}{r^{2}}

  if   r=\infty \: \; \; \; \; \; \; g'=0

g'\rightarrow Value of acceleration due to gravity

r\rightarrow height above earth's surface

- wherein

No effect of earth gravitational pull at infinite distance.


if d<g      g= \frac{Gm}{K^2}d

d=R,           g=\frac{Gm}{R^2}

d> R,           g=\frac{Gm}{d^2}

i.e. g \propto \frac{1}{d^2}

Option 1)


Option 2)


Option 3)


Option 4)


Posted by

Sabhrant Ambastha

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