Question

Show that if a Body is taken to height "h",the acceleration is decreased by the factor $\rm{\dfrac{R^2}{(R+h)^2}}}$,where R is the Radius of the earth.

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Answer 1

Hi there!

Nice Question !

Your Answer :-

We know the formula that :-

$\rm\red{g = \dfrac{Gm}{ {r}^{2} } }$

But height is also given, So Acceleration will become :-

$\rm\blue{g = \dfrac{Gm}{{r + h}^{2} } }$

It will become :-

$\rm\green{g = G \dfrac{ {r}^{2} }{ {r + h}^{2} } }$

$\rm\pink{ \dfrac{g}{G} = \dfrac{ {r}^{2} }{ {r + h}^{2} } }$

Hence, Proved.

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Answer 2

We know that,

Acceleration due (g):

 $g=\frac{GM}{R^{2} }$  where g= Acceleration due to gravity

                           G= Universal gravitational constant

                           M= Mass

                           R= Radius

Or,

$GM=gR^{2}$                           --------(1)

Let g' be the new acceleration due to gravity at earth's surface at height 'h.' Therefore,

$g'=\frac{GM}{R+h^{2} }$

$g'=\frac{gR^{2} }{R+h^{2} }$                        ------ From (1)

As we know that g is constant for earth i.e 9.81 m/s². Therefore,

$g'=\frac{R^{2} }{R+h^{2} }$

Hence proved.