The acceleration due to gravity at height h a | vedclass

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(A) For a point at a height $h$ above the surface of the earth,the acceleration due to gravity is given by:$g(h) = \frac{GM}{(R+h)^2}$where $G$ is the

Vedclass · Accepted Answer

$g^{\prime}=g\left(1-\frac{2 h}{R}\right)$

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(A) For a point at a height $h$ above the surface of the earth,the acceleration due to gravity is given by:$g(h) = \frac{GM}{(R+h)^2}$where $G$ is the gravitational constant,$M$ is the mass of the earth,and $R$ is the radius of the earth.We can rewrite this expression as:$g(h) = \frac{GM}{R^2(1 + \frac{h}{R})^2}$$g(h) = \frac{GM}{R^2} (1 + \frac{h}{R})^{-2}$Since $g = \frac{GM}{R^2}$ is the acceleration due to gravity at the surface of the earth,we have:$g(h) = g (1 + \frac{h}{R})^{-2}$Given the condition $h \ll R$,we can use the binomial approximation $(1 + x)^n \approx 1 + nx$ for $|x| \ll 1$:$(1 + \frac{h}{R})^{-2} \approx 1 - \frac{2h}{R}$Substituting this back into the equation:$g(h) \approx g (1 - \frac{2h}{R})$Thus,the acceleration due to gravity at height $h$ is $g^{\prime} = g(1 - \frac{2h}{R})$.

Column-$I$	Column-$II$
$(1)$ Maximum value of acceleration due to gravity $g$	$(a)$ At the center of the Earth
$(2)$ Minimum value of acceleration due to gravity $g$	$(b)$ At the poles
$(3)$ Zero value of acceleration due to gravity $g$	$(c)$ At the equator

The acceleration due to gravity at height $h$ above the earth if $h \ll R$ (radius of earth) is given by

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Column-$I$ Column-$II$

$(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth

$(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles

$(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator

Assuming the density of the Earth is constant, which graph correctly represents the variation of acceleration due to gravity $(g)$ with the distance $(r)$ from the center of the Earth (radius of the Earth $= R$)?

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