Question

PLEASE ANSWER URGENTLY:
Discuss the variation of "g" due to altitude from the surface of earth. Also find the acceleration due to gravity at a height equal to the radius of earth from the surface of earth.

Answer 1

Consider earth to be a sphere of mass M, radius R with centre at O. Let g be the value of acceleration due to gravity at a point A on the surface of earth.

$\sf{g=\dfrac{GM}{R^2}}$.....(1)

If g' is the acceleration due to gravity at a point B, at a height h above the surface of earth, then the force on the body at B is due to earth whose mass M is concentrated at the centre O of earth.

$\sf{g'=\dfrac{GM}{\left(R+h\right)^2}}$.....(2)

Dividing Eq. (2) by Eq. (1), we get

$\sf{\dfrac{g'}{g}=\dfrac{GM}{\left(R+h\right)^2}\times\dfrac{R^2}{GM}}\\\sf{=\dfrac{R^2}{\left(R+h\right)^2}}$

$\sf{=\dfrac{R^2}{R^2\left(1+\dfrac{h}{R}\right)^2}}$

$\sf{=\left(1+\dfrac{h}{R}\right)^{-2}}$.....(3)

If h<<R, then $\sf{\dfrac{h}{R}}$ is very small compared to 1. Expanding the R.H.S. of the above equation by Binomial theorem and neglecting the square and the higher powers of $\sf{\dfrac{h}{R}}$, we get

$\sf{\dfrac{g'}{g}=1-\dfrac{2h}{R}}$

$\sf{g'=g\left(1-\dfrac{2h}{R}\right)}$.....(4)

From this we can say that value of acceleration due to gravity decreases with height.

At a height equal to the radius of the earth, from (3),

$\boxed{\sf{g'=\dfrac{gR^2}{\left(R+R\right)^2}=\dfrac{g}{4}}}$