Question

The value of acceleration due to gravity on the surface of earth is 9.8m/s2.Calculate the value of g at a height of 200 km above the surface of earth given radius of earth is 6400km.

Answer 1

Let the acceleration due to gravity at a height 'h' above the surface of the earth be g'.

Well, the acceleration due to gravity on the surface of the earth is g, and we know the expression for it.

$g=\dfrac {GM}{R^2}$

From this,

$GM=gR^2\quad\longrightarrow\quad (1)$

But, at the point which is at the height 'h' above the surface, the distance between that point and the center of the earth changes from R to (R + h). Then, acceleration due to gravity will be,

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

since G and M are constants. From this,

$GM=g'(R+h)^2\quad\longrightarrow\quad (2)$

Comparing (1) and (2),

$g'(R+h)^2=gR^2\\\\\\g'=g\left [\dfrac {R^2}{(R+h)^2}\right]\\\\\\g'=g\left [\dfrac {R}{R+h}\right]^2\\\\\\g'=g\left [\dfrac {R+h}{R}\right]^{-2}\\\\\\g'=g\left [1+\dfrac {h}{R}\right]^{-2}$

So, here,

R = 6400 km

h = 200 km

No need to convert R and h into metres, since $\dfrac {h}{R}$ is a ratio.

So,

$g'=9.8\left [1+\dfrac {200}{6400}\right]^{-2}\\\\\\g'=9.8\left [1+\dfrac {1}{32}\right]^{-2}\\\\\\g'=9.8\left [\dfrac {33}{32}\right]^{-2}\\\\\\g'=9.8\left [\dfrac {32}{33}\right]^2\\\\\\\mathbf {g'=9.2\ ms^{-2}}$

Answer 2

$answer \: by \: manisha$