Question

At what height from the surface of earth will the value of g be reduced by 36% from the value on the surface? Take radius of earth R = 6400 km.

Answer 1

Answer:

$\sf{1.6\times\:10^6\:m}$

Formula:

% decrease in the value of g is

$\sf{\left(g-g'\right)\times\:100=\dfrac{2h}{R}\times\:100}$

Explanation:

Since the acceleration due to gravity reduces by 36%, the value of acceleration due to gravity there is 100-36=64%. It means, $\sf{g'=\dfrac{64}{100}g}$ If h is the height of location above the surface of earth, then

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

$\sf{\dfrac{64}{100}g=g\dfrac{R^2}{(R+h)^2}}$

$\sf{\dfrac{8}{10}=\dfrac{R}{R+h}}$

$\sf{8R+8h=10R}$

$\implies\sf{h=\dfrac{2R}{8}=\dfrac{R}{4}=\dfrac{6.4\times\:10^6}{4}}$

$\sf{h=1.6\times\:10^6\:m}$

Answer 2

Answer:

hope it helps..

Above 4266.66 KM value of g will reduced to it's 36% . r (radius of earth) is 6371km, so using h = 2r/3 we get an altitude of 4,247km. So that is your answer. Gravity at that altitude is 36% of that on the surface.