Question

at what height from the surface of the earth will the value of g be reduced by 36 % from the value at the surface radius of earth is 6400 km

Answer 1

We know that acceleration due to gravity from the surface of Earth at height "h" is given as,
g' = g [1- 2h/R] ........... (1)
where h is the height from the surface of Earth
The radius of the Earth = R = 6400 km.
We have to calculate the height from the surface of the earth at which the value of g is reduced by 36% from the value at the surface of Earth.
i-e, g' = 36% of g .......... (2)
Comparing equations () and (), we get:
36% of g = g [1- 2h/R]
(36/100) g = g [1- 2h/R]
0.36 = 1 - 2h/R
2h/R = 1 - 0.36
2h/R = 0.64
2h = (0.64) x R
h = (0.64)/2 x R
h = 0.32 x R
h = 0.32 x 6400
h = 2048 km
which is the required height at which the value of "g" is 36% from the value at the surface of Earth.