Question

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

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.

Answer 2

At the surface of the earth acceleration due to gravity is given by the relation

g=GM/R^2

Here

G =Universal gravitational constant

M =mass of the earth

R=radius of the earth

At the height h from the earth surface acceleration the to gravity g’ is

g’=GM/(R+h)^2

From above two relations we have

g’/g=[R/(R+h)]^2

Given

g’/g=36/100=[R/(R+h)]^2

Or. R/R+h=6/10

Or. 1+h/R=5/3

Or. h/R=2/3

h =2R/3=(2×6400/3)km

(since R=6400km)

=4266 km