AT A PLACE, VALUE OF g IS LESS BY 1% THAN ITS VALUE ON THE SURFACE OF THE EARTH(RADIUS OF - Science - Gravitation

Question

AT A PLACE, VALUE OF g IS LESS BY 1% THAN ITS VALUE ON THE
SURFACE OF THE EARTH(RADIUS OF THE EARTH=6400KM)
THE PLACE IS
(A)64KM BELOW THE EARTH'S SURFACE (B)63KM ABOVE THE EARTH'S
SURFACE
(C)30KM ABOVE THE EARTH'S SURFACE (D) 32 KM BELOW THE SURFACE OF
THE EARTH

nitish.gupta · Accepted Answer

g, at the surface of the earth = GM/R2 ------------(1)
and lets take g', at some distance R' from the centre of the
earth: g' = GM/R'2 -----------(2)
Now g' = 1% less than the previous value ie g' = 99g/100
Now (2) becomes,
99g/100 = GM/R'2 ---------------(3)
Dividing the equation (1) & (3) and solving it further by
putting the given value of R ie radius of the earth, we get
R' = 6432 24 km
which means distance of the place from the surface of the earth
is: 32 24 â‰ˆ 30 KM ABOVE THE EARTH 'S SURFACE

AT A PLACE, VALUE OF g IS LESS BY 1% THAN ITS VALUE ON THE SURFACE OF THE EARTH(RADIUS OF THE EARTH=6400KM). THE PLACE IS (A)64KM BELOW THE EARTH'S SURFACE (B)63KM ABOVE THE EARTH'S SURFACE (C)30KM ABOVE THE EARTH'S SURFACE (D) 32 KM BELOW THE SURFACE OF THE EARTH

AT A PLACE, VALUE OF g IS LESS BY 1% THAN ITS VALUE ON THE SURFACE OF THE EARTH(RADIUS OF THE EARTH=6400KM).

THE PLACE IS

(A)64KM BELOW THE EARTH'S SURFACE (B)63KM ABOVE THE EARTH'S SURFACE

(C)30KM ABOVE THE EARTH'S SURFACE (D) 32 KM BELOW THE SURFACE OF THE EARTH