Question

Give me detail solution of gravitation chapter 10 of NSO workbook. (class 09)
1 ) At what place, value of 'g' is less by 1% than its value on the surface of earth (Radius of Earth = 6400 km ). The place is 
(A) 64 km below the surface of the Earth
(B) 64 km above the surface of the Earth
(C) 30 km above the surface of the Earth
(D) 32 km below the surface of the Earth

Answer 1

32 km above surface and 64 km below the surface.   Both.

$g = \frac{G\ M_e}{R_e^2}$

     g = acceleration due to gravity at surface of Earth
     Re = radius of Earth
     Me = Mass of Earth

Acceleration due to gravity " g' " at a height h above the surface is :

$g'=\frac{G\ M_e}{(R+h)^2}=\frac{G\ M_e}{R^2(1+\frac{h}{R})^2}=\frac{g}{(1+\frac{h}{R_e})^2}\\\\g'= \frac{g}{1+\frac{2h}{R}}=g(1-\frac{2h}{R}),\ as\ \frac{h}{R}<<1\\\\1-\frac{g'}{g}=\frac{2h}{R_e}=1\%=0.01\\ \\h=0.01*6400/2=32km$

Hence gravity is less by 1% at a height of 32 km above the surface of Earth.
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Below the surface of Earth:

Let us assume that the density of Earth is uniform below surface.  Then
Density of Earth = d = Me / (4πRe³/3)
Let us take a depth of h below the surface of Earth. 
Mass M of Earth enclosed within the radius of r = (R-h)  contributes towards gravity at the depth h.  The volume of Earth in the spherical shell of width h does not contribute to the gravity.  This is because all the mass around the shell attracts the mass from all sides and the net effect is zero.

$M = \frac{M_e}{\frac{4}{3} \pi R_e^3}*\frac{4}{3} \pi r^3=\frac{M_er^3}{R_e^3}\\\\g'=\frac{GM}{r^2}=\frac{GM_er}{R_e^2*R_e}=g\frac{r}{R}\\\\\frac{g'}{g}=\frac{R-h}{R}=1-\frac{h}{R}\\\\1-\frac{g'}{g}=\frac{h}{R_e}=1\%=0.01\\\\h=0.01*R_e=64km$

Hence, gravity is 1% less at a depth of 64 km under the surface of Earth.