Question

At what height above the surface of the earth, the value of g becomes 64% of its value on the surface is the earth. Radius of earth = 6400 km.

Answer 1

We just use the Newton's law of gravitation.  that is simple.
 
Let the Force of gravitation on an object of mass m at an altitude h from Earth's surface distance  be equal to F.

let  Radius of Earth = Re = 6400 km
      and mass of Earth be Me.

$F = G \frac{M_e \ m}{(Re + h)^2}= mg'\\\\So\ g'=\frac{GM_e}{(R_e+h)^2}\\\\g=\frac{GM_e}{R_e^2}\\\\\frac{g'}{g}=(\frac{R_e}{R_e+h})^2=64/100\\\\So\ \frac{R_e}{R_e+h}=0.8$

So the altitude = h=$\frac{R_e}{4}=1600\ km$

at a height of 1/4 th of Earth's radius..

Answer 2

We just use the Newton's law of gravitation. that is simple.

Let the Force of gravitation on an object of mass m at an altitude h from Earth's surface distance be equal to F.

let Radius of Earth = Re = 6400 km

and mass of Earth be Me.

\begin{gathered}F = G \frac{M_e \ m}{(Re + h)^2}= mg'\\\\So\ g'=\frac{GM_e}{(R_e+h)^2}\\\\g=\frac{GM_e}{R_e^2}\\\\\frac{g'}{g}=(\frac{R_e}{R_e+h})^2=64/100\\\\So\ \frac{R_e}{R_e+h}=0.8\end{gathered}

F=G

(Re+h)

2

M

e

m

=mg

′

So g

′

=

(R

e

+h)

2

GM

e

g=

R

e

2

GM

e

g

g

′

=(

R

e

+h

R

e

)

2

=64/100

So

R

e

+h

R

e

=0.8

So the altitude = h=\frac{R_e}{4}=1600\ km

4

R

e

=1600 km

at a height of 1/4 th of Earth's radius..