Question

At what height above the earth's surface ,acceleration due to gravity is reduced by 36% from the value on the surface of earh.(R=6400 km)?

Answer 1

use formula
g'=g/(1+h/r)^2
now g is reduced by 36%
so, g'=g-36 % g=64/100g
=64/100 g
now
64/100g=g/(1+h/r)^2

=> 64/100=1/(1+h /r)^2

=>(10/8)^2=(1+h/r)^2

=> 10/8=1+h/r

=> 10/8-1=h/r

=> 2/8=1/4=h/r

h=r/4=6400km/4=1600km
hence at 1600km above height acceleration due to gravity reduced by 36%

Answer 2

3 answers · Physics 

Answers

from the equation F = G m1 m2 / r^2 
we see that the force of gravity is inversely proportional to the square of the distance between the centers of the objects attracting each other. 
Since F also equals m a it is a short step to see that g is inversely proportional to r^2 
so 
g_h / g = [R / (R+h)]^2 
you can put in g and the radius of the Earth then solve for h 
but this leads to a quadratic equation. 
An easier way is to solve for (R+h) first then subtract R. 

From the given g_h = 0.99 g 
so 
0.99 g/ g = [R/(R+h)]^2 
g's cancel and take sqrt of both sides 
0.9945 = R/(R+h) 
(R+h) = R / 0.9945 = 6.371E6 m / 0.9945 = 6.4062E6 m 
h = 6.40+3E6 - 6.371E6 = 3.52E4 m or 35.2 km above the surface of the Earth 

this for 1 percent similarly solve for 36 percent