Question

the height above surface of earth where the value of gravitational acceleration is one fourth of that surface will be​

Answer 1

g = G*M/(r)^2

g_prime = G*M/(r+h)^2

g/g_prime = (r+h)^2/(r)^2

sqrt(g/g_prime) = (r+h)/r

substituting the given values in the above equation:

sqrt(g/g/4)=sqrt(4)=2=(r+h)/r

2*r = r+h

h = r

Therefore at a height equal to radius of the earth(6400 km), earth’s acceleration due to gravity will be one fourth of it’s value on the surface of the earth.

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Answer 2

Explanation:

according to second equation of motion

the hieght will be,

-2(v^2-u^2)/g

and according to third equation of motion

the hieght will be,

ut-g^2t/32