Question

Find the height at which acceleration due to gravity becomes 1/4th of its value on the surface of the earth

Answer 1

"G" is a constant of proportionality

"M" and m are the two masses exerting the forces

"r" is the radius of earth.

"g'" is gravitational acceleration at height H from surface of the earth.

g'= GM/(r+H)^2 ------equation2

Now the calculation goes as:

Divide the equation1 by equation2

g/g' = (r+H) ^2/ r^2

4/1 = (r+H) ^2/ r^2

2/1 = r+H/ r ===>> r=H

The height above the surface when the acceleration due to gravity will be 1/4th of its value on the surface of earth is equal to “r”(radius of the earth).

Answer 2

see answer

correct answer