at what height above the earth surface acceleration due to gravity becomes 1/4 of what it is at earth surface.radius=6400km
Answers
g = GM/r^2 ------ equation1
"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).
"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).