A body of mass is taken from earth surface to the height h equal to twice the radius of earth (Re), the increase in potential energy will be : (g = acceleration due to gravity on the surface of earth)
Question
A body of mass is taken from earth surface to the height h equal to twice the radius of earth (Re), the increase in potential energy will be : (g = acceleration due to gravity on the surface of earth)
Solution
The potential energy (PE) of a body of mass m at a height h from the surface of the earth is given by the formula:
PE = -GMm/r
where G is the gravitational constant, M is the mass of the earth, and r is the distance from the center of the earth to the body.
At the surface of the earth, r = Re (radius of the earth), so the potential energy is:
PE_surface = -GMm/Re
At a height h = 2Re, r = Re + h = 3Re, so the potential energy is:
PE_height = -GMm/3Re
The increase in potential energy is the difference between these two values:
ΔPE = PE_height - PE_surface = -GMm/3Re - (-GMm/Re) = GMm/Re - GMm/3Re = 2GMm/3Re
So, the increase in potential energy when the body is taken to a height equal to twice the radius of the earth is 2GMm/3Re.
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