If $$'g\,'$$ is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of mass $$'m\,'$$ raised from the surface of the earth to a height equal to the radius $$'R\,'$$ of the earth is-
A.
$$\frac{1}{4}mgR$$
B.
$$\frac{1}{2}mgR$$
C.
$$2\,mgR$$
D.
$$mgR$$
Answer :
$$\frac{1}{2}mgR$$
Solution :
$$\eqalign{
& \therefore U = \frac{{ - GmM}}{{2R}} + \frac{{GmM}}{R}; \cr
& \Delta U = \frac{{GmM}}{{2R}} \cr
& {\text{Now}}\,\,\frac{{GM}}{{{R^2}}} = g; \cr
& \therefore \frac{{GM}}{R} = gR \cr
& \therefore \Delta U = \frac{1}{2}mgR \cr} $$
Releted MCQ Question on Basic Physics >> Gravitation
Releted Question 1
If the radius of the earth were to shrink by one percent, its mass remaining the same, the acceleration due to gravity on the earth’s surface would-
If $$g$$ is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth, is-
A geo-stationary satellite orbits around the earth in a circular orbit of radius $$36,000 \,km.$$ Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface $$\left( {{R_{earth}} = 6400\,km} \right)$$ will approximately be-