The potential energy of a rock, having mass $$m$$ and rotating at a height of $$3.2 \times {10^6}m$$ from the earth surface, is
A.
$$ - 6\,mg{R_e}$$
B.
$$ - 0.67\,mg{R_e}$$
C.
$$ - 0.99\,mg{R_e}$$
D.
$$ - 0.33\,mg{R_e}$$
Answer :
$$ - 0.67\,mg{R_e}$$
Solution :
Mass of the rock = $$m$$ and height of rock from earth $$\left( h \right) = \frac{{{R_e}}}{2} = 3.2 \times {10^6}m$$
We know that gravitational potential energy of the rock rotating at height
$$h = - \frac{{G{M_e}m}}{{{R_e} + h}} = - \frac{2}{3}mg{R_e}\,\,\left( {{\text{where,}}\,G{M_e} = g{R_e}\,{\text{and}}\,h = {R_e}} \right)$$
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-