Question
A body attains a height equal to the radius of the earth. The velocity of the body with which it was projected is
A.
$$\sqrt {\frac{{GM}}{R}} $$
B.
$$\sqrt {\frac{{2GM}}{R}} $$
C.
$$\sqrt {\frac{5}{4}\frac{{GM}}{R}} $$
D.
$$\sqrt {\frac{{3GM}}{R}} $$
Answer :
$$\sqrt {\frac{{GM}}{R}} $$
Solution :
Energy at surface of the earth = energy at maximum height
or $$\left( {K + U} \right)$$ at the earth's surface $$= \left( {K + U} \right)$$ at maximum height
$$\therefore \frac{1}{2}m{u^2} - \frac{{GMm}}{R} = \frac{1}{2}m \times {\left( 0 \right)^2} - \frac{{GMm}}{{R + h}}$$
At maximum height it has only potential energy
or $$\frac{1}{2}m{u^2} = \frac{{GMm}}{R} - \frac{{GMm}}{{R + R}}\,\,\left( {\because h = R} \right)$$
$$\eqalign{
& {\text{or}}\,\,{u^2} = \frac{{2GM}}{R} - \frac{{2GM}}{{2R}} \cr
& {\text{or}}\,\,{u^2} = \frac{{GM}}{R} \cr
& \therefore u = \sqrt {\frac{{GM}}{R}} \cr} $$
Alternative
The expression for the speed with which a body should be projected so as to reach a height $$h$$ is $$u = \sqrt {\frac{{2gh}}{{1 + \left( {\frac{h}{R}} \right)}}} $$
Here, $$h = R$$ (given)
$$u = \sqrt {\frac{{2gR}}{{1 + \left( {\frac{R}{R}} \right)}}} = \sqrt {\frac{{2 \times \frac{{GM}}{{{R^2}}} \times R}}{2}} = \sqrt {\frac{{GM}}{R}} $$