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A particle of mass $m$ is thrown upwards from the surface of the earth, with a velocity $u.$ The mass and the radius of the earth are, respectively, $M$ and $R.$ $G$ is gravitational constant and $g$ is acceleration due to gravity on the surface of the earth. The minimum value of $u$ so that the particle does not return back to earth, is
${\left( {\frac{{GM}}{R}} \right)^{\frac{1}{2}}}$
$\;{\left( {\frac{{8GM}}{R}} \right)^{\frac{1}{2}}}$
$\;{\left( {\frac{{2GM}}{R}} \right)^{\frac{1}{2}}}$
$\;{\left( {\frac{{4GM}}{R}} \right)^{\frac{1}{2}}}$
Solution
According to law of conservation of mechanical energy
$\frac{1}{2}m{u^2} – \frac{{GMm}}{R} = 0\,\,or\,\,{u^2} = \frac{{2GM}}{R}$
$u = \sqrt {\frac{{2GM}}{R}} = \sqrt {2gR} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {g = \frac{{GM}}{{{R^2}}}} \right)$