- Home
- Standard 11
- Physics
The masses and radii of the earth and the moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are distance $d$ apart. The minimum speed with which particle of mass $m$ should be projected from a point midway between the two centres so as to escape to infinity is
$v = \sqrt {\frac{{4g({M_1} + {M_2})}}{d}} $
$v = \sqrt {\frac{{4G({M_1} + {M_2})}}{d}} $
$v = \sqrt {4G({M_1 M_2})} $
$v = \sqrt {4Gd({M_1} + {M_2})} $
Solution
$PE$ of particle of mass $m$ placed at a point $mid-way$ between the earth and the moon,
$\mathrm{U}=-\left(\frac{\mathrm{GM}_{1} \mathrm{m}}{\left(\frac{\mathrm{d}}{2}\right)}+\frac{\mathrm{GM}_{2} \mathrm{m}}{\left(\frac{\mathrm{d}}{2}\right)}\right)=-\frac{2 \mathrm{Gm}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)}{\mathrm{d}}$
The particle will escape if its total energy becomes zero, i.e.
$\mathrm{U}+\mathrm{K}=0$
$\Rightarrow-\frac{2 \mathrm{Gm}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)}{\mathrm{d}}+\frac{1}{2} \mathrm{mv}^{2}=0$
$\Rightarrow \mathrm{v}=\sqrt{\frac{4 \mathrm{G}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)}{\mathrm{d}}}$
