Two particles of masses 10 kg and 20 kg are placed 10 cm apart find gravitational force between them

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Two small bodies of masses 10 kg and 20 kg are kept a distance 1.0 m apart and released, assuming that only mutual gravitational forces are acting find the speeds of the particles when the separation decreases to 0.5 m.

Solution

The linear momentum of 2 bodies is 0 intially. since gravitational force is internal, final momentum is also zero.
So, $\left(10kg\right)v_1=\left(20kg\right)v_2$
or$v_1=2v_2$ Since P.E.is conserved,

Initial P.E. $=\frac{-6.67\times {10}^{11}\times 10\times 20}{1}=-13.34\times {10}^{-9}J$

When separation is 0.5 m

$\Rightarrow 13.34\times {10}^{-9}+0=\frac{-13.34\times {10}^{-6}}{\frac{1}{2}}+\left(\frac{1}{2}\right)\times 10v_1^2+\left(\frac{1}{2}\right)\times 20v_2^2\Rightarrow -13.34\times {10}^{-9}=26.68\times {10}^{-9}+5v_1^2+10v_2^2\Rightarrow -13.34\times {10}^{-9}=26.68\times {10}^{-9}+30v_2^2\Rightarrow v_2^2=\frac{-13.34\times {10}^{-9}}{30}=4.44\times {10}^{-10}\Rightarrow v^2=2.1\times {10}^{-5}m/sSo,v_1=4.2\times {10}^{-5}m/s$