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Displacement

Two hypotheti...

Question

Two hypothetical planets of masses $m_{1}$ and $m_{2}$ are at rest when they are infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centres. What is their speed if their separation is 'd'?
(Speed of $m_{1}$ is $v_{1}$ and that of $m_{2}$ is $v_{2}$ )

v_{1} = v_{2}

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v_{1} = m_{2} \sqrt{\frac{2 G}{d (m_{1} + m_{2})}}

v_{2} = m_{1} \sqrt{\frac{2 G}{d (m_{1} + m_{2})}}

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v_{1} = m_{1} \sqrt{\frac{2 G}{d (m_{1} + m_{2})}}

v_{2} = m_{2} \sqrt{\frac{2 G}{d (m_{1} + m_{2})}}

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v_{1} = m_{2} \sqrt{\frac{2 G}{m_{1}}}

v_{2} = m_{1} \sqrt{\frac{2 G}{m_{2}}}

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Solution

The correct option is D $v_{1} = m_{2} \sqrt{\frac{2 G}{d (m_{1} + m_{2})}}$
$v_{2} = m_{1} \sqrt{\frac{2 G}{d (m_{1} + m_{2})}}$

Since this is a system with no external forces, both energy and momentum will be conserved.

From momentum conservation, $m_{1} v_{1} + m_{2} v_{2} = 0$

$⟹ v_{1} = \frac{- m_{2}}{m_{1}} v_{2}$ --- eq.1

From Energy conservation, $\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} = G \frac{m_{1} m_{2}}{d}$ --- eq.2

Substituting $v_{1}$ from eq. $1$ in eq. $2$ , we get $G \frac{m_{1} m_{2}}{d} = \frac{1}{2} \frac{m_{2}^{2}}{m_{1}} v_{2}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

Which gives $v_{2}^{2} = \frac{2 G m_{1}^{2}}{d (m_{1} + m_{2})}$

$\Rightarrow v_{2} = m_{1} \sqrt{\frac{2 G}{d (m_{1} + m_{2})}}$ and $v_{1} = m_{2} \sqrt{\frac{2 G}{d (m_{1} + m_{2})}}$

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Two hypothetical planets of masses m1 and m2 are at rest when they are infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centres. What is their speed if their separation is 'd'?(Speed of m1 is v1 and that of m2 is v2)