Answer:
[tex]|\Delta v |=\sqrt{\frac{4Gm}{d} }[/tex]
Explanation:
Consider two particles are initially at rest.
Therefore,
the kinetic energy of the particles is zero.
That initial K.E. = 0
The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are
[tex]\mu= \frac{m_1m_2}{m_1+m_2}[/tex]
now, since both the masses have mass m
therefore,
[tex]\mu= \frac{m^2}{2m}[/tex]
= m/2
The final K.E. of the particles is
[tex]KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2[/tex]
Distance between two particles is d and the gravitational potential energy between them is given by
[tex]PE_{Gravitational}= \frac{Gmm}{d}[/tex]
By law of conservation of energy we have
[tex]KE_{initial}+KE_{final}= PE_{gravitaional}[/tex]
Now plugging the values we get
[tex]0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}[/tex]
[tex]|\Delta v |=\sqrt{\frac{4Gm}{d} }[/tex]
[tex]=\sqrt{\frac{Gm}{d} }[/tex]
This the required relation between G,m and d
