Question

two bodies of masses m1 & m2 are initially at rest at infinite distance apart.they are thn allowed to move towards each othr undr mutual gravitational attraction.their relative velocity of approch at a separation distance r between them is
1.[2G(m1-m2)/r]^1/2
2.[2G(m1+m2)/r]^1/2
3.[2Gr/(m1m2)]^1/2
4.[2Gm1m2/r]^1/2

Answer 1

here the total energy (potential energy + kinetic energy) has to be conserved
Now at initial position, the kinetic energy is zero, also the potential energy is zero, since the relative distance is infinite
At final position, when the relative distance is \(r\), the potential energy is:
\[\Large {U_f} = - G\frac{{{m_1}{m_2}}}{r}\]
whereas the kinetic energy is:
\[\Large KE = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2\]
where \(v_1,\;v_2\) are the magnitudes of the velocities with respect to an inertial system
so we can write:
\[\Large 0 = - G\frac{{{m_1}{m_2}}}{r} + \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2\qquad (1)\]
we need another condition, and such condition comes from the conservation of total momentum, since the system is an isolated system:
\[\Large {m_1}{v_1} = {m_2}{v_2}\qquad (2)\]
the requested magnitude of the relative velocity, is:
\[\Large {v_1} + {v_2} \qquad (3)\]
please solve the system of equations \((1),\) and \((2)\), for \(v_1\) and \(v_2\), then find the magnitude of relative speed according to equation \( (3)\)