First slide

Gravitaional potential energy

Question

Two masses m₁ and m₂ are initially at rest and are separated by a very large distance. If the masses approach each other subsequently, due to gravitational attraction between them, their relative velocity of approach at a separatation of d is

Difficult

A

$\sqrt{\frac{2 {Gm}_{2}^{2}}{(m_{1} + m_{2}) d}}$
B

$\sqrt{\frac{2 G (m_{1} + m_{2})}{d}}$
C

$\sqrt{\frac{2 G}{d (m_{1} + m_{2})}}$
D

zero

Solution

As there is no external force acting on the system, linear momentum of the system is conserved.
$0 = m_{1} \vec{V_{1}} + m_{2} \vec{V_{2}} magnitude V_{2} = \frac{m_{1} V_{1}}{m_{2}} - - - (1)$
the relative velocity of approach is $|\vec{V_{2}} - \vec{V_{1}}| = V_{2} + V_{1} = (\frac{m_{1} + m_{2}}{m_{2}}) V_{1} - - - - (2)$
Applying conservation of mechanical energy for the system
$0 = - \frac{{Gm}_{1} m_{2}}{d} + \frac{1}{2} m_{1} {V_{1}}^{2} + \frac{1}{2} m_{2} {V_{2}}^{2} substituting (1) in the above we get V_{1} = \sqrt{\frac{2 {Gm}_{2}^{2}}{(m_{1} + m_{2}) d}}$
Substitute V₁ in (2) we get
relative velocity of approach $= \sqrt{\frac{2 G (m_{1} + m_{2})}{d}}$

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