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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 when their separation is ‘d’ ?

(Speed of m1 is v1 and that of m2 is v2)

Option: 1

v_{1}=v_{2}


Option: 2

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

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


Option: 3

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

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


Option: 4

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

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


Answers (1)

best_answer

Initial energy of the system = 0

Final energy = \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}\:-\frac{Gm_{1}m_{2}}{d}

From conservation of energy

\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}\:=\frac{Gm_{1}m_{2}}{d}-------- 1

From conservation of Linear Momentum

m_{1}v_{1}^{2}+m_{2}(\frac{-m_{1}v_{1}}{m_{2}})^{2}\:=\:\frac{2Gm_{1}m_{2}}{d}\\\frac{m_{1}m_{2}v_{1}^{2}+m_{1}^{2}v_{1}^{2}}{m_2}=\frac{2Gm_{1}m_{2}}{d}=v_{1}=m_{2}\sqrt{\frac{2G}{d(m_{1}+m_{2})}}

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

Posted by

Ritika Jonwal

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