Question

two masses m1=2kg and m2=8kg infinite distance apart are initially at rest . under their mutual grav attaraction they start moving. When separation becomes 1m , velocity of m2 will be???
ans in terms of G?

Answer 1

Answer on Question #41163, Physics, Mechanics | Kinematics | Dynamics

Question:

two masses m1=2kg and m2=8kg infinite distance apart are initially at rest. under their mutual grav attraction they start moving. When separation becomes 1m, velocity of m2 will be???

ans in terms of G?

Answer:

The law of conservation of momentum:

$$m _ {1} \overrightarrow {v _ {1}} + m _ {2} \overrightarrow {v _ {2}} = 0$$

Therefore:

$$v _ {1} = \frac {m _ {2}}{m _ {1}} v _ {2}$$

The law of conservation of energy:

$$T + U = \text{const}$$

where $T = \frac{mv^2}{2}$ is kinetic energy, $m$ -mass, $v$ -speed, $U = -\frac{Gm_1m_2}{r}$ is potential energy.

Therefore:

$$0 = \frac {m _ {1} v _ {1} ^ {2}}{2} + \frac {m _ {2} v _ {2} ^ {2}}{2} - G \frac {m _ {1} m _ {2}}{r}$$ $$\frac {m _ {1} \left(\frac {m _ {2}}{m _ {1}} v _ {2}\right) ^ {2}}{2} + \frac {m _ {2} v _ {2} ^ {2}}{2} = G \frac {m _ {1} m _ {2}}{r}$$ $$v _ {2} ^ {2} = \frac {2 G m _ {1}}{r \left(1 + \frac {m _ {2}}{m _ {1}}\right)}$$ $$v _ {2} = \sqrt {\frac {2 G m _ {1}}{r \left(1 + \frac {m _ {2}}{m _ {1}}\right)}} = 7.3 \cdot 10 ^ {- 6} \frac {m}{s}$$

Answer: $7.3 \cdot 10^{-6} \frac{m}{s}$

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