Question

Two point masses are $m_1$ and $m_2$ are initially at rest and at infinite distance apart. They start moving towards one another under mutual gravitational field. Their relative speed when they are at a they are at a distance of apart is. . .?
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Answer 1

Step-by-step explanation:

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Two point masses are $\sf m_1$ and $\sf m_2$ are initially at rest and at infinite distance apart. They start moving towards one another under mutual gravitational field. Their relative speed when they are at a they are at a distance of apart is. . .?

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${\huge{\rm{\underline{\underline{Answer\checkmark}}}}}$

As , no other forces are present , leaving aside the internal Gravitational attraction , the Total mechanical energy and linear momentum of system remains conserved .

So , We can write

$\rm \sum {p _{i}} = \sum{p _{f}}$

$\mapsto \rm \: 0 = m _{1}v _{1} - m _{2}v _ {2} \: \: \: \: \: \: \: \: ....(i) \:$

Negative sign has been chosen to keep v1 and v2 as magnitude of velocities only .

Also ,

$\rm U_i + K_i = U_f + K_f \:$

${\rm{\mapsto\;0+0=\dfrac{-Gm_1m_2}{r}+\dfrac{1}{2}m_1{v_1}^2+\dfrac{1}{2}m_2{v_1}^2\;\;\;\;....(ii)}} \:$

From (i) , we get

$\mapsto \rm \: m _{1}v _{1} = m _{2}v _{2} = p \: (say) \:$

From (ii) , we get

$\rm\dfrac{-Gm_1m_2}{r} = \dfrac{p^2}{2m_1}+\dfrac{p^2}{2m_2}\;\;\{as\;k=\dfrac{p^2}{2m}\} \:$

$\mapsto\rm p=m_1m_2\sqrt{\dfrac{2G}{(m_1m_2)r}} \:$

Now ,

Relative speed = $\rm v_1+v_2$

$= \rm \: \dfrac{p}{m _{1}} + \dfrac{p}{m _{2}}$

$\underline{ \boxed{ \rm{ \purple{Relative \: Speed = \sqrt{ \frac{2G(m _{1} m _{2})}{r} } }}}} \:$

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${\large{\rm{\underline{\star\; Kepler's\;Law:-}}}}$

# Kepler's First Law :

All planets move in elliptical orbits , with the sun at one foci of the ellipse .

# Kepler's Second Law :

The line that joins any planet to the sun sweeps out equal area in equal intervals of time .

# Kepler's Third Law :

The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet .

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Answer 2

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