IIT JEE , AIEEE Forum - Physics , Mechanics

cool_anu5

Two bodies of masses m1 and m2 are initially at rest at infinite distance apart. Thay are then allowed to move towards each other under mutual gravitational attraction. What is their relative velocity of approach at a separation distance r between them?

djdylan2000

I think 0. If they are at infinite distance, they wont come together under mutual gravitation force.

Decoder

look force is GmM / R^2..
so potential enery is -GmM/R..at a distance R b/w them..

now velocitty gained by m1..till then by WET..
is root(Gm2/R)..
similarly root(Gm1/R) for m2.....

add them u get relative velocvity of approach..

elastiboysai

ayshwarya

hey elastiboysai can u plz say how v vud get mu=m1.m2/m1+m2

anchitsaini

wait i'll post the proof

anchitsaini

v1 = velocity of m1
v2 = velocity of m2

conserving momentum--

m1 v1 = m2 v2 -----------1

conserving energy --

m1 v1 ² / 2 + m2 v2 ² / 2 = Gm1 m2 / r ---------2

putting value of v2 from 1 in 2 we get --

v1 ² ( m1 + m2) / m2 = 2G m2 / r

v1 = m2

[ 2G / (m1 + m2) r ]

similarly putting value of v1 from 1 in 2 we get --

v2 = m1

[ 2G / (m1 + m2) r ]

relative velocity = v1 + v2 since they are in opposite directions
v _r =

[ 2G / (m1 + m2) r ] * (m1 + m2)

=

[ 2G (m1 + m2) / r ]

anandghegde

$\text{Conserve Energy}\\ \\ \frac{G.m_1.m_2}{r^2} = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

$\text{Conserve momentum}\\ \\ m_1v_1 = m_2v_2$

now solve and get v_1+v_2

(relative velocity of approach)

anandghegde

oh sorry anchit...dint see you post.

elastiboysai

Wait
Posting the proof for reduced mass .
Im posting it in a different context btw

elastiboysai

Consider 2 blocs m1 and m2 attached to each other by a massless spring
of force constant k.
Supposer l is the free length of spring,
consider ur system 2 b placed on x axis, with 1 mass m2 at origin.
if m1 and m2 are displaced and have coordinates by x1 and x2
extension of spring= x =(x1-x2)-l --->1
now
m₁d²x₁/dt² =-kx
m₂d²x₂/dt² =kx
from the abov 2,

m₁m₂d²(x₁-x₂)/dt²=-kx(m₁+m₂)
m₁m_2/(m₁+m₂)--=y say
differentiating 1 twice we get
yd²x/dt²=-kx
so
d²x/dt²= -k/y *x
here y plays the role of mass.
We can straightaway use d standard equations by treating y as the mass.

I agree my proof is not rigorous.
Reduced mass is just a technique .
If no external force acts on the system and only internal conservative forces are in play,
mechanical energy of the system will remain conserved.
in such cases , u cn play wid reduced mass

The abov problem is such an example
To the system of 2 masses, gravitational force between them is an internal conservative froce n u can use d same technique..

elastiboysai

my apologies for the bad drawing

edison

In such problems, apply conservation of energy and momentum both

1) As here gravitational potential energy is getting converted into kinetic energy of the two bodies

2) Also the momentum is conserved and total momentum throughout the motion of the two bodies = 0

Now solve the two equations formulated as above, it will be linear equations in two variables, and its solution gives desired results.

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