Deriving the Relation for Escape Velocity from a Proto-Star Cloud

AI Thread Summary
The discussion focuses on deriving the escape velocity from a proto-star cloud with uniform density, mass M, and radius R. The initial calculation for the velocity when falling freely from the surface to the center is noted as "root over" GM/R, while the escape velocity is calculated as "root over" 2GM/R. Participants agree that the falling velocity and escape velocity should be symmetric, implying they are equal when considering motion from infinity to the center and vice versa. Clarification is requested on how to derive the original formula, emphasizing the relationship between potential energy and mass enclosed within a radius r. The conversation highlights the connection between falling and escape velocities in gravitational systems.
trina1990
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: Derive a relation for the escape velocity of an object, launched from the center
of a proto-star cloud. The cloud has uniform density with the mass of M and radius R...
Ignore
collisions between the particles of the cloud and the launched object. If the object were
allowed to fall freely from the surface, it would reach the center with a velocity equal to
"root over" GM/R...

i ended up getting the velocity to be "root over" 2GM/R...
where's my fault?
& how to relalte the escape velocity with this falling velocity?
 
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trina1990 said:
i ended up getting the velocity to be "root over" 2GM/R...
where's my fault?
& how to relalte the escape velocity with this falling velocity?

I imagine that it is symmetric -- the speed achieved when "falling-in" to the centre from infinity (assuming starting from rest) is the same as the speed needed to escape to infinity from the centre. So you have the answer, you just have to derive it.

I think that the potential energy of the system when the object is at radius r just depends on the mass enclosed within radius r. Does that help?
 
yes...it makes sense to me that escape velocity & falling velocity is the same for the system...
but unfortunately i didn't get the 2nd para of your answer..
can you please provide me some more hints to derive the original formula given in the question?
 

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