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

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SUMMARY

The discussion focuses on deriving the escape velocity for an object launched from the center of a proto-star cloud with uniform density, mass M, and radius R. Participants confirm that the escape velocity is equivalent to the falling velocity from the surface to the center, which is calculated as "√(2GM/R)". The relationship between potential energy and the mass enclosed within radius r is emphasized as a key factor in understanding the derivation. Clarifications are requested regarding the derivation process and the symmetry of the velocities involved.

PREREQUISITES
  • Understanding of gravitational potential energy
  • Familiarity with the concepts of escape velocity and free fall
  • Knowledge of basic physics equations involving mass (M), gravitational constant (G), and radius (R)
  • Ability to manipulate square root equations and algebraic expressions
NEXT STEPS
  • Study the derivation of gravitational potential energy in spherical mass distributions
  • Learn about the concept of escape velocity in astrophysical contexts
  • Explore the mathematical relationship between kinetic and potential energy
  • Investigate the implications of uniform density in celestial mechanics
USEFUL FOR

Astronomy students, astrophysicists, and educators seeking to deepen their understanding of gravitational dynamics and escape velocity in proto-star environments.

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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