Calculating the Escape Velocity from a Neutron Star

Click For Summary

Homework Help Overview

The discussion revolves around calculating the escape velocity from the surface of a neutron star, given its radius and mass in relation to Earth. Participants are exploring the implications of gravitational potential energy and kinetic energy in this context.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the formula for escape velocity and expresses uncertainty about their calculations and the correct application of energy equations. Some participants clarify the definition of gravitational potential energy and discuss the conditions for escape velocity.

Discussion Status

Participants are actively engaging with the concepts of energy conservation and gravitational potential. Clarifications regarding the signs in the potential energy equation have been provided, and there is a constructive exchange of ideas about the underlying physics.

Contextual Notes

The original poster has acknowledged a potential error in their calculations and the omission of a negative sign in the potential energy formula. There is an emphasis on understanding the conditions for escape velocity without reaching a definitive conclusion.

cougar_21
Messages
15
Reaction score
0
Question:

The radius of a neutron star is 750 times smaller than the Earth's radius, and its mass is 1.8 times larger than the Earth's mass. What is the escape velocity from the surface of a neutron star? (Ignore the fact that, at high speeds, one should not really use mv^2/2 for the kinetic energy.)

If my calculations are correct:
Rn=-4.7711x10^9 m
Mn= 1.0757x10^30 kg

I know that E= K + U in whick K is mv^2/2 and U = GmM/R but don't know how to advance, please help :redface:

Does vesc= square root of (2GMn/Rn) ?
 
Last edited:
Physics news on Phys.org
cougar_21 said:
Question:

The radius of a neutron star is 750 times smaller than the Earth's radius, and its mass is 1.8 times larger than the Earth's mass. What is the escape velocity from the surface of a neutron star? (Ignore the fact that, at high speeds, one should not really use mv^2/2 for the kinetic energy.)

If my calculations are correct:
Rn=-4.7711x10^9 m
Mn= 1.0757x10^30 kg

I know that E= K + U in whick K is mv^2/2 and U = GmM/R but don't know how to advance, please help :redface:

Does vesc= square root of (2GMn/Rn) ?

You have omitted the minus sign from potential energy. The condition for escape is K + U > 0 ([itex]U \rightarrow 0[/itex] as [itex]R \rightarrow \infty[/itex]).

AM
 
First, as Andrew said, U = -GmM/R. It is defined that way so that U = 0 when the distance between
the objects is "infinite".

Assuming your object is given some initial velocity and thereafter no force acts on it
other than gravity from the star, the total energy is constant, and at any given moment
E = K + U = (mv^2)/2 - GmM/R

Escape velocity means just enough to get to such a distance from the star that it's gravity
can no longer affect the object; i.e. theoretically the distance approaches
infinity, all of the initial kinetic energy is converted to potential energy and
therefore the final velocity = 0.

Holding total energy constant,
(1/2)mvi2 - GmM/Ri = (1/2)mvf2 - GmM/Rf
but since vf = 0 this becomes
(1/2)mvi2 - GmM/Ri = - GmM/Rf
rearranging terms:
(1/2)mvi2 = GmM/Ri - GmM/Rf
vi2 = 2GM(1/Ri - 1/Rf)
and since Rf -> ∞ the second fraction approaches 0 so this gives your equation:
vi2 = 2GM/Ri
 
thank you andrew and gnome, i would get the square of a negative answer and knew that was not right, but indeed I missed the minus sign. Thanks for the explanation gnome it helped me understand what i was doing ! Thank you both bunches XD!
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
4
Views
2K
Replies
1
Views
3K