Calculating the Escape Velocity from a Neutron Star

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SUMMARY

The escape velocity from the surface of a neutron star, which has a radius 750 times smaller than Earth's and a mass 1.8 times greater, is calculated using the formula vesc = √(2GM/R). The correct values for the neutron star's radius (Rn = -4.7711 x 10^9 m) and mass (Mn = 1.0757 x 10^30 kg) are essential for accurate calculations. The discussion clarifies the importance of including the negative sign in the gravitational potential energy equation, U = -GmM/R, to ensure the total energy condition for escape is met.

PREREQUISITES
  • Understanding of gravitational potential energy (U = -GmM/R)
  • Familiarity with kinetic energy (K = mv²/2)
  • Basic knowledge of escape velocity concepts
  • Proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study the derivation of escape velocity in astrophysics
  • Learn about gravitational potential energy and its implications in celestial mechanics
  • Explore the properties and characteristics of neutron stars
  • Investigate the effects of relativistic physics on escape velocity calculations
USEFUL FOR

Astronomy students, astrophysicists, and anyone interested in the physics of neutron stars and gravitational dynamics will benefit from this discussion.

cougar_21
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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) ?
 
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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 (U \rightarrow 0 as R \rightarrow \infty).

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!
 

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