Calculating the Escape Velocity from a Neutron Star

AI Thread Summary
To calculate the escape velocity from a neutron star, the correct formula is vesc = sqrt(2GM/R), where G is the gravitational constant, M is the mass of the neutron star, and R is its radius. The discussion highlights that the potential energy should include a negative sign, as it approaches zero when the distance is infinite. The total energy must remain constant, leading to the conclusion that the escape velocity is derived from the balance of kinetic and gravitational potential energy. Clarifications were made regarding the signs in the equations, which helped the original poster understand their mistake. The conversation emphasizes the importance of correctly applying gravitational concepts in astrophysics calculations.
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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