Velocity Problem of neutron star

In summary: Remember, G is the gravitational constant, which is 6.673x10-11 N/m2). So if we wanted to solve for a, we could use the following equation: a=G*M/r^2.
  • #1
Cursed
39
0

Homework Statement


A neutron star has a mass of 2.0 x 1030 and a radius of 5.0 x 103. Suppose an object falls from rest near the surface of the star. How fast would it be moving after it had fallen a distance of 0.010 m? (Assume that the gravitational force is constant over the distance of the fall, and that the star is not rotating.)

Homework Equations



(a) F=(G*m1*m2)/r2
(b) F=ma
(c) V2=Vo2 + 2ax


The Attempt at a Solution



Using equation (a), I found the force to be about 1.068 x 1043 N

Using equation (b), I found the acceleration to be 5.338 x 1012

Using equation (c), I found the velocity to be 231042.68 m/s after the object had fallen a distance of 0.01 m.
 
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  • #2
What's the question? b) looks fine. I don't see how you got a) since you didn't give a mass for the object. c) doesn't look correct numerically.
 
  • #3
Dick said:
What's the question? b) looks fine. I don't see how you got a) since you didn't give a mass for the object. c) doesn't look correct numerically.

Yeah, (c) didn't look right.

For (a), I squared the mass because I only had one mass given, and that was 2 x 1030. Then I divided it by the distance squared, which I used 5 x 103 for, and multiplied it by G.

F = (G*M2)/r2
 
  • #4
If you did that then the force in a) is the force between two neutron stars with centers 5000m apart. For c), I don't mean that it's REALLY wrong, it just looks like somebody forgot a 2.
 
  • #5
Yeah you don't want to be doing any mass squaring on this one. What happens is you set F=ma in the Gm1m2/r^2 equation and the m of the falling object cancels out, so it's not needed to solve the problem.
 
  • #6
Now I'm confused. :s

So what equation(s) would I use to go about solving this?
 
  • #7
If M is the neutron star mass and m is the mass of the object, F=G*M*m/r^2. (If they don't give you a value for m, that's all you can say). Since we also have F=m*a, a=G*M/r^2.
 

1. What is the velocity problem of neutron stars?

The velocity problem of neutron stars refers to the discrepancy between the observed velocities of neutron stars in our galaxy and the predicted velocities based on their formation mechanisms. Neutron stars are formed from the collapse of massive stars, and it is expected that they would receive a "kick" of velocity during this process. However, the observed velocities are much higher than predicted, leading to the velocity problem.

2. Why is the velocity problem of neutron stars important?

The velocity problem of neutron stars is important because it can provide insight into the formation and evolution of these mysterious objects. It also has implications for our understanding of gravitational waves and the distribution of matter in the galaxy.

3. What are the proposed explanations for the velocity problem of neutron stars?

There are several proposed explanations for the velocity problem of neutron stars, including asymmetric supernova explosions, interactions with binary companions, and interactions with the interstellar medium. However, none of these explanations can fully account for the observed velocities.

4. How do scientists study the velocity problem of neutron stars?

Scientists study the velocity problem of neutron stars by observing the velocities of neutron stars in our galaxy and comparing them to theoretical predictions. They also use computer simulations and models to better understand the formation and evolution of neutron stars and the factors that may contribute to their high velocities.

5. What are some potential future research directions for the velocity problem of neutron stars?

Future research on the velocity problem of neutron stars may involve more detailed observations of neutron stars and their surrounding environments, as well as improved computer simulations and models. Additionally, studying the properties of neutron stars in other galaxies may provide further insights into their formation and velocities.

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