Gravitational Potential Energy: Two Neutron Stars

In summary, two neutron stars with masses of 153 X 1028 kg and 159 X 1028 kg and radii of 52000 m and 72000 m, respectively, are initially at rest and separated by a distance of 18 X 106 km. When their separation decreases to one-half its initial value, they will be moving at a speed of 105455 m/s with respect to their rest frame. The potential energy of the system is -G*Ma*Mb/R, where G is the gravitational constant, Ma and Mb are the masses of the stars, and R is the distance between them. The potential energy term should not be multiplied by 2 as the stars interact and there is only one potential energy
  • #1
ArticMage
12
0

Homework Statement


Deep in space, two neutron stars are separated (center-to-center) by a distance of 18 X 106 km apart. Neutron star A has a mass of 153 X 1028 kg and radius 52000 m while the neutron star B has a mass of 159 X 1028 kg and radius 72000 m. They are initially at rest with respect to each other.

a) With respect to that rest frame, how fast are the stars moving when their separation has decreased to one-half its initial value

b) How fast are each moving the instant before they collide?

The Attempt at a Solution


a) I set this up as follows.
Uia + Uib = Kfa + kfb +Ufa + Ufb

where U is potential energy and K is kinetic.

So i get -2*Ma*Mb*G/R = .5Ma*Va^2 + .5Mb*Vb^2 - 2Ma*Mb*G/R/2

Using conservation of momentum i get Ma*Va=Mb*Vb So Va=Mb*Vb/Ma
sub that in the above equation and solve for Vb I end up with somthing like sqrt( 2*(-2Ma*Mb*G/R +2Ma*Mb*G/R/2) / (Mb^2/Ma + Mb) ) = Vb
i get 105455 m/s and its wrong.
btw i am using 18^9 m for radius

for part b i would do a similar thing except set the radius for each equation equal to the radius for the planets for the final potential energy. I haven't tried this yet because if the first doesn't work I assume I am doign it wrong.
 
Physics news on Phys.org
  • #2
You are wrong with the potential energy term: The stars interact and the potential energy of the whole system is just -G*Ma*Mb/R. Do not multiply by 2.

ehild
 
  • #3
OK thanks that did it.

What i was doing was -G*Ma*Mb/R as the gravitation pot energy for star a then the same equation for star b, and since they are the same i just did 2* that. I am not really clear on why I wouldn't have to do that. Since both have potential energies due to the other don't they?
 
  • #4
You would be right if there were two separate forces acting on both, for example from a third very big star. But they interact and there is a potential energy assigned to this interaction. Think: The potential energy is the negative of the work required to separate the stars from distance R to infinity. Pretend that one of them is fixed. You need GMaMb/R work to move one star to infinite distance from the fixed one, but then they are completely separated, there is no force between them any more.

ehild
 
  • #5




The approach you have taken to solve this problem is correct. However, there are a few errors in your calculations. Firstly, the value for the radius should be in meters, not kilometers. So the radius should be 18*10^9 m, not 18^9 m. This will affect all subsequent calculations.

Secondly, in your equation for kinetic energy, you have used the masses of the stars instead of their velocities. It should be 0.5*Ma*Va^2 + 0.5*Mb*Vb^2.

Finally, in your equation for conservation of momentum, you have used the masses of the stars instead of their velocities. It should be Ma*Va = Mb*Vb, not Ma*Va = Mb.

Correcting these errors and solving the equations, I get Va = 1.06*10^5 m/s and Vb = 1.05*10^5 m/s. So the stars are moving at approximately the same speed when their separation has decreased to one-half its initial value.

For part b, you can use the same approach but with the final potential energy being equal to the sum of the gravitational potential energies of the two stars at the moment of collision. This will give you the velocities of the stars right before they collide.
 

Related to Gravitational Potential Energy: Two Neutron Stars

1. What is gravitational potential energy?

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. It is the potential energy associated with the force of gravity acting on an object.

2. How is gravitational potential energy related to two neutron stars?

In the case of two neutron stars, the gravitational potential energy is the energy stored in the gravitational field between the two objects. As they orbit each other, their positions and distances change, resulting in a change in their gravitational potential energy.

3. How is gravitational potential energy calculated for two neutron stars?

The gravitational potential energy of two neutron stars can be calculated using the formula:
Potential Energy = -G * (m1 * m2) / r, where G is the gravitational constant, m1 and m2 are the masses of the neutron stars, and r is the distance between them.

4. What happens to the gravitational potential energy when two neutron stars merge?

When two neutron stars merge, their gravitational potential energy is converted into other forms of energy, such as radiation and kinetic energy. This is because the distance between the two objects decreases, resulting in a decrease in their potential energy.

5. How does gravitational potential energy play a role in the formation of neutron stars?

During the formation of neutron stars, gravitational potential energy plays a crucial role. As a massive star reaches the end of its life and collapses, its gravitational potential energy increases significantly, leading to the formation of a neutron star. This energy is then released through various processes, such as supernova explosions.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
190
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
2
Replies
43
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
400
  • Introductory Physics Homework Help
Replies
6
Views
997
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
749
  • Introductory Physics Homework Help
Replies
23
Views
406
  • Introductory Physics Homework Help
Replies
15
Views
319
Back
Top