# Gravitation (just to check if calculation is right)

1. Jul 24, 2006

### Hollysmoke

I was wondering if someone could tell me if I did this right:

Two neutron stars are separated by a distance of 10^10m. They each have a mass of 10^30kg and a radius of 10^5m. They are initially at rest relative to each other. How fast are they moving when they collide?

R = 2r + d = 2(10^5)+10^10

v = sqrroot (2Gm2/R^2), sub in all the numbers and v = 1.2x10^5m/s

2. Jul 24, 2006

### HallsofIvy

Staff Emeritus
Where did you get "v= sqrroot(2Gm2/R2"? Shouldn't that be v= sqrroot(2Gm2/R)?

3. Jul 24, 2006

### Hollysmoke

Hmm I think I rearranged the equation wrong then. Thanks.

4. Jul 24, 2006

### Hollysmoke

I did Fg = Gm1m2/r^2 = 1/2mv^2

5. Jul 24, 2006

### Andrew Mason

The total work done is the integral of the force from initial r to final r:

$$\int_{r_i}^{r_f}Fdr = \int_{r_i}^{r_f}\frac{Gm^2}{r^2}dr = Gm^2\left(\frac{1}{r_i} - \frac{1}{r_f}\right)$$

This becomes the kinetic energy of both stars, each of which has half this energy:

$$KE_{star} = \frac{1}{2}mv^2 = \frac{1}{2}Gm^2\left(\frac{1}{r_i} - \frac{1}{r_f}\right)$$

$$v = \sqrt{Gm\left(\frac{1}{r_i} - \frac{1}{r_f}\right)}$$

AM

6. Jul 24, 2006

### Hollysmoke

We haven't done integrals in this unit so I don't know how to solve that method =(

7. Jul 24, 2006

### Andrew Mason

All you need to know is that the potential energy is $U = -GmM/r$. The change in potential energy between two positions is just $\Delta U = -GmM(1/r_f - 1/r_i)$. This is the change in potential energy of the system. In this case, the system consists of two stars which move toward each other with equal speed.

AM