# Gravitational potential energy problem

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1. Oct 30, 2016

### Buffu

1. The problem statement, all variables and given/known data

Question :-
Two stars, each of a solar mass and radius $10^7 m$ are at a distance of $10^{12} m$ from each other. Find the speed of each star before collision if initial speed is negligible.

2. Relevant equations

$$V(r) = {-G Mm \over r}$$

3. The attempt at a solution

Initial total energy of a star would be $${-G m^2 \over 10^{26}}$$
and final energy would be $${-G m^2 \over 10^{14}} + {1 \over 2 }mv^2$$

use conservation of energy i got

$${-G m^2 \over 10^{26}} = {-G m^2 \over 10^{14}} + {1 \over 2 }mv^2$$

Solving for $v$ i got, $3.6 \times 10^6 m/s$. But, the given answer is $2.6 \times 10^6 m/s$.

Now i can get the given answer by solving of $v$ in equation $${-G m^2 \over 10^{26}} = {-G m^2 \over 10^{14}} + mv^2$$

I am not getting the concept that why i should drop the half from equation ? Please help me.

2. Oct 30, 2016

### BvU

How many stars are moving at impact time ?

3. Oct 30, 2016

### Buffu

Two stars. But how does that help ?

4. Oct 30, 2016

### BvU

So how many times ${1\over 2}mv^2$ ?

5. Oct 30, 2016

### PeroK

Can you explain this fomula? What is $V(r)$ in this case?

6. Oct 30, 2016

### Buffu

Gravitational potential energy.

7. Oct 30, 2016

### PeroK

... of what?

8. Oct 30, 2016

### Buffu

Of any one star.

9. Oct 30, 2016

### PeroK

10. Oct 30, 2016

### Buffu

But why do we add the Kinetic energy of other star, i am only considering Potential energy of one star.

11. Oct 30, 2016

### Buffu

I think yes but i am getting a feel that i am wrong. But why ?

12. Oct 30, 2016

### PeroK

If $m \ne M$ do both stars have the same PE? Think about $M >> m$.

13. Oct 30, 2016

### Buffu

The bigger one will have less P.E, right ? But that does not answer why $V(r)$ is P.E of both stars combined.

14. Oct 30, 2016

### PeroK

You can derive it by looking at the force on each object and how that varies with the distance between them. That's a different exercise.

For the time being:

$$V(r) = {-G Mm \over r}$$

Is the combined total PE of both stars.

15. Oct 30, 2016

### Buffu

So if we have 3 stars of equal mass, then total PE would still be $V(r) = {-G m^2 \over r}$ or $V(r) = {-3G m^2 \over r}$ or $V(r) = {-3G m^2 \over 2r}$. I think it is the third one.

16. Oct 30, 2016

### PeroK

What's $r$ if you have 3 stars?

17. Oct 30, 2016

### Buffu

Distance between them. Like a vertices of a equilateral triangle. I just thought of this problem now.

18. Oct 30, 2016

### PeroK

Derivation of PE for two masses, $m_1$ and $m_2$:

$F_1 = F_2 = \frac{Gm_1 m_2}{r^2}$

$dKE_1 + dKE_2 = \frac{Gm_1 m_2}{r^2}(-dr_1 - dr_2) = -\frac{Gm_1 m_2}{r^2}dr$ (where $r = r_1 + r_2$)

$dV = \frac{Gm_1 m_2}{r^2}dr$

$V(r) = -\frac{Gm_1 m_2}{r}$

19. Oct 30, 2016

### PeroK

By symmetry, the PE of each star due to each of the others is $-\frac{Gm^2}{2r}$, so yes the total is $-\frac{3Gm^2}{2r}$

20. Oct 30, 2016

### Buffu

Ok i got this.

Ok imagine three objects like $$\underbrace{\star}_{m_1} \overbrace{------}^{r} \underbrace{\star}_{m_2}\overbrace{------}^{r} \underbrace{\star}_{m_3}$$
What should be the total PE of middle star ?

it is not $$- {Gm_1m_2\over r} - {Gm_3m_2\over r}$$.
We can't use symmetry argument here because masses are different.

So how can i know the PE of mid star w.r.t to first star and w.r.t to third star separately ?