# Homework Help: 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 ?

21. Oct 30, 2016

### PeroK

First, you are forgetting that the formula is for the total PE of a two-body system. You would need to calculate the share of the total PE that each mass has.

Second, you have to be careful to note the difference between $r$ as a variable for which you can define a function and $r$ as some fixed value at a particular time, when you are better using the notation $r_0$. In this case, the masses might start in that configuration and you can, indeed, calculate a function $V(r)$ for the total PE of every such starting position. But, that does not represent the PE function for the system, as the distance between the masses will not remain equal. In this case, $V(r)$ would not describe the motion of the system, as $r$ is not a valid variable for the system.

Instead, you must have a function of two variables $V(r_1, r_2)$ and $V_0 = V(r_0, r_0)$.

If $m_1 = m_3$ then by symmetry you know that the masses remain the same distance from $m_2$ and you do have a function $V(r)$ that describes the motion of the system.

22. Oct 30, 2016

### Buffu

Yes how would i calculate the share of each mass in the total potential energy of the system of two particles ?

My go that it :-

$$m_1 = km_2$$
$$\therefore \text{contribution of }m_1\hspace{1 mm}{ is} - {Gm_2m_1\over kr}$$
Am i correct ?

23. Oct 30, 2016

### Buffu

Delete this message

24. Oct 30, 2016

### PeroK

Well, that fails for $k = 1$

Calculate PE_1 in terms of the distance $r_1$ from the centre of mass. Then, relate $r_1$ to $r$ using conservation of momentum.

25. Oct 30, 2016

### Buffu

So for simplicity we will consider no movement.
Let CM be at $r^\prime$
So, contribution of $m_1$ is
$$\large{-Gm_1m_2\over r\large{r^\prime \over r}} \implies {-Gm_1m_2\over r^\prime}$$