Gravitational potential energy problem

[Buffu]

Homework Statement

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.

Homework Equations

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

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.

 

[BvU]

How many stars are moving at impact time ?

 

[Buffu]

How many stars are moving at impact time ?

Two stars. But how does that help ?

 

[BvU]

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

 

[PeroK]

Homework Equations

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

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

 

[Buffu]

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

Gravitational potential energy.

 

[PeroK]

Gravitational potential energy.

... of what?

 

[Buffu]

... of what?

Of anyone star.

 

[PeroK]

Of anyone star.

Are you sure about that?

 

[Buffu]

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

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

 

[Buffu]

Are you sure about that?

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

 

[PeroK]

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

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

 

[Buffu]

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

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

 

[PeroK]

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

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.

 

[Buffu]

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.

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.

 

[PeroK]

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.

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

 

[Buffu]

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

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

 

[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 $r = r_1 + r_2$)

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

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

 

[PeroK]

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

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

 

[Buffu]

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

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

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

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

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

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 ?

 

[PeroK]

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 potential 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 second star separately ?

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.

 

[Buffu]

You would need to calculate the share of the total PE that each mass has.

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 ?

 

[Buffu]

Delete this message

 

[PeroK]

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 :-

$$m_1 = km_2$$
$$\therefore \text{contribution of}m_1{ is} - {Gm_2m_1\over kr}$$

Well, that fails for $k = 1$

Instead, try this:

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.

 

[Buffu]

Well, that fails for $k = 1$

Instead, try this:

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.

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

 

[PeroK]

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

That's not right at all.

If we take $r = r_1 + r_2$, then, by conservation of momentum (assuming the two masses are at rest initially), we have:

$v_2 = \frac{m_1}{m_2}v_1$

Hence:

$r_2 = \frac{m_1}{m_2}r_1$

Hence:

$r = \frac{(m_1 + m_2)r_1}{m_2} = \frac{(m_1 + m_2)r_2}{m_1}$

Now, we have:

$F_1 = \frac{Gm_1m_2}{r^2} = \frac{Gm_1m_2^3}{(m_1 + m_2)^2 r_1^2}$

This is the force on $m_1$ in terms of $r_1$, the distance from the fixed centre of mass. Therefore:

$PE_1 = -\frac{Gm_1m_2^3}{(m_1 + m_2)^2 r_1} = -\frac{Gm_1m_2^2}{(m_1 + m_2)r}$

And:

$PE_2 = -\frac{Gm_1^2 m_2}{(m_1 + m_2)r}$

You can check that in the limit $m_1 >> m_2$, then these reduce to $PE_1 = 0$ and $PE_2 = -\frac{Gm_1m_2}{r}$

And, of course, $PE_1 + PE_2 = PE$

 

[PeroK]

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

Here's a better derivation using KE.

As above, by conservation of momentum, $m_1v_1 = m_2v_2$, hence at any time:

$KE_1 = \frac12 m_1v_1^2 = \frac{m_2}{m_1} KE_2$

As the KE is in this proportion, the PE must be in the same proportion, so we have:

$m_1PE_1 = m_2 PE_2$

And:

$PE_1 = \frac{m_2}{m_1 + m_2}PE = -\frac{Gm_1 m_2^2}{(m_1 + m_2)r}$

 

[Buffu]

Here's a better derivation using KE.

Amazing derivation. I learned a lot from you today. Thank you very much for your time.