# I have a problem where two stars are at a certain time r0 apart from

1. Oct 16, 2012

### assaftolko

I have a problem where two stars are at a certain time r0 apart from each other. They both has the mass of m, and at that time one of them is at rest and the other has the speed of v0 in the opposite direction of the gravity force that the other star exerts upon him. I need to find the maximal distance between them. now it's quite easy using energy and momentum conservation, but what I don't understand is why in the solution they say that the potential gravitational energy of this system at the beginning is -Gm*m/r0. this is the potential energy of each one of the stars, so isn't this suppose to be doubled for the whole system? I'm confused...

2. Oct 16, 2012

### Simon Bridge

Re: Gravity

The gravitational potential energy of a mass is that mass multiplies by the total gravitational field due to all the other masses.
If those are the only two masses, then the gpe for the whole system is going to be zero.

In Newtonian gravity, masses do not feel their own gravity... so as far as each star is concerned, there is only one gravitational field present.

3. Oct 16, 2012

### assaftolko

Re: Gravity

I don't understand - if as far as each star is concerned, there is only one gravitational field present - why is the total gpe is 0 and not 2* -G*m*m/r0... can you show the calculations?

4. Oct 16, 2012

### Simon Bridge

Re: Gravity

If I say that U is the gravitational PE of mass m, and V is the gravitational potential at the place that m is at, then U=mV.

Thus, each star has a potential that varies with radius around it ... that would be V(r).
Each star only sees that one potential ... so it's potential energy is U=mV(d) where d is the distance separating the stars.

The GPE of the entire system, however, would be the total mass of the two stars 2m times the potential due to any other masses around. Since there are no other masses around, that potential is zero. U=0x2m=0. Hence the GPE of the binary star system is zero. IRL. a star system will be part of a galaxy so it would experience a V from all the other stars in that galaxy and the GPE would not be zero.

Explicitly:
$$U_{tot}=2mV_{tot}=2m\sum_{i=1}^N V(m_i,\vec{r}_i)$$... where $m_i$ is the ith mass present, $\vec{r}_i$ is the vector pointing from the ith mass to the center of mass of the star system, and $N$ is the number if masses besides the stars that are present. N=0 means there are no terms in the sum:$$N=0 \Rightarrow \sum_{i=1}^N V(m_i,\vec{r}_i) = 0 \Rightarrow U_{tot}=0$$

You do know what gravitational potential energy is right?
It is the amount of work you have to do to move a mass a long way away (mathematically: to infinity).

There is no work against gravity in moving the entire binary system unchanged to another location because there are no other masses around to make gravity to work against. But it does take some work against gravity to separate the two stars.

Last edited: Oct 16, 2012
5. Oct 16, 2012

### assaftolko

Re: Gravity

I read some more and saw that the defenition of U as -Gm1m2/r is the gravitational potential energy of a system which consists of m1 and m2 which are at distance r from each other. Thus I now think that they were right with their answer. I still don't quite understand how you determined the gpv is 0 :) I mean I understand what you wrote - that it's 0 because the absence of other masses aside from these 2, but it's a bit contradicting to what I read... maybe I suppose to assume these two stars aren't the only thing out there?

6. Oct 16, 2012

### Simon Bridge

Re: Gravity

Each star by itself has GPE due to the other star's gravity.
This is the amount of work you have to do to remove one of the stars completely.

If you have a spherical mass $m_1$ at position $\vec{r}_1$ (by which we mean that is the position of it's center of mass) then it's gravitational potential at some arbitrary position $\vec{r}$ is given by: $$V_1(\vec{r})=\frac{Gm_1}{|\vec{r}-\vec{r}_1|}$$

Some mass $m_2$ at position $r_2$ will have a potential energy due to $m_1$ of $$U_2=m_2V_1(\vec{r}_2)=\frac{Gm_1m_2}{|\vec{r}_2-\vec{r}_1|}$$ $|\vec{r}_2-\vec{r}_1|=|\vec{r}_{21}|=|\vec{r}_{12}|$ is just the distance between the masses.

Similarly, $m_2$ has a potential$$V_2(\vec{r})=\frac{Gm_2}{|\vec{r}-\vec{r}_2|}$$ and $m_1$ has a gpe due to that given by:$$U_1=m_1V_2(\vec{r}_1)=\frac{Gm_1m_2}{|\vec{r}_1-\vec{r}_2|}$$ $|\vec{r}_1-\vec{r}_2|=|\vec{r}_{21}|=|\vec{r}_{12}|$ is just the distance between the masses.

That's how the book got the definition it did by putting $m_2$ at the origin so that $\vec{r}_2=(0,0,0)$ and $\vec{r}_1=\vec{r}$.

What if you have more than two masses ... then you just add up the potentials of all the masses, and the gpe of $m_1$ due to all the others would be given by:$$U_1=m_1\sum_{i=2}^NV_i(\vec{r}_1)$$

The reason we only sum over the other masses is that the force of gravity at the center of each mass is zero... and it is the center-of-mass that counts.

Bear in mind that U is not the GPE of the system, it is the GPE of one of the masses in the system.

(Caution: in all this gets more complicated if the masses are large in volume and/or not spherical. But for this discussion the main points are not affected.)

7. Oct 17, 2012

### assaftolko

Re: Gravity

Ok thanks a lot!

8. Oct 18, 2012

### Simon Bridge

Re: Gravity

Neatoh.

Just for completeness ... a sphere of mass $M$ and radius $R$ and uniform mass density $\rho=M/V=3M/4\pi R^3$ centered at the origin, has a gravitational potential given by:$$V(r)=\frac{GM}{R^2}r + \left ( \frac{GM}{r} - \frac{GM}{R^2}r \right ) \text{u}(r-R)$$... where u(r) is the unit-step function: u(r)=0 if r<0 else u(r)=1.

If you look at that, if r>R you get the familiar "inverse-distance" result, but if r<R (inside the sphere ye ken?) you get a linear result with distance.

The GPE of the sphere due to itself would be $U=MV(r=0)=0$ ... which is why you only consider the other masses.

some observations:
... the work to remove a small mass $m$ from radius $r$ to infinity would be $U=mV(r)$ ... the work to change the radial distance of the mass from $r_1$ to $r_2$ is
$W=\Delta U=m\Delta V = mV(r_2)-mV(r_1)$

$GM/R^2 \equiv g$ is the acceleration of gravity at the surface of the sphere. For the Earth $$g_\oplus=\frac{GM_{\oplus}}{R_{\oplus}^2}=\text{9.8ms}^{-2}$$

I can rewrite V in terms of g as:$$V(r)=gr + \left ( \frac{gR^2}{r} - gr \right ) \text{u}(r-R)$$... the work to lift a small mass $m$ from $r_1=R$ to $r_2=R+h$ can be obtained from $W=m\Delta V = mV(R+h)-mV(R)$ ... you should be able to work this out for the case that h<<R (hint: binomial approximation) to get the familiar $W=mgh$.

And now, everything should click into place.