Potential energy of a two-body system

[AlonsoMcLaren]

If two masses, m and 2m, are separated by a distance r, what is the potential energy of mass m? What is the potential energy of mass 2m? What is the potential energy of the system?

 

[Doc Al]

The gravitational PE is a property of the system, not of the individual masses.

 

[AlonsoMcLaren]

The gravitational PE is a property of the system, not of the individual masses.

So what is the value of PE of the system?

 

[Doc Al]

So what is the value of PE of the system?

The gravitational PE of two masses is given by:
$$U = - \frac{G m_1 m_2}{r}$$
Where m₁ and m₂ are the masses and r the distance between them.

 

[AlonsoMcLaren]

The gravitational PE of two masses is given by:
$$U = - \frac{G m_1 m_2}{r}$$
Where m₁ and m₂ are the masses and r the distance between them.

How to derive it from F=Gm1m2/(r^2)

 

[K^2]

$$\Delta U = \int_{r_1}^{r_2}\vec{F}\cdot d\vec{r}$$

In this case, it's easiest to integrate along the radial, and taking potential at infinity to be zero, you get this.

$$U = \int_{\infty}^{r}G\frac{m_1 m_2}{r^2}dr = -G\frac{m_1 m_2}{r}$$

 

[Doc Al]

How to derive it from F=Gm1m2/(r^2)

In general:
$$F = - \frac{dU}{dr}$$
So, to go from F to U, integrate.

(Oops... K^2 beat me to it.)

 

[AlonsoMcLaren]

$$\Delta U = \int_{r_1}^{r_2}\vec{F}\cdot d\vec{r}$$

In this case, it's easiest to integrate along the radial, and taking potential at infinity to be zero, you get this.

$$U = \int_{\infty}^{r}G\frac{m_1 m_2}{r^2}dr = -G\frac{m_1 m_2}{r}$$

So are you using m1 (or m2) as frame of reference? If so, the frame is non-inertial.

 

[K^2]

Depends. If m₁>>m₂, then the acceleration of m₁ is negligible, and we can use it as an inertial frame of reference. Or vice versa. But if m₁ and m₂ are comparable, you are right, and we have to use center-of-mass coordinates. In that case, r is the distance from center of mass, and either m₁ or m₂ is the reduced mass.

The idea behind reduced mass is that with central potential, which gravity happens to be, instead of looking at two bodies orbiting each other, you can consider bodies one at a time, and treat them as if each orbits the center-of-mass point as if it was immovable point with gravitational attraction. The strength of attraction is determined by reduced mass, which you compute to be sufficient to generate same force as in the original setup.

It's a bit messy, but for two bodies it works fine. Throw in a third body, and it goes from messy to near-impossible.

P.S. Gravitational potential works out to be exactly the same, by the way, so it doesn't matter.