# Two particles interacting with each other through their mutual gravitational field

LCSphysicist
Homework Statement:
Two particles of mass m1 and m2 are released in rest, they are separated by a distance ro, and all motion is by the mutual gravitational attraction. Calc their velocity when their distance is r < ro.
Relevant Equations:
All below.
This problem is very easy to solve considering that the two particles belong a closed system under action of conservatives force.

My doubt is if it is possible to solve the problem by consider one particle by time, that is:

Suppose that we know the particle m one is under gravitational force, at first its energy is = -Gm1m2/ro

After certain time, the distance to the central force will be r, and so -Gm1m2/r + m1v²/2.

If we apply E1=E2, it will be wrong! The question is why, since all forces are conservative.

For example, we can say that m2 is like the earth, when we apply to m1 E1=E2 is right, but in this case no.

This make me wonder that E1 = E2 under the gravitational field of the Earth is actually a approximation (yes, good).

## Answers and Replies

Suppose that we know the particle m one is under gravitational force, at first its energy is = -Gm1m2/ro

After certain time, the distance to the central force will be r, and so -Gm1m2/r + m1v²/2.

Remember that the potential energy ##-\frac{Gm_1 m_2}{r}## is a property of the system (pair) of masses, and cannot be attributed to anyone of the particles individually.

In the special case that one of the masses is significantly larger than the other, you can make the approximation that zero work is done by gravity on the large mass (i.e. it doesn't move) and you can "attribute" all of the potential energy to the smaller mass. This is what we do when considering the "GPE of an object" above the Earth.

If you want to solve it one particle at a time, you might consider looking into the two-body problem. Specifically, try and see if you can form an equation of motion of one body w.r.t. the position of the other body (which occupies a non-inertial frame).

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LCSphysicist
LCSphysicist
Remember that the potential energy ##-\frac{Gm_1 m_2}{r}## is a property of the system (pair) of masses, and cannot be attributed to anyone of the particles individually.

In the special case that one of the masses is significantly larger than the other, you can make the approximation that zero work is done by gravity on the large mass (i.e. it doesn't move) and you can "attribute" all of the potential energy to the smaller mass. This is what we do when considering the "GPE of an object" above the Earth.

If you want to solve it one particle at a time, you might consider looking into the two-body problem. Specifically, try and see if you can form an equation of motion of one body w.r.t. the position of the other body (which occupies a non-inertial frame).
Woll, that's truth, by working most of the time with two bodies with masses very different, i just forget that the potential energy is property of the system :S, thx

You can try this. Let ##\vec{F}_{12} = m_1\ddot{\vec{x}}_1## and ##\vec{F}_{21} = m_2\ddot{\vec{x}}_2## be equations of motion in the inertial (space) frame. Now $$\vec{a}_{21} = \ddot{\vec{x}}_2 - \ddot{\vec{x}}_1 = \frac{1}{m_2}\vec{F}_{21} - \frac{1}{m_1}\vec{F}_{12} = (\frac{1}{m_1} + \frac{1}{m_2})\vec{F}_{21}$$ You end up with $$\vec{F}_{21} = \frac{m_1 m_2}{m_1 + m_2}\vec{a}_{21} = \mu \vec{a}_{21}$$So now you can let ##\vec{F}_{21} = -\frac{Gm_1 m_2}{r^3}\vec{r}## and solve for the relative motion!