How can Kepler's laws help solve tough gravity equations?

[bjon-07]

Tough Gravity equations

Two 1 kg objects are floating in free space 10 meters away from each other(and are not subjected to any other forces other than the gravitonal attraction they have for each other)

How long will take for the two objects to come in contact with each other?


This is tough because acceleration is constantly changing as they get closer together.

I orginally tryed using the universal gravity equation to solve this, and ended up with the following 2nd order non-linear diff eq, which I can't solve

x''=Gm(x^-2) ; want to solve for

I was talking with my professor and he said there was an simplier way to solve this using Kelpers laws. I have no idea how to use kelpers laws to solve this problem sense the objects are rotating around each other, rather they are moving in a straight line towards each other.

Can some one please point me in the right path to solving this problem using kelpers laws, or help me sovle the diff equation.

 

[OlderDan]

Two 1 kg objects are floating in free space 10 meters away from each other(and are not subjected to any other forces other than the gravitonal attraction they have for each other)

How long will take for the two objects to come in contact with each other?


This is tough because acceleration is constantly changing as they get closer together.

I orginally tryed using the universal gravity equation to solve this, and ended up with the following 2nd order non-linear diff eq, which I can't solve

x''=Gm(x^-2) ; want to solve for

I was talking with my professor and he said there was an simplier way to solve this using Kelpers laws. I have no idea how to use kelpers laws to solve this problem sense the objects are rotating around each other, rather they are moving in a straight line towards each other.

Can some one please point me in the right path to solving this problem using kelpers laws, or help me sovle the diff equation.

Some thoughts to explore:

Reduce the problem from a two body problem to the equivalent one body problem in the usual way. The force center is then the CM of the two masses, and the equivalent one body has the reduced mass.

Assume the one body is in elliptical orbit, and think about the period of the orbit in relation to the dimensions of the ellipse (Kepler's laws). What happens in the limit as the minor axis of the ellipse goes to zero?

 

[bjon-07]

Thanks for the quick response,


If I turn this into a one body problems; to find the center of mass 1 kg at -5x and 1kg at 5 x then the center of mass of the system is at 0 (5(1)+-5(1))/10

It is also intuitively obvious that the center of mass of the two objects has to be at zero, but from this point on I am lost

In regards to kelper’s laws could I set C, the curvature, to 0 so that we have a straight line. Am I on the right track, I remember talking abour C in my physics class but I can't seem to find anything about it an my notes.

 

[OlderDan]

Thanks for the quick response,


If I turn this into a one body problems; to find the center of mass 1 kg at -5x and 1kg at 5 x then the center of mass of the system is at 0 (5(1)+-5(1))/10

It is also intuitively obvious that the center of mass of the two objects has to be at zero, but from this point on I am lost

In regards to kelper’s laws could I set C, the curvature, to 0 so that we have a straight line. Am I on the right track, I remember talking abour C in my physics class but I can't seem to find anything about it an my notes.

You could talk about the radius of curvature of an elliptical orbit at different points, but I don't think that is particularly helpful.

The center of mass is, as you have seen, the point midway between the two masses, and each mass will behave as if it were a very small mass orbiting a very large mass located at the center of mass. The very large mass is the sum of the two masses. The period of orbit turns out to depend only on the major axis of the ellipse and the masses. You can easily calculate the period for a circular orbit, where the major axis is a diameter. From that you can deduce the period of an elliptical orbit that starts the same distance from the CM as a circular orbit, and from that you can deduce the time to impact in the limit as the ellipse becomes a line.

 

[bjon-07]

Thanks again for all the help but I am still at a lost to sovle this problem. Which on of Kelper's laws should I be using?

 

[OlderDan]

Thanks again for all the help but I am still at a lost to sovle this problem. Which on of Kelper's laws should I be using?

The third law that relates the period of the orbit to the semimajor axis of the ellipse.

Compare two cases. You will be working on the reduced mass problem of one very small mass orbiting the center of mass (CM) as if there were an object of mass 2M located there. Suppose you launched the mass with a tangential velocity sufficient to put it into circular orbit. You know that the orbit is independent of the small mass, and you can calculate the orbital period. A circle is a special case of an ellipse, so this orbit has a semimajor axis equal to the radius of the orbit.

If you launched the little mass with a greater tangential velocity (not of interest to this problem, but in general) it would go into elliptical orbit and you would have some work to do to figure out the dimensions of the ellipse based on energy and angular momentum considerations. If you launched the small mass with a tangential velocity that is less than that required for circular orbit it would go into elliptical orbit, gaining velocity as it moved closer to the CM, swinging around and eventually returning to the starting point with its initial velocity. It would again require some work to figure out the dimensions of the orbit, but at least you would know the maximum distance from the CM for this orbit, so you would know how far you are from the one focus of the ellipse.

Keplers third law would allow you to derive the period of any such orbit by comparison to the circular orbit if you knew the dimensions of the ellipse. You don't know the dimensions in general, but what you do know is that as you reduce the tangential velocity the ellipse gets flatter and flatter, with the turning point on the far side of the CM getting closer and closer to the CM. When the velocity is really small, the orbiting mass barely misses going through the CM and whips around before returning to the starting point. Think of the most extreme orbit of a comet. These things enter the planetary region and spend only a couple of days near the sun, then get whipped back out to the farthest regions of the solar system. Your little mass will behave that way. The limiting case is that you start the mass with essentially no velocity, and if it doesn't run into anything it turns on a dime at the CM and gets whipped back out to where it came from.

Now, what are the dimensions of the ellipse in the limiting case where the mass starts with no velocity? Compare that to the dimensions of the circular orbit and use the relative dimensions of those orbits with Kepler's third law to derive the orbital period. What part of that period corresponds to the time of flight of the mass into the center of mass, where it will in fact collide with its sister mass?