Understanding the Three Body Problem in Space

[JANm]

Suppose a body in a circular orbit around a very heavy one. The body itself has a very light one moving in an small orbit in the same plane.

1 What is the orbit of the very light one in space?

very= 1000 times, so the heaviest= 1 million times heavier than the "smallest".

small=1/10, so the smallest distance of the very light one 0,9 times the radius of the orbit of the body around the heaviest and the largest distance of the very light one is 1,1 times the radius of the orbit of the body around the heaviest one, approximately.

2 do these numbers 0,9 and 1,1 need corrections?

 

[Janus]

The largest orbit a one body can have around another body when the 2nd body orbits a third body is limited by the Hill sphere.

This is found by:

$$R = a \sqrt[3]{\frac{m}{3M}}$$

In this case, we define "a" as the orbit of the middle sized object around the largest object and assign it a value of 1.

m is the mass of the middle sized object and also will equal 1
M is the mass of the largest object and therefore = 1000.

so:

$$R = 1 \sqrt[3]{\frac{1}{3(100)}} = 0.069$$

Which is smaller than the 0.1 you are trying to use for the smallest object's orbit around the middle sized one. IOW, the smallest object cannot orbit at that distance from the middle sized one. It would be pulled away into an independent orbit around the largest object.

 

[Chronos]

The three body problem in orbital mechanics has, to my knowledge, never been completely solved. It is a humbling reminder of how much we have yet to learn about the universe. We can, however, approximate solutions to problems such as this, as well as much more complicated systems with amazing accuracy.

 

[D H]

The largest orbit a one body can have around another body when the 2nd body orbits a third body is limited by the Hill sphere.

This is found by:

$$R = a \sqrt[3]{\frac{m}{3M}}$$

More or less. This is a three body problem, after all. An alternative to the Hill sphere is Laplace's sphere of influence:

$$R = a\left(\frac m M\right)^{2/5}$$

With M=1000*m, the sphere of influence is 6.3% of the distance between the primary and secondary bodies (compare to 6.9% for the Hill sphere).

Interestingly, neither the Hill sphere nor the sphere of influence is a sphere.

 

[JANm]

The three body problem in orbital mechanics has, to my knowledge, never been completely solved. It is a humbling reminder of how much we have yet to learn about the universe. We can, however, approximate solutions to problems such as this, as well as much more complicated systems with amazing accuracy.

Hello Chronos
Thanks for this poetical way of understanding three body problems.
I don't quite understand the Lagrange like boarders for posing three-body problems... I have taken care that my problem didnot counteract geometrical possibility by defining radii, I have sayd things about masses, must say have defined them relatively absolute, but the impossibilities of the posing of the question gathers around...
greetings Janm