Why couldn't Newton figure out the gravitational pull of more then two bodies?
Who says he couldn't? It would just be tedious work. Or did you expect a closed-form formula for their motions?
>>7744345
Why couldn't OP figure out the correct version of than/then?
Also, the *pull* (force), rather than the motions, is trivial to work out.
The earth is not a globe.
I'm fairly certain you actually can't calculate for 3+ moving bodies. You have to use an approximation. Someone who actually knows physics care to elaborate?
>>7744443
if they're moving can't you just integrate their velocity functions to find their position?
>>7744459
With only thinking about the problem for a few minutes it looks like the problem will be integration of three equations with six unknowns, twice.
>>7744465
6 equations, since equal and opposite forces is 3 + 3 equations
>>7744345
How is that even supposed to work? Like an apple falls towards a bigger apple that's falling on your head?
>>7744345
You can easily work out the force from an number of bodies. What is hard is to find an analytic solution to the paths that result from that.
This is called the 3 body problem.
>>7744459
>can't you just integrate their velocity functions
Contrary to the false impression you gained at school, the vast majority of differential equations are intractable ie no closed form analytic solution.
https://en.wikipedia.org/wiki/Three-body_problem
>>7744475
It's not hard, it's actually impossible and this was proven by Poincare.
>>7744411
it's an oblate spheroid
Gravity is center-seeking, so when you have three bodies orbiting around each other, the force vectors have to keep rotating to "catch up" with the centers of the bodies.
Note: I've only ever learned to handle vector algebra with Clifford algebra, so that's how I'll explain this.
To apply a vector rotation about the n_hat direction by an angle theta, you'll have to left-multiply by e^(-i*n_hat*theta/2) and right-multiply the complex conjugate of that exponential. The result from this multiplication mathematically represents the direction that your force vector takes (we're using polar coordinates so theta is a coordinate).
Now that you have rotating force vectors, multiply by a magnitude of G*m_1*m_2/r^2, and you're all set.
Take one of the three bodies. Plot the force vectors it feels from the other two bodies (using the above method). Take the vector sum.
That's your net force. You'll have to integrate once to get momentum. Because of the steps you took (multiply two exponentials with 2-vector exponents, multiply magnitude that relies on a vector difference), you'll have an expression that simply no one knows how to integrate.
Because it was 600 years ago and that shit was complicated enough?
>>7745599
same anon here
I've never tried doing this problem with Lagrangian or Hamiltonian formalisms. It certainly looks like it's easier (just need to add gravitational potential energies from the two other bodies), but I've never tried.
>>7744443
If I recall there seems to be a numerical solution involving an infinite series of cubes of time, some chinese mathematician I think
Gif related
