Deriving Orbital Trajectories from Universal Gravitation

In summary, the individual is trying to derive the equations for a simple circular orbit or for the more realistic, but more complex elliptical orbits, and is having difficulty doing so due to the recursive nature of the equation for gravitational acceleration.
  • #1
Vorde
788
0
For a long time I've wondered how you derive the equations of orbital motion from the law of universal gravitation. Recently I've been feeling like my mathematics knowledge has caught up to the point where I can attempt to do this myself, but I keep getting stuck at the same point and I was hoping someone could point me in the right direction in terms of where to proceed.

I've tried the problem from about half a dozen different angles so I won't list step-by-step my work unless someone asks me to, but the problem is that I am trying to work out the parametric equations for the orbital path of objects and I get stuck in a recursive loop.

I am trying to describe the radial distance of an object from a gravitational source but the acceleration that this object is feeling is dependent on radial distance, so I get a equation for the radial distance that depends on the radial distance.

Can anyone point me towards a way to get around this? Or to clarify what I'm doing?
 
Physics news on Phys.org
  • #2
I think you'll need to give some detail. No one will be able to find your mistakes if they can't see your work.
 
  • #3
Fair enough, the reason I hesitated is that I think it's less that I'm making mistakes and more that I'm going about this the wrong way.

I've been starting by assuming that there is a point mass at (0,0), and an object of negligible mass at ##(r_0,θ)##, using polar coordinates obviously. Ignoring theta for now and postulating zero tangential and zero radial initial velocity, I say that the radial position at time t will be ##r_0+ x(t)## where x(t) is the distance traversed by the smaller object being accelerated by the larger object's gravitational pull.

Now if this were a simple kinematics problem with an acceleration that depended on time, I would just double-integrate the equation for acceleration to get displacement and plug that in for x(t). The problem is for me is that the equation for gravitational acceleration is a function of distance, not time. I need a way to express the acceleration due to gravity as a function of time, or a way to parametrize the universal gravitational law in terms of time.

Or at least that's what I'm trying to do, it hasn't been working obviously so if anyone has any suggestions on a better way to do this, I'm all ears.
 
  • #4
OK. Let's take a step back now. Are you trying to derive the equations for a simple circular orbit or for the more realistic, but more complex elliptical orbits?
 
  • #5
Because I had been trying to derive the path for a body with zero tangential motion, whereby I assumed the path would just be osculating back and forth, I had planned to add tangential motion into it after I'd figured out the radial motion. I would like to be able to do it for elliptical orbits.
 
  • #6
I would simplify your notation a bit. Since the gravitational acceleration depends on the distance from ##(0,0)##, measuring the distance ##x## from ##(r_0,0)## is messy.

Why not just call the position of the particle ##r(t)## measured from the origiin, and let ##r(0) = r_0## (the starting position) and ##\dot r(0) = 0## (the starting ve;ocity).
 
  • #7
Vorde said:
Because I had been trying to derive the path for a body with zero tangential motion, whereby I assumed the path would just be osculating back and forth, I had planned to add tangential motion into it after I'd figured out the radial motion.
That's the wrong approach to solving the general problem. Zero tangential motion (zero angular momentum) means a one dimensional trajectory. The math is quite different from that for a system with non-zero angular momentum.
 
  • #8
D H said:
That's the wrong approach to solving the general problem. Zero tangential motion (zero angular momentum) means a one dimensional trajectory. The math is quite different from that for a system with non-zero angular momentum.

So it won't be easier to solve the one dimensional problem then generalize it? But regardless, I think even in two dimensions my problem will persist. I tried approximating the acceleration based off discrete points but I it didn't give me any clues to the more general continuous solution.
 
  • #9
D H said:
That's the wrong approach to solving the general problem. Zero tangential motion (zero angular momentum) means a one dimensional trajectory. The math is quite different from that for a system with non-zero angular momentum.

Vorde said:
So it won't be easier to solve the one dimensional problem then generalize it? But regardless, I think even in two dimensions my problem will persist. I tried approximating the acceleration based off discrete points but I it didn't give me any clues to the more general continuous solution.

D H's point is that the initial conditions you are imposing do not describe an 'orbit' in the sense that you are using the word. Your initial conditions imply that the orbiting body is starting at rest (no initial tangential or radial motion). Thus, the 1/r^2 force will just pull the body towards the center of mass of the system.

Is this the problem you want to solve mathematically? If you're trying to tackle elliptical orbits you've probably already solved problems like this in introductory mechanics.
 
  • #10
A little simulation...

I think the problem here is that this is an example of nonlinear system. What you are looking for is called analytical solution(closed form expression). Nonlinear systems rarely have them, unless it's for special case or if it fits nicely into some other function. It may not have a generic analytical solution, but I think it still should be possible to express it as some sort of infinite sum. What made me think about this is, I ran it as a little simulation, just plugged a Newton's law of gravity, and gave one of the points mass of earth, the smaller mass immediately started describing a precise elliptical path! I actually expected it to break down due to some numerical error, but it keeps a very stable orbit. What i noticed is that, the simulation itself, the way i made it is essentially an infinite sum, it computes little pieces and adds them, forever, but some pieces repeat and form elliptical path(very pretty). So it must be possible to express this as parametric equation in form of a sum. Simulation is performed in steps:

Code:
DirectionToEarth = Normalize(Earth.Pos - Vehicle.Pos)
DistanceToEarth = Length(Earth.Pos - Vehicle.Pos)

Vehicle.Accel.x = DirectionToEarth.x*G*Earth.Mass/DistanceToEarth^2
Vehicle.Accel.y = DirectionToEarth.y*G*Earth.Mass/DistanceToEarth^2
 
Vehicle.Velocity.x = Vehicle.Velocity.x + dt*Vehicle.Accel.x
Vehicle.Velocity.y = Vehicle.Velocity.y + dt*Vehicle.Accel.y

Vehicle.Position.x = Vehicle.Position.x + Vehicle.Velocity.x*dt
Vehicle.Position.y = Vehicle.Position.y + Vehicle.Velocity.y*dt

Direction to Earth was a special vector that helps to guide acceleration. Notice the pattern, new is being added to the old that was there before. Old position and velocity cannot be ignored, even if you want to compute at some specific time you still have to compute all the positions and velocities before that. Next clue, trajectory is ellipse in simple two body case. Ellipse is defined using trigonometric functions. Trigonometric functions numerically can be approximated with Tailor power series, which is also an infinite sum. Coincidence? Also the whole sequence of operations looks very much like some kind of crude manual integration. It looks like a derivative of something that isn't yet integrated. Even delta t is in there, an infinitely small time-step(i didn't put it there on purpose, i just had no other choice, it wouldn't work with straight time). I'm not really sure how to write this down mathematically, but it works as a step-wise simulation. So what you are doing must be possible, but it won't be a simple function that you can just plug a number to and get answer, it will rather be a precise definition of such a function. This is still a worthy goal, could be useful for generating special case closed-form functions.

So far from above we have:

p(t) = p(t-dt) + v(t)*dt,

v(t) = v(t-dt) + a(t)*dt

a(t) = G*M/||(o - p(t-dt))||^2

t-dt, here means something like t-1, or previous time, just trying to keep it consistent with dt.

"o" stands for position of mass M, || o - p || - means distance to o from p.

Combining it together into formidable mess:
p(t) = p(t-dt) + (v(t-dt) + G*M/||(o - p(t-dt))||^2*dt)*dt

I still lack experience with infinite sums, but I'm definitely seeing something carving out there, i need more time to figure this out in complete form, i hope that at least gives you some ideas.

Good luck!
 

What is universal gravitation?

Universal gravitation is a physical law that describes the force of attraction between any two objects with mass.

How does universal gravitation explain orbital trajectories?

Universal gravitation explains orbital trajectories by stating that the gravitational force between two objects is inversely proportional to the square of the distance between them. This force causes objects to orbit around each other in elliptical paths.

What factors affect orbital trajectories?

The factors that affect orbital trajectories include the mass of the objects, the distance between them, and the speed and direction of their initial velocities.

What is the formula for calculating orbital trajectories using universal gravitation?

The formula for calculating orbital trajectories using universal gravitation is F = (Gm1m2) / r2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

How is universal gravitation used in space exploration?

Universal gravitation is used in space exploration to calculate the trajectories of spacecraft and satellites. It also helps scientists understand the movements of planets, moons, and other celestial bodies in our solar system.

Similar threads

Replies
3
Views
757
Replies
14
Views
1K
Replies
1
Views
790
Replies
2
Views
836
Replies
3
Views
494
Replies
2
Views
1K
Replies
7
Views
2K
Replies
3
Views
2K
Replies
2
Views
1K
Back
Top