# What are the forces in an elliptical orbit, broken down into x and y factors?

I know that, if I wanted to map the circular orbit of an object by an iteration of time, I know that I can use the relationship between centripetal force and gravitational force...this is how we can find our orbital velocity.

But I also know that the second we move away from a circular orbit an into an elliptical one, I can't use the relationship between centripetal and gravitational forces to map out my path. Why not?

What would the force vectors look like when broken down into x and y vectors, where y is a line that runs through the object and planet (i.e. always perpendicular), and x is consistently tangent to the planet?

What would the force vectors look like when broken down into x and y vectors, where y is a line that runs through the object and planet (i.e. always perpendicular), and x is consistently tangent to the planet?

Exactly the same as in the circular case. The force in the tangential direction is zero, and the force in the normal direction is $GMm\over r^2$ directed towards the center of the planet. $M$ is the mass of the planet, $m$ is the mass of the object, and $r$ is the distance between the center of the object and the center of the planet.

What's different from a circular orbit is that $r$ varies in time in the elliptical orbit.

• In the case of an elliptical orbit, the tangential component of force is zero only at the apsides. Moreover, the normal to the curve points to the planet only at the apsides. Jul 28 '15 at 15:38
• You need to read his definition of x and y in the question. You are talking about the velocity vector. He is talking about the position vector. I even copied the relevant part of the question in my answer. Jul 28 '15 at 15:50
• I see. He's using polar coordinates. Jul 28 '15 at 15:54

The truth is, there is only 1 force acting on a satellite at any time, and that is the gravitational force. The gravitational force pulls the object straight towards the body around which it is orbiting.

What ends up happening is you have a movement in one direction, and a velocity in a different direction. For a circular orbit, and certain parts of an elliptical orbit, the pull is 90 degrees from the velocity direction. In this case, the direction of the velocity will change.

However, in the case of a highly elliptical orbit, sometimes the velocity vector will not be perpendicular to the gravity vector. In fact, for a very elliptical orbit, they can almost be in the same direction! I think that is the mistake that you are making in your analysis.

There is a great image found at this site, that I can't find the copyright to use here, unfortunately. It shows how these vectors are not at 90 degrees for an elliptical orbit.

• I agree, but I would still be able to break my velocities down into vcos(phi) and vsin(phi) where phi is the angle of the orbit path to the planet surface. If my only force is gravity (which consistently points to the planet), what acceleration vectors would be keeping the craft from plummeting to the surface of the planet? As much as I want to say centripetal, I would be arriving right back to my initial problem - it breaks down in elliptical orbits. When broken down into x and y components, I don't see a y acceleration fighting my gravitational acceleration...do see what I'm saying? Jul 28 '15 at 15:06
• I don't see what you're saying. First, you don't need an acceleration against gravity to stay in orbit. You stay in orbit because as you fall you're going sideways so fast, you keep missing the planet. Note that your coordinates $x$ and $y$ are not inertial, and are rotating with the object. So your velocity vector has to rotate as well, changing the components of the velocity in your $x$ and $y$ directions, even with no acceleration at all. It's a terrible coordinate system in which to try to solve the problem. Jul 28 '15 at 16:07
• @MarkAdler - It's actually a fairly standard coordinate system used to solve the problem. Jul 28 '15 at 16:29
• @MarkAdler - Even by computer? It would essentially re-solve for each component every second and then reset to it's new calculated location...I mean, I'm sure it would be easier to calculate it in an intertial frame, but it is possible to do in polar coordinates...yes? Jul 28 '15 at 16:30
• @MarkAdler: Actually, you do need acceleration to stay in orbit. If you didn't have it, you'd keep going straight, and no longer orbit the planet. Both factors are key, the velocity and the pull of gravity. I won't disagree however that it's a poor coordinate system by which to solve the problem. Jul 28 '15 at 16:51

What would the force vectors look like when broken down into x and y vectors, where y is a line that runs through the object and planet (i.e. always perpendicular), and x is consistently tangent to the planet?

A better name for those vectors are $\boldsymbol e_r$ and $\boldsymbol e_\theta$ (or $\hat r$ and $\hat \theta$) in place of your y and x (in that order; it's standard to represent $r$ first and $\theta$ second). These are the unit vectors for polar curvilinear coordinates, which is what you are using.

The displacement vector from the center of the planet to the orbiting object is given by $\boldsymbol r = r \boldsymbol e_r$. Differentiating this with respect to time from the perspective of a non-rotating planet-centered frame yields the velocity vector, $\boldsymbol v = \dot r \boldsymbol e_r + r \dot {\boldsymbol e}_r = \dot r \boldsymbol e_r + r \dot \theta \boldsymbol e_\theta$. Note that the velocity vector is orthogonal to the position vector only in the case of circular motion.

What about acceleration? Gravitation is a central force (note well: not centripetal force). In central forces, the forces are directed along or against the line connecting the two interacting bodies. The gravitational acceleration vector in the two body point mass problem is directed against the displacement vector: $\boldsymbol a = - \frac {GM}{r^2} \boldsymbol e_r$. Since the velocity vector in general is not orthogonal to the position vector, it in general also is not orthogonal to the acceleration vector.