# Is it possible to predict the orbital shape with the given kinetic and potential energy?

Is it possible to predict whether the orbital of the object is circular, elliptical or hyperbolic only with the help of potential and the kinetic energy of the object is given.

Yes you can, and it's really easy. Assuming that you have taken the reference level for 0 gravitational potential at infinity (this is quite standard), then, if KE+PE is positive, the pathis hyperbolic. If it is negative, it is elliptical (or circular1). If it is zero, you are at escape velocity and the path is parabolic.

In a hyperbolic path, the object has enough total energy to reach "infinity". So, KE+PE>0. When it is equal to zero you get a parabolic path, because it still has enough energy to reach infinity, but just enough (and a parabola can be looked at as a limiting case of a hyperbola lobe). If it is less than zero, it does not have enough energy and will eventually "turn back", giving a elliptical orbit.

Here's a simulation (I've taken some of the quantities to be equal to 1 for simplicitly):

(Note: The image seems to become a parabola before it really does at -0.04 total energy because there isn't enough space to show the large ellipse)

1. If you wish to determine whether or not the orbit is circular, you need to know the mass ratio and the angle the velocity makes with the radius vector

• Does being in the vicinity of a black hole change this? – Deer Hunter Aug 9 '13 at 19:16
• @DeerHunter Well, the diagram was drawn for free space. Putting even another star nearby would change this as the shapes won't be so simple. Also I was considering the system to be Newtonian. With a black hole, you're considering a relativistic situation, which means that potential isn't exactly defined. Though I think you can define it knowing the metric, and with that definition the "escapes"/"doesn't escape"/"just escapes" thing will still hold, however the shapes won't be conic sections (and will also depend on your reference frame) – Manishearth Aug 9 '13 at 19:31

the difference between the kinetic and the potential energy is H $$H = T - \left. {\left| \phi \right.} \right| = - {1 \over 2}{{{k^4}m} \over {{h^2}}}(1 - {e^2})$$ solving for e: $$e = \sqrt {1 + {{2{h^2}H} \over {m{k^4}}}}$$ which shows that if H

• negative then e<1 (the trajectory is ellipse)(kinetic energy < potential energy)
• positive then e>1 (the trajectory is hyperbola)(kinetic energy > potential energy)
• equal then e=1 (the trajectory is parabola)(kinetic energy = potential energy)

Okay, so let's take a look at what you actually have with kinetic and potential energy. For simplicity's sake, let's assume you only have gravitational potential energy.

• Kinetic Energy = $\frac{m v^2}{2}$
• Potential Energy = $m*g*height$.

Okay, so given those two terms, what are you missing? What you really need to know to determine an orbit is a position and velocity, having both will give you the orbit. Alternatively, multiple positions can give you the same thing. So, what do you still need?

1. You need to know the gravity to get the height.
2. You need to know the direction to get velocity (Energy is scalar, it doesn't have a direction)
3. You need the mass of the object.
4. You need position other than height.

You could form a partial orbit with two bits of information:

1. Gravity (Specifically, g(height)) If orbiting Earth, this is easily known.
2. The mass of the object.

This will give you some, but not all, of the Keplerian elements. Specifically, you could get eccentricity, Mean Motion, and Mean Anomaly, but you can't get inclination, Right Ascension Node, or Argument of the Perigee.

If you are considering only two objects and want to determine whether one object is in an orbit with respect to the other, you need to know:

1. The combined mass of the two objects in question
2. The position, at some point in time, of the object in question, with respect to the combined center of mass of the two objects
3. The velocity (vector) of the object in question, with respect to the combined center of mass, at the same point in time you determined its position.

It's not really about knowing the kinetic and/or potential energy of an object as such.