Why must an orbit become smaller when an orbiting body loses velocity?

[Canuck156]

I'm just curious as to why, when an orbiting body loses velocity (slows down) it's orbit must become more shallow, or smaller. I know this can be explained logically, but I would appreciate if someone could prove mathematically that it should happen.

Thanks.

 

[Sirus]

$$v=\sqrt{\frac{Gm_{E}}{r}}$$
where v is orbital velocity, G is the universal gravitational constant, and E stands for Earth ($m_{E}$ represents the body around which something orbits, not necessarily the Earth). As you can see,
$$v^{2}~\alpha~\frac{1}{r}$$
so when velocity decreases, the object should actually begin to move away from whatever it was orbitting around, meaning orbital radius should increase. Why do you think it decreases?

 

[BobG]

The object's trajectory is determined by it's kinetic energy vs. it's potential energy (pull of gravity). Potential energy is negative, since it's pulling the object towards the center of the Earth (the center of the Earth is normally your point of reference).

If the total energy (kinetic and potential) is less than zero, the object follows a closed orbit (either an ellipse or a circle). If it's equal to or greater than zero, the object will follow a parabola and hyperbola and you won't have to worry about it for long (it won't return). The less total energy, the closer the object comes to the Earth (since total energy is negative for objects in orbit, this means they're becoming more negative).

$$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}$$

Epsilon is your total specific energy per unit of mass (since you're really concerned about motion instead of the actual amount of energy, you can work with one unit of mass - since all the mass of th object is presumably connected, if you figure out the path of one piece of the mass, you've figured out the path of all the pieces of the mass).
mu is the geocentric gravitational constant (substitute the gravitational constant of the object you're orbiting around, if not Earth) It's equal to the mass of the Earth times the universal gravitational constant.
r is the radius
v is the velocity

If you're at some given distance from the Earth (r) and reduce your velocity, the specific energy has to get smaller (more negative). You've reduced your kinetic energy, which means your overall energy must have decreased.

The semi-major axis, average radius of an orbit, is determined by the total specific energy.

$$a=\frac{-\mu}{2\epsilon}$$

The more negative the specific energy (the larger its absolute value), the smaller the average radius gets.

 

[Canuck156]

Sirus, I realize that if you are in a perfectly circular orbit, then the further away from the central body you are, the slower your velocity must be to stay in that orbit. But thinking logically, if you are in a circular orbit, and the satellite (for some reason) loses velocity, surely you would move towards the central body, rather than away from it?

Thanks BobG, I think that explains my question quite well. I was trying to solve the problem using GPE and KE, but I could only find the equations for a circular orbit.

 

[Sirus]

In that case, I stand corrected. My knowledge on orbits does not extend as far as BobG's, so you had better stick to his explanation.

 

[BobG]

If you rearranged that second equation that used specific energy to determine semi-major axis, you could get:

$$\epsilon=\frac{-\mu}{2a}$$

In other, words, for a given orbit (elliptical or circular):

$$\frac{-\mu}{2a}=\frac{v^2}{2}-\frac{\mu}{r}$$

If you rearrange that to solve for velocity, you get:

$$v=\sqrt{\mu*(\frac{2}{a}-\frac{1}{r})}$$

In a circular orbit, r is always equal to the semi-major axis (r = a), which gives you the equation Sirus posted for velocity in a circular orbit.

$$v=\sqrt{\frac{\mu}{r}}$$