# Orbital Energies And Force Perpendicular to Motion

1. Jan 7, 2006

### G01

Hi. I'm having some trouble trying to grasp a concept. I know that mathmatically Force perpendicular to motion is zero work. I can't seem to grasp this conceptually. First It seems That is an object is moving right and a force acts straight up on it, it will cause the object to move in a up/right path, thus causing work. This came up when my book started to talk about orbital energies and escaping an orbit. It talked about how you must apply a force in the tangential direction to cause an object in orbit to go to a higher orbit. The book said that you can't apply a force in the radial direction because it is perpendicular to the motion of the object. Again, here I don't se why it won't do work. It should still cause the craft to move farther out and into a higher elliptical orbit. Can someone hlp me understand this?

2. Jan 7, 2006

### .:JimmY:.

Lets see, Force is applied to an object in some direction - do not make an assumption where the object will move- if the object moves or displaces in any direction perpendicular to the direction of the applied force, it is obvious that our force has nothing to with the motion of the object or the work done by the force is zero. Lets say, You are pushing a block- forward- but rather than moving in the direction of the force it starts to move either to your left or right, in that case the force you are applying has nothing to do with the motion of the object.

Well the concept behind applying force in the tangential direction is to increase its tangential velocity.
Previously- before the force was applied- the satelite was in a state of equilibrium where
the centrifugal force, (mv^2)/r = gravitational force. (GMm)/r^2
and r is the distance of the satelitle from the center of the earth.
But when the velocity of the satelite is changed- in case the tangential force is applied- the equilibrium is disturbed, in order to maintain the equilibrium the hieght of the satelitle will also change.

Last edited: Jan 7, 2006
3. Jan 7, 2006

### G01

OK i see what your saying. For the second part its not that a radial force would do absolutely no work on the satalite its just that a tangential force would be much more useful and do more work? I think what I'm asking here is wouldn't a perpendicular force also change the satalites velocity, disturbing equilibrium and thus making the satalite's height change?

4. Jan 7, 2006

### .:JimmY:.

Yes, Instead of going against the gravitational pull- while applying force in perpendicular direction of the motion- it is much easier to do.
Sure it will.
the equation yields
(v^2) = GM/r .... r= R+H (R = radius, H = Hieght above the surface)
G is a constant asn so is M- in case mass of a planet. Change in "v" or "r" even in "M"- if somehow- will affect each other.

Last edited: Jan 8, 2006
5. Jan 7, 2006

### G01

I think it was the way my textbook explained the issue that caused the problem. Thanks.

6. Jan 8, 2006

### Janus

Staff Emeritus
There are two reasons changing an orbit by applying a force along the radial is not perferable.
The first is that it is less efficient.
For example if you want to reach escape velocity from an orbital velocity you need to increase the magnitude of your velocity by a factor of $\sqrt{2}$.
If you apply this velocity change in the direction of motion you only have to add about 41% of your present velocity to your present velocity.
Now what if you add your velocity along the radial? The additional velocity is added at an right angle to the orbital velocity so the magnitude of you final velocity will be gotten by $V_f = \sqrt{V_o^2+V_r^2}$. It turns out that you would have make an velocity change equal to your 100% of orbital speed to get the desired 41% increase in the final velocity you need.
Thus it is more efficient to apply the velocity change along the line of motion.
The same is true if you just want to change an orbit.
The total energy of an orbiting object is given By $E = \frac{mv^2}{2}-\frac{GMm}{R}$ where R is the current radial distance of the object.
it can also be given by $E = -\frac{GMm}{2A}$, where A is the semimajor axis of the orbit (average radial distance for an elliptical orbit). For a circular orbit R always equals A, for an elliptical orbit R doesn't. Since the total energy of the orbit never changes we can say that for any orbit.
$$-\frac{GMm}{2A}=\frac{mv^2}{2}-\frac{GMm}{R}$$
or
$$\frac{GMm}{2A}=\frac{GMm}{R}-\frac{mv^2}{2}$$
note that as we increase v, the right side of the equation decreases, and the A on the left must increase to keep the sides balanced.
The upshot is that an that the average radial distance increases with an increase with the magnitude of the velocity.
But, as before, you get the most increase in velocity magnitude when you apply this velocity along the direction of movement.
Also, when you apply the velocity change along the radial you make ither orbital changes that might not be desirable.
If you apply a velocity increase to a circular orbit along the line of motion you create an new elliptical orbit with a perigee equal to the present orbital height and a new higher apogee.
If you apply the velocity change along the radial, you increase the apogee, but at the same time create a new lower perigee. The new orbit will swing in closer to the planet than the orginal orbit. If you are in low Earth orbit to start with, you are about as close as you want to get already; any lower and you'll start to dip into the atmosphere.

7. Jan 10, 2006

### rcgldr

If the force is only applied perpendicular to motion, then you only change the shape of the orbit, and there's no increase in energy. You need some component of force in the direction of motion to change the energy (sum of potential and kinetic energy).