# Orbital mechanics question

When in orbit, a satellite is 'falling' with an accelerating velocity as per the simple gravity formula. Yet the earth is 'dropping' away from the satellite at a constant rate since it is spherical. This means that the 'left over' value between those two rates should cause the satellite to eventually 'gain' on the earth and come nearer, rather than remain at a constant distance. Therefore the only other force that can compensate for that left over velocity is centrifugal force.

Do I have this correct? If not, can someone explain it to me please.

tex

## Answers and Replies

Jonathan Scott
Gold Member
When in orbit, a satellite is 'falling' with an accelerating velocity as per the simple gravity formula. Yet the earth is 'dropping' away from the satellite at a constant rate since it is spherical. This means that the 'left over' value between those two rates should cause the satellite to eventually 'gain' on the earth and come nearer, rather than remain at a constant distance. Therefore the only other force that can compensate for that left over velocity is centrifugal force.

Do I have this correct? If not, can someone explain it to me please.

tex

I don't understand exactly what you're saying, but it seems quite mixed up. If you move in a straight line past a spherical surface, it drops away at a steadily increasing rate, not at a constant rate, in that the angle of dropping away is proportional to the circumferential distance travelled. Similarly, a satellite being accelerated by gravity has a downwards component of velocity which increases at a steadily increasing rate. When those rates match, you get a circular orbit (when you also take into account the fact that the downwards direction changes as you progress around the orbit).

The curve that represents the path taken by a falling object (the satellite) is a parabola I believe. The curve representing the 'fall' of the surface of the earth is a circle (the sphere).

[PLAIN]http://www.longviewbluegrass.com/parabola.jpg [Broken]

The two curves will never match. In fact if a rifleman shoots a bullet, regardless of the speed of the bullet, the only way it wont hit the ground if the ground falls away at or greater than the rate at which the bullet is falling.

[PLAIN]http://www.longviewbluegrass.com/parabola2.jpg [Broken]

Im not saying Im right. On the contrary, I know very little about orbits. Think of this as layman's confusion and lack of understanding. But it does seem than since you can never match the two curves that, considering only curve-matching, either the satellite will eventually hit the earth or go farther into orbit.

Wait.......

Wait......

Ive just had a eureka moment. If it IS true that the curves can never match THEN it IS true that either the satellite will eventually hit the earth or, if too fast, go out beyond it, this being determined by speed. THEREFORE..... there must be some value of velocity where, even though the two curves dont match, the satellite neither gains on the surface or retreats from it. At that point you have a sustained orbit.

Right?

tex

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Janus
Staff Emeritus
Gold Member
The curve that represents the path taken by a falling object (the satellite) is a parabola I believe. The curve representing the 'fall' of the surface of the earth is a circle (the sphere).

[PLAIN]http://www.longviewbluegrass.com/parabola.jpg [Broken]

The two curves will never match. In fact if a rifleman shoots a bullet, regardless of the speed of the bullet, the only way it wont hit the ground if the ground falls away at or greater than the rate at which the bullet is falling.

[PLAIN]http://www.longviewbluegrass.com/parabola2.jpg [Broken]
Actually, the path taken by a satellite or the bullet is an ellipse. The parabola that we normally associate with falling objects is just an approximation based on the assumption that the Earth is flat and gravity always pulls perpendicular to the bullet's original trajectory. Over small distances, this approximation is close enough.

The actual shape of this ellipse is called it eccentricity, the closer the eccentricity is to 1, the flatter the ellipse and the closer it is to 0, the rounder the ellipse. A circle is a special case of ellipse where the eccentricity is exactly 0 The eccentricity of the path of a falling object depends on the strength of gravity and its initial speed and trajectory.

For a satellite, these combine to give the satellite a path that is has a eccentricity that is very nearly 0 and is almost a perfect circle. However, circles or near circles are not the only stable orbit shape. As long as the ellipse does not actually intersect the body being orbited, you can have orbits that are very eccentric (flat). Comets are examples of this. Many have very eccentric orbits that only bring them in close to the Sun for short periods.

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Do I have this correct? If not, can someone explain it to me please.

tex

All bodies in the universe exert forces on one another. The vast majority of them can be discounted, at least in our lifetime. :)

There are a couple of planet-killers out there, but we're tracking them. The key is to ensure we blast them from their ability to smack us a million years in the past before they can.

Can it happen? Sure. Actually, it's happened on Earth several times before.

Let's not let it happen again.

Ken G
Gold Member
Ive just had a eureka moment. If it IS true that the curves can never match THEN it IS true that either the satellite will eventually hit the earth or, if too fast, go out beyond it, this being determined by speed. THEREFORE..... there must be some value of velocity where, even though the two curves dont match, the satellite neither gains on the surface or retreats from it. At that point you have a sustained orbit.
Your Eureka moment brought you very close to the answer-- your problem is in thinking that the curves can't match (as was pointed out, the problem there is your belief the curve must be a parabola, which gets less and less true as the speed of the bullet increases). Your analysis was completely correct-- at low speed, the curve mismatch causes the bullet to fall and hit the surface, and at high speed, the mismatch causes it to rise away from the surface. What you missed is that if there is a perfect speed where neither happens-- it doesn't fall down nor fly upward-- then it must maintain a constant height above the surface. Since that situation keeps replicating itself (we are neglecting air resistance of course), it means the height is maintained, which further means that there is no curve mismatch at all in that situation. That's the sustained circular orbit speed, and a circle is not a parabola-- you would have been fine had you known that the curves only look parabolic when the speed is low.