Falling in orbit

[Fez]

Really? I suspected as such, but I didn't look at it relative to the surface. I focused my view on kerbin in the map view, then watched the two ship icons go, checking their vertical position on my monitor. I don't see how that would be an issue, but of course it could be

[K^2]

11 minutes ago, Fez said:

Also, I think the reason higher orbits are slower IS that there's slightly less gravity, since there's also only a slight difference in orbital velocity

Yes. If the force was constant, higher orbits would have to be actually faster.

The fundamental problem you are having with picturing this is due to the frames of reference. I'm sorry I can't explain this better without throwing a bunch of math at you. But I can give you another analogy that might help you a little.

Picture a rock that you've thrown at an angle. We'll ignore air resistance. Then rock travels along an almost perfect parabola until it strikes the surface again. This is what you normally picture as falling. Now imagine that earth surface wasn't there, and all of Earth's mass was concentrated at the center. (Neutron star densities would suffice, actually.) What would happen to the rock? Well, it'd still travel along the same curve where you've thrown it, but then it would drop down bellow, take a turn around the center, and come back up, completing an ellipse. In fact, it would be traveling along a classical, albeit very elliptical orbit. When you see something thrown into the air, and as it drops down, it does in fact merely traverses a very small section of that orbital motion.

[Fez]

Yeah, that makes sense. The way I prefer to think of it is that it's falling, but because its moving so fast, in the time it takes to fall a little, it goes forward a lot, so the trajectory is gentle. But thanks

[Fez]

22 minutes ago, mikegarrison said:

Yes, this is correct. The horizontal speed does not actually "counteract" gravity in any way.

The reason your experiment is confusing you is that you have not properly considered the different reference frames involved.

I think I solved my experiment issue. I tried it again, this time launching a rocket straight up to meet a orbiting craft, so that the craft I launched straight up would fall from an initial speed of 0 m/s. I think the problem in the earlier experiment was that when i was comparing how fast the two were falling, the craft that was re entering was already flying downwards WITH gravity, so it started falling faster

[Bill Phil]

7 hours ago, mikegarrison said:

Yes, this is correct. The horizontal speed does not actually "counteract" gravity in any way.

The reason your experiment is confusing you is that you have not properly considered the different reference frames involved.

The horizontal speed (only horizontal at any given point, relative to the current axis) does counteract gravity, in that it pushes the craft outward, in a balance of forces.

The trajectory is curved due to gravity's hold and the horizontal speed. If you draw a vector from an origin on a Cartesian plane, you can easily craft a oarabola based upon the acceleration vector. Gravity changes the vector's direction, but the horizontal aspect is the same. Nothing gives it a curve, it's just how it works out. But for orbits, the horizontal vector is so huge that the gravity vector only curves it a tiny amount. 

It counteracts gravity because the vector goes outward (at a small angle) and gravity pulls it back in, curving it towards the surface.

The vis viva equation, used in many calculations regarding orbits, is actually very similar to the equation for acceleration of a rotating object.

[K^2]

1 hour ago, Bill Phil said:

The horizontal speed (only horizontal at any given point, relative to the current axis) does counteract gravity, in that it pushes the craft outward, in a balance of forces.

Not in any inertial frame of reference.

[mikegarrison]

2 hours ago, Bill Phil said:

The horizontal speed (only horizontal at any given point, relative to the current axis) does counteract gravity, in that it pushes the craft outward, in a balance of forces.

The trajectory is curved due to gravity's hold and the horizontal speed. If you draw a vector from an origin on a Cartesian plane, you can easily craft a oarabola based upon the acceleration vector. Gravity changes the vector's direction, but the horizontal aspect is the same. Nothing gives it a curve, it's just how it works out. But for orbits, the horizontal vector is so huge that the gravity vector only curves it a tiny amount. 

It counteracts gravity because the vector goes outward (at a small angle) and gravity pulls it back in, curving it towards the surface.

The vis viva equation, used in many calculations regarding orbits, is actually very similar to the equation for acceleration of a rotating object.

No, it doesn't. Do a free body diagram on the object in orbit. There is no "push" from the horizontal speed. There is no force, so no push.

The forces are not balanced. The object is always accelerating.

Consider an airplane in level flight. It experiences four forces: lift, drag, weight, and thrust. Those forces are balanced. The object in orbit experiences only one force: weight. If there are no other forces, what do you think balances it? How could it be moving in a curved path if it's not accelerating, and how could it be accelerating if the forces are balanced on it?

Edited February 4, 2016 by mikegarrison

[K^2]

22 minutes ago, mikegarrison said:

No, it doesn't. Do a free body diagram on the object in orbit. There is no "push" from the horizontal speed. There is no force, so no push.

Again, depends on the coordinate system. If you write out equations of motion in polar coordinates, there is a centrifugal term. I'm trying to avoid this discussion for Fez' benefit, but you should be aware of it.

[mikegarrison]

2 minutes ago, K^2 said:

Again, depends on the coordinate system. If you write out equations of motion in polar coordinates, there is a centrifugal term. I'm trying to avoid this discussion for Fez' benefit, but you should be aware of it.

Well, yeah, if you make a new coordinate system you can make the forces on the object be anything you want, but then you just move the problem to the relationships between coordinate systems.

[K^2]

In terms of intuitive understanding, this is about relationship between coordinate systems, though. People keep talking about "vertical" and "horizontal", which are local concepts. Hence confusion. And yes, we should try and limit this particular discussion to inertial frames for that very reason, but I still caution about telling someone they are flat out wrong about centrifugal effect being a factor. Choice of frames is an important caveat and should be pointed out.

There is also the fact that you cannot always choose a frame that is inertial everywhere, but that is a separate story.

[mikegarrison]

9 hours ago, K^2 said:

In terms of intuitive understanding, this is about relationship between coordinate systems, though. People keep talking about "vertical" and "horizontal", which are local concepts. Hence confusion. And yes, we should try and limit this particular discussion to inertial frames for that very reason, but I still caution about telling someone they are flat out wrong about centrifugal effect being a factor. Choice of frames is an important caveat and should be pointed out.

There is also the fact that you cannot always choose a frame that is inertial everywhere, but that is a separate story.

I'm trying to remember back about 30 years ago now, so bear with me.

Let's take a simple case. You are standing directly underneath a geostationary satellite. In your body-centered reference frame, the satellite is not accelerating. So all forces must be balanced. There is a force from gravity, which must be balanced by a centrifugal force. But the thing is, your reference frame is itself accelerating relative to the source of the gravity (the center of the earth). [Actually, the gravity comes from all the earth, but for most purposes it can be treated as if it comes from a point right at the center of mass.]

Now let's take a new reference frame, one that is centered on the center of the earth but does not rotate with the earth. In this frame, both you and the satellite are whizzing around in a circle. (I'll just assume it's a circle, anyway, for simplicity.) Since you are not moving in a straight line, you must be accelerating, so you must be feeling unbalanced forces. Those are due to gravity. In this unaccelerated reference frame, there is no centrifugal force. Instead, the satellite is constantly being accelerated toward the center of the earth, giving it a curved path. However, the height and speed at which it starts makes that curved path always stay the same radial distance from the center of the earth. You, on the other hand, feel the same gravity. But you are moving at a slower speed, so your curve would not keep you at a constant radial distance. However, you also feel a reaction force from the ground pushing you up and balancing (partially) gravity. So that's why your curve also stays at the same radial distance from the earth's center.

Of course, from the point of view of the sun, that new reference frame is also accelerating, because it is in orbit around the sun. Really, there are no inertial reference frames relative to the whole universe. People talk about "sidereal reference frames" but that only works because the stars are so far away that our accelerations relative to them are tiny enough that we lose them in round-off.

Edited February 4, 2016 by mikegarrison