Falling in orbit

[Fez]

Would two spacecraft that are orbiting "fall" vertically at the same rate?  When i say fall, I don't mean relative to the ground, but just in the sense that orbiting objects fall while orbiting, but go fast enough sideways to keep from hitting the ground.  I tried an experiment in KSP, by getting 2 spacecraft in slightly different orbits close to each other, then slowing one down to have a steep re-entry.  I watched as the one re-entering fell vertically slightly faster than the one in orbit.  But I thought they would have the same force of gravity acting on them? 

[K^2]

Both craft are going to accelerate towards the planet at exactly the same rate. But it's hard to gauge by just looking at, because instinctively, you're looking at altitude above ground as reference, and that inherently puts you into a different frame of reference, where you have to start thinking about inertial forces.

Anyways, if you actually draw velocity vectors for both craft, and look at how velocities change, you'll see that the rate of change is the same.

[Fez]

So they would both fall vertically at the same speed?

Also, if they're accelerating down at some rate, wouldn't that mean the crafts vertical speed would be increasing until it hit the ground?

[Bill Phil]

Can you elaborate on the scenario? I'm confused, and from what I understood one craft is on a reentry trajectory and one isn't?

[Fez]

Yeah, I was trying to see if both craft, despite differences in horizontal speed, would fall vertically at the same rate (since acceleration due to gravity would be near equal). Another problem I just thought of, though, is if the craft is accelerating down at say 7.8 m/s/s, then wouldnt its vertical speed keep increasing? Thanks

[Bill Phil]

I'm quite sure that the vertical rate is very different. 

First of all, the gravity of Earth is mostly counteracted by the horizontal speed. Very little vertical acceleration occurs between apo and peri, if you're in a near circular orbit. On a steep orbit, you fall much faster, but we need the amount of time and the difference in altitudes.

[Fez]

I thought that gravity still pulls the craft down, but the horizontal speed gives it a curved trajectory. How would the horizontal speed counteract gravity?

[Temstar]

2 minutes ago, Fez said:

How would the horizontal speed counteract gravity?

Because at a very special horizontal speed the surface of the Earth curves away from you at the exact same rate as the 9.82m/s acceleration downwards acceleration is pulling you down by.

If you have less speed than this, then you fall down faster than the Earth is curving away from you, hence why you loose altitude.

[Fez]

I know that, but if acceleration means its vertical velocity, the rate at which it falls down, is increasing by 9.8 m/s every second, why doesn't the trajectory become very steep?

[Temstar]

14 minutes ago, Fez said:

why doesn't the trajectory become very steep?

It does though, After 1/4 of an orbit the direction of your motion is now 90 degrees different from your starting position. If the Earth still has its normal CoM but is instead a flat plan you will hit the ground vertically at orbital velocity. 

Edited February 3, 2016 by Temstar

[Fez]

I mean, wouldn't the trajectory not be a circle, but exponentially get steeper till it crashed? If its downward speed is constantly increasing, while its horizontal speed is constant

[Temstar]

12 minutes ago, Fez said:

I mean, wouldn't the trajectory not be a circle, but exponentially get steeper till it crashed? If its downward speed is constantly increasing, while its horizontal speed is constant

Horizontal speed will not be constant (from surface of Earth reference frame) because the direction of "down" is constantly changing. Remember acceleration is towards CoM of earth (assuming spacecraft of negligible mass relative to earth), and that acceleration's direction is constantly changing (again from earth reference frame) because the spacecraft is moving.

Horizontal speed and direction of acceleration are only constant from spacecraft reference frame.

Edited February 3, 2016 by Temstar

[K^2]

1 hour ago, Fez said:

So they would both fall vertically at the same speed?

Also, if they're accelerating down at some rate, wouldn't that mean the crafts vertical speed would be increasing until it hit the ground?

No, if you look at height and vertical velocity, you are inherently looking at it from a polar coordinate system. If you insist on doing it that way, besides the force of gravity, you must include the centrifugal term. Math gets a little tricky really fast if you are using anything other than Cartesian coordinates.

In Cartesian coordinates, just picture a velocity vector for each space craft. The arrow for the faster one is way longer. Now shift the tips of these arrows down by the same amount. Clearly, the shorter one is more sloped downwards. That's what causes the slower satellite to drop while the faster one remains in orbit.

[Fez]

Alright, thanks. Quick question, at what rate is the spacecraft falling? I don't mean relative to the surface, but just in space. Because as its falling, its also moving forward

[andrewas]

'Downward' changes direction during the orbit. The ship is constantly accelerating towards the planet, but the speed remains constant (in a perfectly circular orbit, anyway).

[K^2]

8 minutes ago, Fez said:

Alright, thanks. Quick question, at what rate is the spacecraft falling? I don't mean relative to the surface, but just in space. Because as its falling, its also moving forward

At the rate equal to the strength of gravitational field, which is going to be slightly weaker than surface gravity in the low orbit. In other words, a little less than g, unless your orbit is really high.

[Fez]

So its falling toward the CoM at about 9.8 m/s?

[K^2]

A little bit less, but yeah. The actual formula is g*R²/(R+h)², where g is surface gravity, R is radius of the planet (i.e. distance of surface from CoM) and h is the altitude. You can compute the exact value for h = 80km, or whatever orbit you have in mind.

[pincushionman]

If you're in a circular orbit, your vertical (radial) speed is zero, and will stay zero. It is under continuous acceleration, but the result if that acceleration is a constantly-changing direction, but no change in speed.

I'm not sure what quanity you're looking at to compare the "falling speeds"; looking at altitude won't help because that won't change for the circular orbit. Comparing the Cartesian coordinates in the falling direction won't be helpful either, because the acceleration vector changes at the same rate as the velocity vector.

[Temstar]

3 hours ago, Fez said:

So its falling toward the CoM at about 9.8 m/s?

No, you're accelerating towards CoM at 9.8m/s² (or whatever the value is for your local gravity), but your distance to CoM remains constant for a circular orbit, which by definition means you have zero velocity towards the CoM at any instant.

Remember, gravity is an acceleration and speed is... speed. Acceleration is measured in m/s² and speed in m/s.

[Fez]

The way I had orbits explained to me is that it's falling, but moving so fast that it's trajectory equals the curvature of the surface. What I want to know, is how much is "falls".

[sevenperforce]

12 minutes ago, Fez said:

The way I had orbits explained to me is that it's falling, but moving so fast that it's trajectory equals the curvature of the surface. What I want to know, is how much is "falls".

Depends entirely on where the orbit is.

Try this: Draw a scale graphic of the Earth. Now, draw a circle around it that is 3" larger than the Earth. This is one possible orbit. Now, draw another circle around it that is 6" larger. This is another possible orbit.

The first circle bends much more tightly than the second circle, but the gravity of Earth tugs on objects at the first circle with approximately the same force as it does objects on the second circle. Thus, objects on the first circle must be "falling" much faster than objects on the second circle in order to avoid hitting the Earth.

If you start out without any speed, you'll start falling vertically quite fast, but that's another matter entirely.

For Earth, an object just outside of the atmosphere needs to be "falling" sideways at around 7,800 meters per second.

[Fez]

I'm sorry, I'm even more confused now. 

[Fez]

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

[mikegarrison]

21 hours ago, Fez said:

I thought that gravity still pulls the craft down, but the horizontal speed gives it a curved trajectory. How would the horizontal speed counteract gravity?

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.

Edited February 3, 2016 by mikegarrison