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Noclevername
2014-Nov-06, 11:21 PM
Suppose that an object in LEO were to suddenly come to a halt relative to the surface of the Earth immediately under* it, and the energy of its movement dispersed harmlessly.

Would it fall straight down* to the surface? What would the atmosphere do to it?


*in this case, it means the part of the surface directly in line between the orbiting object, and the center of the Earth.

pzkpfw
2014-Nov-06, 11:27 PM
As the atmosphere moves with Earth (the reason you can't just go up in a balloon and come back down somewhere else), I'd assume some "sideways" velocity would be given to the falling thing. How much? Would it even be noticeablee for a given object? Dunno. Mass, volume, shape, polar versus equatorial drop, local weather conditions, ... lots of factors.

Edit:

Ah phooey. John Mendenhall is correct (except for the burn), in that I missed a major part of the question and jumped to a conclusion. Sorry.


...

Van Rijn
2014-Nov-06, 11:43 PM
It will free fall until it hits significant atmosphere. If it is falling over a pole, it will continue pretty much straight down. At the equator, the atmosphere will be hitting it in the side at about a thousand miles an hour. Those are the extremes, and either way, it will be going down pretty quick if it's been free falling for one or two hundred miles before it hits significant air.

Van Rijn
2014-Nov-06, 11:50 PM
Oh, of course, if it doesn't match velocity with the atmosphere first, and if it happens to come down by a mountain, it could slam into the side at up about a thousand miles an hour.

grapes
2014-Nov-07, 12:25 AM
g is pretty close to 9.8m/s/s up to LEO, and it would experience a work of gM (force) time h (height), which would be converted to heat and energy once it impacts. So, without friction, a velocity of sqrt of (200km * 9.8m/s/s ), 1400m/s, four times the speed of sound.

Noclevername
2014-Nov-07, 02:10 AM
What would it take for a manned capsule to be able to land from this condition?

John Mendenhall
2014-Nov-07, 02:11 AM
Suppose that an object in LEO were to suddenly come to a halt relative to the surface of the Earth immediately under* it, and the energy of its movement dispersed harmlessly.

Would it fall straight down* to the surface? What would the atmosphere do to it?


*in this case, it means the part of the surface directly in line between the orbiting object, and the center of the Earth.

Guys, the OP very specifically states "stationary with respect to the surface of the Earth directly underneath it". I interpret that as a (momentarily) qeosynchronous satellite with way inadequate orbital velocity. It would then fall straight down with negligible velocity wrt to the surface or the atmosphere other than downward gravity induced vertical acccelerstion.

It's gonna burn up.

Jeff Root
2014-Nov-07, 02:34 AM
This is exactly the situation with rockets launched vertically to
altitudes of a few hundred miles, which is not uncommon.

-- Jeff, in Minneapolis

Noclevername
2014-Nov-07, 03:21 AM
Guys, the OP very specifically states "stationary with respect to the surface of the Earth directly underneath it". I interpret that as a (momentarily) qeosynchronous satellite with way inadequate orbital velocity. It would then fall straight down with negligible velocity wrt to the surface or the atmosphere other than downward gravity induced vertical acccelerstion.

It's gonna burn up.

Can you give me some numbers?

Senor Molinero
2014-Nov-07, 04:21 AM
It won't be going any where near fast enough to burn up. Things only do that when they enter the atmosphere at ORBITAL velocities. 1400m/s is nowhere near 11000m/s.

tony873004
2014-Nov-07, 05:29 AM
Gravity is a bit weaker up there. Where the ISS orbits, its about 8.6 m/s2, but Grape's numbers are still pretty good. Assuming the top of the atmosphere to be 100 km, and orbital altitude at about the ISS height of about 429 km (6800 km from center of Earth), it would hit the top of the atmosphere at sqrt(-2GM[(1/6800000^2)-(1/6471000^2)]) = 1.34 km/s. If air didn't slow it down, it would hit the ground at 1.5 km/s. I don't think this would cause a burnup.

John Mendenhall
2014-Nov-07, 07:47 AM
I think the no burn up guys are correct. As much as I harp on the problems of accelerating and decelerating to and from orbital speed, I neglected it here. I also like Jeff's V2 style explanation better than mine. Essentially, the satellite/rocket will fall straight down through the atmosphere to the ground with little or no lateral motion. So if you look up and see an ever growing object headed for you, "Duck! And cover!"

pzkpfw
2014-Nov-07, 08:23 AM
Geostationary does still mean "has orbital speed". So if the thing is still with respect to the Earths surface - at the distance it's initially at - is it still going to be still with respect to the surface as it falls? (Ignoring atmosphere for now).

(Much overuse of "still").

Jens
2014-Nov-07, 09:38 AM
The result might also depend on the height. Apparently, LEO can range between 160 and 2000 km, so it might make a difference.

Noclevername
2014-Nov-07, 12:15 PM
Geostationary does still mean "has orbital speed". So if the thing is still with respect to the Earths surface - at the distance it's initially at - is it still going to be still with respect to the surface as it falls? (Ignoring atmosphere for now).

(Much overuse of "still").

Geostationary orbit means that. The object in question is at rest with respect to the surface, but at a much lower height, giving it far too little energy to be in orbit.

Noclevername
2014-Nov-07, 12:15 PM
The result might also depend on the height. Apparently, LEO can range between 160 and 2000 km, so it might make a difference.

Assume ISS equivalent height.

Cougar
2014-Nov-07, 02:05 PM
Suppose that an object in LEO were to suddenly come to a halt relative to the surface of the Earth immediately under* it.... *in this case, it means the part of the surface directly in line between the orbiting object, and the center of the Earth.

"The surface directly in line between the orbiting object and the center of the Earth" defines a point. Of course, the object won't fall on that point, since the earth's rotation will carry that point elsewhere* as the object is falling.


Would it fall straight down to the surface? What would the atmosphere do to it?

Yes, the atmosphere question seems a bit tricky. Obviously the atmosphere at the surface is rotating with the earth (otherwise it'd be really windy :) ) I imagine that atmosphere rotation varies with altitude, but I don't really know about that. But certainly the density of the atmosphere varies with altitude, so this would change all the way down (and a little calculus would be required). But basically it depends on that particular object's terminal velocity, i.e., whether the object is a feather or a bowling ball.

ETA: Of course, the effect of frame-dragging would be negligible.

__________
* How far away varies directly with latitude (and time of fall).

Noclevername
2014-Nov-07, 04:49 PM
Yes, the atmosphere question seems a bit tricky. Obviously the atmosphere at the surface is rotating with the earth (otherwise it'd be really windy :) ) I imagine that atmosphere rotation varies with altitude, but I don't really know about that. But certainly the density of the atmosphere varies with altitude, so this would change all the way down (and a little calculus would be required). But basically it depends on that particular object's terminal velocity, i.e., whether the object is a feather or a bowling ball.


Let's say it's a manned space capsule,. A SpaceX's Dragon II will do. Fully loaded, of course.

Jeff Root
2014-Nov-07, 05:10 PM
The easiest way to define the initial conditions without any
ambiguity is to assume the object is dropped from a tall tower.
Unless the tower is at a rotational pole, the top of the tower
has a horizontal speed greater than the speed of the bottom.
So the dropped object will fall to the east of the tower. It is
a vertical Coriolis effect.

-- Jeff, in Minneapolis

Amber Robot
2014-Nov-07, 05:32 PM
Sounding rockets routinely fall back to Earth after reaching altitudes of around 250-300km. They can take quite a beating on the way back down, with deceleration around 10 g or more when "hitting" the atmosphere. Typically, the rockets are put into a bit of a tumble to distribute the heat, but I have seen some good burn damage on the outer skins and canards.

JohnD
2014-Nov-08, 10:55 AM
If the object 'came to a dead stop' then that would be relative to the CoG of the Earth. It would lose all its radial velocity while the Earth would continue to turn. To an observer on Earth, as it fell it would follow a curved path more and more directly downwards as it accelerated under gravity. A hyperbola? BUt it wouldn't fall on the observer who was directly under when it stopped.
To an observer on an object that did not stop, it would appear to fall away from them, at the original orbital velocity, and curve downwards, but I can't see if the observer in orbit would see as different curve.
To an observer on the object, they would appear to fall straight down, with no change in their micro-gravity, free-fall situation, but towards a rapidly turning Earth.

So I don't agree with those above who say that it would come straight down.

And, the Gravity probe B experiment confirmed relativistic frame dragging in Earth's gravity field as it's mass rotates. So there would be a small (very, very small?) tendency for the object to follow the Earth's rotation as it fell. Oops, Cougar said that. How much tendency?

John

grapes
2014-Nov-08, 12:25 PM
And, the Gravity probe B experiment confirmed relativistic frame dragging in Earth's gravity field as it's mass rotates. So there would be a small (very, very small?) tendency for the object to follow the Earth's rotation as it fell. Oops, Cougar said that. How much tendency?

Almost none. Not even the width of the object.

If it were that easy to observe, we wouldn't have needed Gravity Probe B (which took, by the way, over forty years to implement and execute, according to wiki it was first funded in 1964).

Githyanki
2014-Nov-11, 01:00 AM
A dead stop relative to what? The Earth? Or an absolute zero motion? That would depend on how fast the Earth is orbiting the sun: How fast the Sun is orbiting the galaxy and how fast the galaxy is moving: they might not even fall back to Earth but find them selves, flying away at a odd direction, despite the fact they are not moving.

From the moment they stop, gravity will take hold of them; whether or not the Sun's or Earth's gravity is able to capture them, due to their high relative-speeds, is another matter.

Jeff Root
2014-Nov-11, 01:52 AM
He said in the first sentence of the question, Githyanki.

-- Me, here

wd40
2014-Nov-12, 09:38 AM
The object does not fall to earth if it is in a geostationary orbit. It falls to earth if it is in a 24-hour
retrograde (west to east instead of the normal east to west) orbit. That's why 24-hour retrograde orbits can only exist from pole to pole. Any other orientation will fall to earth and not straight down; the path will be curved.

Noclevername
2014-Nov-12, 11:24 AM
The object does not fall to earth if it is in a geostationary orbit. It falls to earth if it is in a 24-hour
retrograde (west to east instead of the normal east to west) orbit. That's why 24-hour retrograde orbits can only exist from pole to pole. Any other orientation will fall to earth and not straight down; the path will be curved.

No, not geostationary nor 24 hour. The height of the ISS was specified.

Jeff Root
2014-Nov-12, 11:42 AM
An object moving at the speed of Earth's surface a couple of
hundred miles above the equator is moving at about 6% of
orbital speed for that altitude.

-- Jeff, in Minneapolis

JohnD
2014-Nov-12, 06:32 PM
The object does not fall to earth if it is in a geostationary orbit. It falls to earth if it is in a 24-hour
retrograde (west to east instead of the normal east to west) orbit. That's why 24-hour retrograde orbits can only exist from pole to pole. Any other orientation will fall to earth and not straight down; the path will be curved.

An extraordinary statement, wd40!
Please explain. We have already discussed the infinitesimal effect of frame dragging, so it can't be that bringing down a retrograde satellite.
And how is a pole-to-pole orbit "retrograde"?
John

Hornblower
2014-Nov-12, 07:22 PM
The object does not fall to earth if it is in a geostationary orbit. It falls to earth if it is in a 24-hour
retrograde (west to east instead of the normal east to west) orbit. That's why 24-hour retrograde orbits can only exist from pole to pole. Any other orientation will fall to earth and not straight down; the path will be curved.
You have it backward. West to east is prograde while east to west is retrograde. A satellite in a 24-hour retrograde orbit will not fall in anytime soon. It will stay up just like one in any other high orbit. In theory it eventually would spiral in from tidal interaction but for something as small as a manmade satellite that process would be exceedingly slow.

wd40
2014-Nov-16, 02:03 AM
Of course, the effect of frame-dragging would be negligible.



When the NOAA were asked in 1989 “Is the present movement of a Geostationary Satellite planned
and executed on the basis of a fixed earth or a rotating earth?” the answer returned by the department head of the Office of Satellite Operations was: “Fixed earth".

Is the Fixed Earth Coordinate System still used for launching, positioning and retrieving these satellites?

Hornblower
2014-Nov-16, 05:21 PM
When the NOAA were asked in 1989 “Is the present movement of a Geostationary Satellite planned
and executed on the basis of a fixed earth or a rotating earth?” the answer returned by the department head of the Office of Satellite Operations was: “Fixed earth".

Is the Fixed Earth Coordinate System still used for launching, positioning and retrieving these satellites?
We don't know what the department head thought "fixed" meant. If we model it as stationary center but rotating with respect to an inertial frame of reference, we can do a Newtonian calculation in that frame and get an orbit that works. If we try to do the calculations in a rotating frame in which Earth has no relative rotation, we will make it much harder for ourselves. I don't believe anyone in his right mind would have done it that way.

publiusr
2014-Nov-16, 09:16 PM
Now once I asked about an asteroid that was cut in two, and one end released from a bola at near zero velocity.

You could deposit quite a chunk with little heating at all.

I think one person said that if it were released high enough up, it would still come in like a meteor...

To go O/T for a moment.

Is there a way to use Earths fluids and atmosphere to slow an object in space without heating it directly?

One end of an asteroid bola detaches--the tether coils long enough to attach to a huge underwater structure. This slows the object in space to the point it falls down with little heating as well.

Do-able?

speach
2014-Nov-16, 09:52 PM
Why ask it can't happen.

Jens
2014-Nov-16, 10:57 PM
Why ask it can't happen.

If you're referring to the original question, "what if" questions don't have to be possible.

Jens
2014-Nov-16, 11:02 PM
Is there a way to use Earths fluids and atmosphere to slow an object in space without heating it directly?

One end of an asteroid bola detaches--the tether coils long enough to attach to a huge underwater structure. This slows the object in space to the point it falls down with little heating as well.

Do-able?

Wouldn't that basically like using a rubber band to capture an asteroid? I don't know how fast it would be going or how massive, but you'd have fantastic loads on your bola.

HobartStinson98765
2014-Nov-17, 04:38 AM
This is what would happen if you lifted a mass to orbital height on a space elevator and then detached it from the elevator. It would simply fall back to earth. That is why space elevators will not prove to be useful. If you wanted to orbit payload off of an elevator, you'd have to haul up the elevator an upper stage FILLED with highly explosive propellants that would accelerate your payload to orbital speed: about 17,500 mph. That would be a huge fuel tank that has to ride up the elevator with you.

Noclevername
2014-Nov-17, 04:03 PM
This is what would happen if you lifted a mass to orbital height on a space elevator and then detached it from the elevator. It would simply fall back to earth. That is why space elevators will not prove to be useful. If you wanted to orbit payload off of an elevator, you'd have to haul up the elevator an upper stage FILLED with highly explosive propellants that would accelerate your payload to orbital speed: about 17,500 mph. That would be a huge fuel tank that has to ride up the elevator with you.

No, a space elevator would extend both above and below its center of mass, which is in geosynch. The upper end would be pulled outward, away from Earth, and payloads sent from that end would be flung outward like a stone from a sling.

wd40
2014-Nov-18, 06:39 PM
We have already discussed the infinitesimal effect of frame dragging, so it can't be that bringing down a retrograde satellite.
And how is a pole-to-pole orbit "retrograde"?


Pole-to-pole can be prograde or retrograde. It does not bring down that satellite: the point is that you cannot have a satellite with a 48-hour orbit relative to the stars; it will fall to earth subject to the usual Coriolis and centrifugal forces acting on it.

NEOWatcher
2014-Nov-18, 06:45 PM
the point is that you cannot have a satellite with a 48-hour orbit relative to the stars;
That's the orbit at around 35000 miles.


it will fall to earth subject to the usual Coriolis and centrifugal forces acting on it.
Where are those forces coming from and how do they affect the orbit?

The only thing affecting an orbit of a satellite outside the atmosphere is gravity and other minuscule factors such as collisions with space debris.
Inclination has no effect except for an almost nil effect from the oblate sphere.

And if you really think a retrograde polar orbit can't be done, please explain the MetOp satellite (http://en.wikipedia.org/wiki/MetOp).