# Thread: Forward motion stops...then what?

1. ## Forward motion stops...then what?

I'm not sure if this is the right place for this question; if not, feel free to move to the appropriate forum.
I was watching the movie "The Astronaut Farmer" yesterday, a movie about a civilian astronaut who builds his own ship and flies an orbital mission around the earth, blah, blah. The movie took a lot of liberties with reality, but it did move me to ask my question.
In the film, as the astronaut is getting ready to make his reentry, he fires his rockets to brake his forward motion. When the capsule comes to a (relative?) stop, it starts to drop like a rock toward the earth's atmosphere. In truth, how would a capsule react if it were to stop all forward motion? I know that earth's gravity would bring it down, but I can't believe it would just drop like a stone from 200 miles up in space.
Thanks for any replies.

2. IIRC, gravity isn't all that much different at 200 km away from the surface of the earth. So the rocket would drop at about the same rate that it would drop on earth. The thing is, it wouldn't look like it was dropping that fast, because there would no nearby object to compare it to. On the surface, you see a rock dropping compared to a stationary tree, so it looks fast. If the rocket is in orbit, there aren't clouds or anything nearby. But hypothetically, if you were standing on a very tall tower, and a rocket came by and suddenly decelerated to zero, you would see it drop basically like a rock.

3. Just to add something, the radius of the earth is about 4,000 miles, so at the surface we are 4,000 miles from the center of gravity, whereas the astronaut would be 4,200 miles. So it shouldn't make a huge difference.

4. It would be a lot like the downward half of Scaled Composites' SpaceShipOne -- which went into space, barely, but didn't go into orbit. It went up. It came down.

If you don't remember it, see the video there. There's no third-person view though of the beginning of the fall.

(It would take an enormous amount of fuel to dead-stop an orbiting craft. So instead they just tweak the orbit so that the craft begins an elliptical orbit that passes through earth's atmosphere which takes the energy away.)

5. Originally Posted by 01101001
(It would take an enormous amount of fuel to dead-stop an orbiting craft.
Sure would. And it wouldn't be pleasant for the occupants either. Human pancakes.

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In reality, you only have to reduce the velocity by about 100 meters per second to deorbit.

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Originally Posted by Fendercaster
In the film, as the astronaut is getting ready to make his reentry, he fires his rockets to brake his forward motion. When the capsule comes to a (relative?) stop, it starts to drop like a rock toward the earth's atmosphere. In truth, how would a capsule react if it were to stop all forward motion? I know that earth's gravity would bring it down, but I can't believe it would just drop like a stone from 200 miles up in space.
.

The idea of retro rockets is to reduce the spacecraft velocity by a small amount (few hundreds of feet per second). The reduction of velocity changes the perigee of the orbit to where it intersects the atmosphere. The spacecraft then uses the atmosphere to reduce the remaining velocity.

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## Point of view movie license.

It might help if you look at this in terms of "point of view".

Assume your point of view, the camera, is moving with the capsule. When the capsule fires its retros if the point of view does not slow with the capsule the capsule will appear to fly away from the point of view until its relative speed is a couple of hundred feet per second. The point of view is still moving at 17,500 mph relative to the surface of the earth and the capsule will be moving just a few hundred feet per second slower. The capsule will be so far away in so short a time we can stop this part of the description here.

If however the point of view slows with the retrofiring capsule but does not change its altitude to follow the capsule as another spacecraft would (movie license) then the capsule would slowly drift away from the point of view. It wouldn't drop like a stone. However a combination of motion relative to point of view due to retro away from orbital speed and slow drift away (by movie license) and a particular atitude of the capsule and possible pitch, roll, yaw angle of the point of view could make the capsule appear to drop like a stone away from the point of view.

Off topic

I have made some nifty spacecraft flyby and rendezvous videos in Celestia using the concept of point of view.

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Originally Posted by Fendercaster
In the film, as the astronaut is getting ready to make his reentry, he fires his rockets to brake his forward motion. When the capsule comes to a (relative?) stop, it starts to drop like a rock toward the earth's atmosphere. In truth, how would a capsule react if it were to stop all forward motion?
It would drop like a rock.

I know that earth's gravity would bring it down, but I can't believe it would just drop like a stone from 200 miles up in space.
Sure it would, and it would burn up in the process.

That's why reentry occurs at a shallow angle, so that energy is dissipated over a longer time, rather than all at once.

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That was a good description, aastrotech.

Originally Posted by mugaliens
Sure it would, and it would burn up in the process.
No, the heating would be much less than a normal entry.

In a normal entry the capsule starts out moving at about 7.8 km/s,
and picks up some speed as it falls, before that speed is lost again
due to atmospheric drag (or the equivalent). In this straight-down
entry it starts out moving at zero km/s and picks up a similar amount
of speed as it falls, before reaching terminal speed, which decreases
as the vehicle gets into increasingly denser atmosphere.

Originally Posted by mugaliens
That's why reentry occurs at a shallow angle, so that energy is
dissipated over a longer time, rather than all at once.
Entry occurs at a shallow angle mainly because that is the only
available option. You are right of course that the energy needs to
be dissipated over a long time, but that is because there is so much
energy to dissipate because of the high speed.

-- Jeff, in Minneapolis

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I haven't seen the movie, but I strongly suspect that aastrotech's
description correctly explains what you see. The movie could very well
be perfectly accurate in how it portrays the re-entry. The spacecraft
does not come to a stop at all. It just slows down by maybe 300 mph,
while the camera continues in orbit at the original speed of 17,450 mph.
The relative speed of 300 mph looks really impressive, but it is less than
two percent of the orbital speed. After the retrorockets fire, the capsule
is still moving at more than 98% of its orbital speed. But the difference
is enough for the spacecraft to begin falling like a rock, relative to the
orbiting camera.

-- Jeff, in Minneapolis

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Originally Posted by Jeff Root
That was a good description, aastrotech.
Thanks

Originally Posted by Jeff Root
Entry occurs at a shallow angle mainly because that is the only
available option. You are right of course that the energy needs to
be dissipated over a long time, but that is because there is so much
energy to dissipate because of the high speed.

-- Jeff, in Minneapolis
That's right. To bring it to a full stop prior to entering the apmosphere it would need a retro rocket only a little less powerfull than the one that put it into orbit to begin with.

13. Originally Posted by Jeff Root
No, the heating would be much less than a normal entry.
That may be true, but it is still significant. I've seen the burn marks on sounding rockets after dropping from a couple of hundred miles up. They purposefully put the rockets into a tumble during reentry to help distribute the heat. Remember that both speed and atmospheric density (i.e., dynamic pressure) increase at the same time. Some of the roughest g-forces are felt on a rocket during re-entry.

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Originally Posted by Jeff Root
But the difference
is enough for the spacecraft to begin falling like a rock, relative to the
orbiting camera.

-- Jeff, in Minneapolis
If the retros give one G acceleration it it would be accelerating away from the camera just as much as a falling rock. If the retros produced more then it woud "fall away" faster than a falling rock.

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Originally Posted by Amber Robot
That may be true, but it is still significant. I've seen the burn marks on sounding rockets after dropping from a couple of hundred miles up. They purposefully put the rockets into a tumble during reentry to help distribute the heat. Remember that both speed and atmospheric density (i.e., dynamic pressure) increase at the same time. Some of the roughest g-forces are felt on a rocket during re-entry.
I think those marks could have occured during ascent. I've seen large hobby rockets that only went up a couple thousand feet with significant friction marks.

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The spacecraft *would* drop like a rock, no doubt at all, *if* all forward velocity were suddenly/magically to disappear. An orbiting spacecraft is falling, but its also going forward fast enough to not fall onto the 'center-of-gravity', so to speak. A stable orbit is when horizontal is equal to the vertical; lose the horizontal and then you go vertical only (down!). Wrong words, but hope the concept is seen there, Fendercaster Very much like how a 'fired bullet' falls just as fast a one dropped from hand...physics/gravity is the same on both projectiles.

http://en.wikipedia.org/wiki/Joseph_Kittinger

An excerpt: "On August 16, 1960, he made the final jump from the Excelsior III at 102,800 feet (31,300 m). Towing a small drogue chute for initial stabilization, he fell for four minutes and 36 seconds, reaching a maximum speed of 614 mph [1][2] (988 km/h or 274 m/s) before opening his parachute at 18,000 feet (5,500 m)."

This fellow did just fine, all things considered, from edge of 'space' (except for hand-injury from ripped glove), though not quite at altitude being spoken of. Deceleration is gradual as atmosphere 'thickens'. With *just* vertical/downward motion from a total standstill, you only accelerate 'a bit' compared to orbital speeds and reentry. Going high-Mach speeds is quite different than 'just dropping'

Amber, those 'roughest g-forces' are from going orbital speed into atmosphere, not mere 'couple hundred miles/hour' (or so, relatively speaking). At landing, a spacecraft has to rid itself of ~exactly the same amount of energy used for the launch (within a tiny-% anyways). Something to think of there A 'sounding rocket' probably has a large MaxQ (max pressure of ascent, per se, which is relative to speed/atmospheric-density) which might 'burn it', or maybe its the exhaust/flame marks you see? Not saying I *know* what the sounding's marks are, but most of those things are going a *lot* faster than anything than is simply 'dropped' from 'vacuum' into a slowly thickening atmosphere (from a limited looking up details of 'sounding rockets' just now, fwiw)

HTH,
Alex

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An excerpt: "On August 16, 1960, he made the final jump from the Excelsior III at 102,800 feet (31,300 m). Towing a small drogue chute for initial stabilization, he fell for four minutes and 36 seconds, reaching a maximum speed of 614 mph [1][2] (988 km/h or 274 m/s) before opening his parachute at 18,000 feet (5,500 m)."

Just to clarify, Kittenger hit max speed within 30 seconds of jumping and then began to decelarate gradually. By the time his chute opened at 18,000 feet, he probably wasn't going more than 200 MPH and perhaps considerably less due to the stabilization chute.

It seems possible that some of the scorch marks on sounding rockets comes from the ascent through the atmosphere. It also seems likely that sounding rockets don't have much in the way of heat shielding for the trip back down.

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Originally Posted by Jeff Root
That was a good description, aastrotech.

No, the heating would be much less than a normal entry.

The reason we don't do this is because it would require way too much propellant to arrest all forward motion. It's a lot more efficient to use atmospheric braking during reentry to diminish the orbiting vessel's kinetric and potential energies.

19. Does a sounding rocket come back intact? Being more aerodynamic, it should fall faster than a human from a dead drop at the same height due to air resistance. However, scortch marks should be evidence of hypersonic velocity, which is probably only attained during a powered ascent. Maybe someone could lauch a sounding rocket with a camera pointing out a forward facing window to see if the window gets scortched during ascent or descent.

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Here's a link to NASA's Sounding Rocket Program Handbook (warning, it's a 220 page PDF file). It might have some of the answers you seek. Off hand, it seems likely to me that the payload separates from the rocket in most circumstances.

21. Yes, sounding rockets come back intact -- though as I said they tumble them to distribute the heat. The tough ride is on the way down, not the way up. This is mostly due to the fact that on the way up, speed is increasing while air pressure is decreasing. On the way down they are both increasing. Think of a Shuttle launch and Shuttle reentry. They don't need the heat tiles for the way up.

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Originally Posted by Amber Robot
Think of a Shuttle launch and Shuttle reentry. They don't need the heat tiles for the way up.
Not true. The foam on the tank is also there as an ablative thermal protection system.

All launch vehicles experience heating on the way to orbit. Delta II uses cork in various places. It is just painted blue.

There is an onboard video of a Pegasus flight and you can see the wings char.

23. Originally Posted by Amber Robot
Yes, sounding rockets come back intact -- though as I said they tumble them to distribute the heat. The tough ride is on the way down, not the way up. This is mostly due to the fact that on the way up, speed is increasing while air pressure is decreasing. On the way down they are both increasing. Think of a Shuttle launch and Shuttle reentry. They don't need the heat tiles for the way up.
I'm not quite convinced of this. What if the tiles were presented to the airstream, instead of aerodynamically? Or conversely, how much would the shuttle heat up if it plummeted to the earth without slowing down? I mean, before impact.

I thought returning rockets tumbled so that they weren't aerodynamic--so that they would "drag" and not auger in.

24. I'm not saying that heating doesn't happen on the way up, I'm just saying that I've seen the telemetry from sounding rockets and they have a rougher ride coming down.

25. Originally Posted by Amber Robot
Yes, sounding rockets come back intact -- though as I said they tumble them to distribute the heat. The tough ride is on the way down, not the way up. This is mostly due to the fact that on the way up, speed is increasing while air pressure is decreasing. On the way down they are both increasing. Think of a Shuttle launch and Shuttle reentry. They don't need the heat tiles for the way up.
A tumbling re-entry should make it slow down faster by presenting a larger cross-section to the airstream.

A shuttle doesn't need it on the way up because most of its velocity is attained once it gets up. I would expect sounding rockets to achieve higher speeds at lower altitudes during ascent than the STS during ascent.

26. Originally Posted by Ara Pacis
I would expect sounding rockets to achieve higher speeds at lower altitudes during ascent than the STS during ascent.
That is true. A typical scientific sounding rocket is back on the ground in 15-20 minutes after launch. It takes off like a bat out of hell. A sounding rocket has a boost-phase guidance section that keeps it on trajectory during the ascent. Part of this section are small fins that constantly adjust. Once above a certain altitude they disengage and upon reentry they can take a lot of burn damage, some even breaking under the stress. Also, I've seen the metal on part of a rocket burn completely through when the lack of enough remaining gas in its thrusters could put it into a proper tumble and it mostly returned on one side. I would be surprised if this burn damage occurred on the flight up.

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Drop like a rock from 480,000 feet altitude is not desirable. S= 1/2 at squared. If average a is 12 feet per sec per second, t = 200 seconds: S = 12 times 40,000 = 480,000 feet. You hit the surface of Mercury at very high speed = splat. At Earth not quite as fast, but extremely hot due to air friction. Neil

28. Just in case some numbers for ascent heating would help, the outer skin of the Delta II payload fairing can reach 200–250°C during launch (depending on the size of the fairing). I would be surprised if the reentry heating was this gentle.

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And in case numbers for acceleration due to Earth's gravity at various
altitudes would help:
Code:
```Altitude       g
(m)      (m/s^2)

0     9.8066
10,000     9.7759
20,000     9.7452
30,000     9.7147
40,000     9.6844
50,000     9.6542
60,000     9.6241
70,000     9.5942
80,000     9.5644
90,000     9.5348
100,000     9.5052
110,000     9.4759
120,000     9.4466
130,000     9.4175
140,000     9.3886
150,000     9.3597
160,000     9.3310
170,000     9.3024
180,000     9.2740
190,000     9.2457
200,000     9.2175

300,000     8.9427
400,000     8.6799
500,000     8.4286
600,000     8.1880```
From the CRC Handbook of Chemistry and Physics

-- Jeff, in Minneapolis

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g for Mercury would be 5 to 6 for a similar table, which is why I picked 12 instead of 32.2
19 would probably be closer. Acceleration g or a for Earth is about 31 at 480,000 feet, but a drops to zero at perhaps 20,000 feet, and is negative the rest of the way to the surface, due to air resistance, so 12 may be close to average. The zero a altitude depends somewhat on starting altitude, shape, mass, density and amount of ablation, I think. Neil

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