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Author: Subject: Throwing objects down from satellites - would it work? Why not?
annaandherdad
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[*] posted on 22-5-2015 at 22:21


Quote: Originally posted by smaerd  
annaandherdad - don't objects in low earth orbit still get pulled into the earths atmosphere over time unless they are maintained via thrust?


Yes, but the ISS is well above the atmosphere at 400 km above the earth's surface, and the perturbed orbit is only 1km lower at perigee so it won't make any difference for many, many orbits.




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annaandherdad
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[*] posted on 22-5-2015 at 22:28


Quote: Originally posted by blogfast25  
[
Can you post these formulas?

I agree that angular momentum doesn't change. But why does the radial component of velocity 'disappear'?

I would have thought the object would 'spiral' down, hitting earth at some point.


Yes, I'll post the formulas, but give me time. I just arrived in Europe and am recovering from jet lag. The object doesn't spriral down because orbits in gravitational fields are ellipses. Air resistance would cause the orbit to decay, but it's very small at 400km altitude and wouldn't have an effect for many, many orbits. And the perturbed orbit is an ellipse with perigee and apogee approximately as I quoted, + or - about 1km from the ISS's orbit, so air resistance is negligible for both the ISS orbit and the orbit of the object thrown. I'll post the calculations.




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turd
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[*] posted on 22-5-2015 at 23:34


Quote: Originally posted by blogfast25  
Quote: Originally posted by ahill  
Cheddite Cheese is of course correct - I look at it this way - there is no free lunch in physics - if it took a _lot_ of energy to acheive an orbit, it will take lot lot of energy to de-orbit it.



You may well be right about everything else but what you write there is nonsense.

Actually it makes perfect sense. Suppose you are in free space, no gravity, no friction. You need a certain amount of energy to accelerate to say 0.5 c. You will need the same amount of energy to go back to 0 c (we suppose that you didn't lose any mass by burning fuel).

The same is true in orbit, with the only difference that you are moving along an ellipse (disregarding relativity). By thrusting maneuvers you can change height, eccentricity and inclination. If eccentricity becomes too large, you escape orbit. If you want to enter an orbit, you need the same energy needed to leave it. See: https://en.wikipedia.org/wiki/Delta-v_budget and references therein.
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smaerd
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[*] posted on 23-5-2015 at 05:14


Thanks annaandherdad & turd & chedditecheese I learned something :).



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annaandherdad
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[*] posted on 23-5-2015 at 05:22


Here is an explanation of my conclusions, that the object thrown will never get closer to the earth than about 1km than the ISS itself. (Attached pdf.) It will also rise above the ISS by about the same amount in its subsequent orbit. The more exact figure is something like 800 meters.

Attachment: orbit.pdf (29kB)
This file has been downloaded 199 times





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blogfast25
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[*] posted on 23-5-2015 at 05:47


Quote: Originally posted by annaandherdad  
Here is an explanation of my conclusions, that the object thrown will never get closer to the earth than about 1km than the ISS itself. (Attached pdf.) It will also rise above the ISS by about the same amount in its subsequent orbit. The more exact figure is something like 800 meters.


Thanks AAHD (and turd), will study this. Seems correct at very first glance though. Live and learn! Sapere Aude! :)




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jock88
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[*] posted on 23-5-2015 at 11:12


Quote:


"The same is true in orbit, with the only difference that you are moving along an ellipse (disregarding relativity). By thrusting maneuvers you can change height, eccentricity and inclination. If eccentricity becomes too large, you escape orbit. If you want to enter an orbit, you need the same energy needed to leave it. See: https://en.wikipedia.org/wiki/Delta-v_budget and references therein."


I am probably missing something but if this is true then you would need great big tanks of liquid H2 and + O2 and two great big booster rockets (just like you have on the ground) to get the space shuttle back from the space station to earth?
(I will admit you will have your pay load offloaded but we will pretend that the payload is still on board just for this argument)



The explanation of the thrown object going into in elliptical orbit (from a circular) is fascinating but when it get nearer the earth it has more gravitational pull and when it gets further away it has less therefor the net pull is going to get more and more in favour of the object going to ground. (trust me, I am no rocket scientist).
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turd
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[*] posted on 23-5-2015 at 11:42


You are missing two things:
- At low altitude the atmosphere is doing the braking for you. This is for example used in aerobraking.
[Edit: And vice-versa: you have to overcome the friction of the atmosphere on takeoff.]
- Rockets are not constant mass systems, but expel fuel, eject unused stages, etc. See rocket equations with their exponential terms.

Quote:
The explanation of the thrown object going into in elliptical orbit (from a circular) is fascinating but when it get nearer the earth it has more gravitational pull and when it gets further away it has less therefor the net pull is going to get more and more in favour of the object going to ground.

This makes no sense. The elliptic orbit in the two body system (one fixed) is periodic and stable. For derivation, see here: https://en.wikipedia.org/wiki/Kepler_orbit

[Edited on 23-5-2015 by turd]
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jock88
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[*] posted on 23-5-2015 at 15:33



Are you saying that it takes the same amount of fuel to return the space shuttle back to earth (come out of its orbit with the iss) as it did to put it there in the first place?
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[*] posted on 23-5-2015 at 15:45


If there weren't any atmosphere, that would be true. (The XKCD What If? I posted earlier in this thread does a good job explaining why.)

Almost all of the deceleration is provided by the atmosphere.

[Edited on 24-5-2015 by Cheddite Cheese]




As below, so above.
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blogfast25
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[*] posted on 24-5-2015 at 06:15


Yep, AAHD is completely right. Lovely derivation.

It means also that to 'de-orbit' an object, you can't really just 'throw' it down but it would have to be thrusted down, at least until it hits the high atmosphere where air breaking takes over and kinetic energy is converted to heat.

But I'm still in a fog about something. Take the moon landings (hoaxes all of them, of course! ;-) ) for instance. The lander and command module first orbit the moon, then the lander separates and starts its descent, presumably by means of thrusters. But they did have to use thrusters also to brake the thing so it didn't actually 'break', didn't they?

Where's that 'cockeyed' emoticon when I need it?


[Edited on 24-5-2015 by blogfast25]




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jock88
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[*] posted on 24-5-2015 at 09:58



Are you saying that it takes the same amount of fuel to return the space shuttle back to earth (come out of its orbit with the iss) as it did to put it there in the first place?

My own question above is actually a bit dumb after thinking about it. The space shuttle does not go from one orbit to another orbit. It goes from orbit to actually landing and standing still on the surface. If you were to go from the iss orbit to an orbit 100 yards above the earths surface then you would have to increase the speed to a very very hign speed to stay in orbit at 100 yards.
(IGNORING ALL AIR RESISTANCE OF COURSE).

If we were to keep examples for orbits, landings, ups and downs to the moon it would make things clearer to newbyes liki myself con 'air resistance' is constantly thrown in and this makes things very complicated and unclear as to the nature of gravity/orbits/speeds/ellipsed/circles/up/down etc etc.
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[*] posted on 24-5-2015 at 10:28


Quote: Originally posted by jock88  

Are you saying that it takes the same amount of fuel to return the space shuttle back to earth (come out of its orbit with the iss) as it did to put it there in the first place?



I believe that the energy you put in to get in in orbit, plus the energy to get it back down, are all conserved, minus the energy that's converted to heat (drag resistance).




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SupFanat
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[*] posted on 24-5-2015 at 12:28


[q]where air breaking takes over and kinetic energy is converted to heat.[/q]
Too bad heat and not something better like electromagnetic radiation. :(
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blogfast25
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[*] posted on 24-5-2015 at 13:07


Quote: Originally posted by SupFanat  
[q]where air breaking takes over and kinetic energy is converted to heat.[/q]
Too bad heat and not something better like electromagnetic radiation. :(


Errrrmmm... much that heat IS electromagnetic radiation.




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jock88
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[*] posted on 24-5-2015 at 14:06



So if I were to reask the original question in moon world.
Say we have a station orbiting the moon at 10 km above the surface and going at a speed (I will say speed and not velocity as I mean in a circle around the moon) of 1000km per second. (these figures may not make actual math sense but it does not really matter as we are not looking for a precise mathematical answer, but rather an english explanation of what will happen).
There is no air resistance.
Assum the moon is not rotating.
An small object is thrown straight down towards the surcace at (say) 1 m per second.
Will it hit or reach the surface?
Perhaps it must be thrown at an certain angle towards the surface and also in the direction away from the direction of travel of the station at a certain minimum speed?
Thanks for your time as I may be pissing somebody(s) off with what are probably questions that are too simplistic to have a proper answer.
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[*] posted on 24-5-2015 at 14:12


Whether or not the body as a gravitational force exerted on it, it will move if pushed.

Yes, the small object may well hit the surface.

At 1000km/s it might miss, as 'down' might pass by pretty fast (at that speed you could easily miss).

A better example would be a relative-to-moon stationary object.

Another, much smaller object propelled from there towards the moon would hit it.




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blogfast25
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[*] posted on 24-5-2015 at 14:14


Quote: Originally posted by jock88  

So if I were to reask the original question in moon world.
Say we have a station orbiting the moon at 10 km above the surface and going at a speed (I will say speed and not velocity as I mean in a circle around the moon) of 1000km per second. (these figures may not make actual math sense but it does not really matter as we are not looking for a precise mathematical answer, but rather an english explanation of what will happen).
There is no air resistance.
Assum the moon is not rotating.
An small object is thrown straight down towards the surcace at (say) 1 m per second.
Will it hit or reach the surface?
Perhaps it must be thrown at an certain angle towards the surface and also in the direction away from the direction of travel of the station at a certain minimum speed?
Thanks for your time as I may be pissing somebody(s) off with what are probably questions that are too simplistic to have a proper answer.


No, it would end up in a slightly lower orbit, as evidenced by AAHD's derivation (higher up).

Total kinetic energy of the object would slightly increase, resulting in slightly higher orbital speed (lower orbits have lower orbital periods). The new orbit is calculated from energy (kinetic + potential) and rotational impulse conservation equations.

[Edited on 24-5-2015 by blogfast25]




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[*] posted on 24-5-2015 at 14:19


No, it will orbit closer to the moon.
The best is to throw it not towards the moon, but actually out of the back of the spacecraft in the opposite direction it is moving. Also, you need to throw it so hard that its speed is reduced to below escape velocity. 1 m/s is not enough.

Perhaps it helps to look at this way: the station is already falling towards the moon, but is at the same time moving forward so fast, that it misses the surface. Your rock needs to to slow down enough for it to hit the moon.

Imagine standing on the surface of the moon, and throwing rocks.
The first rock you throw by hand. It simply does what you are familiar with: you throw it, it arcs towards the surface and hits the ground somewhere in front of you.
Now you throw hard. Really, really hard. The rock flies away, arcs towards the surface, but imagine you throw it so hard, it never actually hits the surface, instead it perpetually arcs towards the surface, but its trajectory never intersects the surface, instead it follows the curvature of the moon. You have thrown it into orbit.

If the above confuses you, ignore it.



[Edited on 24-5-2015 by phlogiston]




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[*] posted on 24-5-2015 at 14:25


Phlogiston:

Probably not exactly escape velocity (needed to fully escape gravitational pull altogether). I think throwing it with the same speed as the orbital speed of the spacecraft should do it, in the opposite direction of the spaceship. Non-relativistic speeds add up so speed of object = 0 m/s.

[Edited on 24-5-2015 by blogfast25]




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[*] posted on 24-5-2015 at 14:51


Yes, you are right. It should have been the orbital speed at the surface, not escape velocity. They are related by a factor of sqrt(2). Escape velocity from the surface of the moon is 2.4 km/s, so the maximal orbital speed is 2.4/sqrt(2) = 1.7 km/s. So, anything below that speed and the rock will impact the surface.

So, jock88, in the case of your example, a spaceship in orbit around the moon traveling at 1000 km/s, you would need to slow the rock down to 1.7 km/s by throwing it out of the back of the spaceship with a velocity of 998.3 km/s (relative to the spaceship) for it to hit the moon instead of going round in a lower orbit.

[Edited on 24-5-2015 by phlogiston]




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[*] posted on 24-5-2015 at 14:55


Quote: Originally posted by phlogiston  

So, jock88, in the case of your example, a spaceship in orbit around the moon traveling at 1000 km/s, you would need to slow the rock down to 1.7 km/s by throwing it out of the back of the spaceship with a velocity of 998.3 km/s (relative to the spaceship) for it to hit the moon instead of going round in a lower orbit.



He did make up these numbers. Orbital period and orbital radius are strictly linked by Kepler's Law and the Moon's Mass.

Regarding 'escape velocity': How long does it take to drive to space at 70 mph?


[Edited on 24-5-2015 by blogfast25]




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[*] posted on 28-5-2015 at 02:37


About re-entry... Adiabatic compression can produce high temperatures.
But what about adiabatic expansion which gives low temperatures? How low are the reachable temperatures?
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[*] posted on 28-5-2015 at 03:32


I agree with annaandherdad and ahill with regards to throwing an object from ISS, except that in my opinion 'air' shall we call it is a factor. As we know air pressure decreases with altitude (not sure if it's linear) and it will get to a point as we go higher where air pressure is unmeasurable but this does not mean there is no 'air' present. From there on up it would be more appropriate to term it "moles of gases per volume" and I believe even at ISS altitude that molar value may be very small but not zero.
As long as that molar value is not zero there will be matter interfering with an object's orbiting precession albeit very small, and in doing so energy is transferred from the object's potential energy to the incident matter. This would mean that it's 'horizontal' speed will retard and it's vertical speed will increase and re-commence the process that ahill described in an attempt to find balance.
As the object loses energy to gas molecules it will progressively retard and re-attain lower and lower orbits at an increasing rate until the rate of energy loss to incident matter exceeds the rate of energy gain from acceleration while entering the acute stage of it's elliptical orbit. This is where it would give up and fall to the earth.

Lower orbit objects such as the ISS do so at much higher risk of decay than those at higher orbit such as GPS satellites and the moon.




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SupFanat
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[*] posted on 28-5-2015 at 04:18


Geostationary satellites would have the same speed as the atmosphere at such altitude. Ideally it would orbit the Earth as if it were bound to Earth surface but in the practice the orbit isn't as precious.
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