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Author: Subject: Throwing objects down from satellites - would it work? Why not?
SupFanat
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[*] posted on 22-5-2015 at 08:56
Throwing objects down from satellites - would it work? Why not?


Imagine, an object is thrown straight down from ISS with speed 1 m/s. Initial height 400 km. Would the object continue falling with the speed 1 m/s? If not, why not?
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blogfast25
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[*] posted on 22-5-2015 at 09:18


Quote: Originally posted by SupFanat  
Imagine, an object is thrown straight down from ISS with speed 1 m/s. Initial height 400 km. Would the object continue falling with the speed 1 m/s? If not, why not?


ISIS? They're in space now? Who knew? ;)

No, your object will keep on accelerating according to F = m a

with F the gravitational force exerted by the Earth on the object, m the mass of the object and a the actual acceleration (m/s<sup>2</sup>;)

and F = G (m M)/R<sup>2</sup>, G is Universal Gravitation Constant, M mass of Earth, R distance of object to centre of Earth.

This can all be reworked, then gives you a simple DE to work out speed at impact (but ignoring air drag which will be very significant once the object enters the Earth's atmosphere).


[Edited on 22-5-2015 by blogfast25]




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smaerd
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[*] posted on 22-5-2015 at 09:29


Yea when you're in orbit you're still well influenced by the force of gravity. Astronauts may be "weightless" in orbit but they are in orbit IE bending around the curve of the Earth's mass.

So yea as blogfast said, the object would continue its path of travel and it's velocity would increase until it reached a 'terminal velocity' it should cease. However, depending on the objects composition/geometry and the composition of the atmosphere, it may experience enough "drag" that it burns/breaks apart on re-entry making a 'dynamic' terminal velocity I guess. I dunno I don't physics much anymore.




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blogfast25
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[*] posted on 22-5-2015 at 09:50


Quote: Originally posted by smaerd  
Yea when you're in orbit you're still well influenced by the force of gravity. Astronauts may be "weightless" in orbit but they are in orbit IE bending around the curve of the Earth's mass.

So yea as blogfast said, the object would continue its path of travel and it's velocity would increase until it reached a 'terminal velocity' it should cease. However, depending on the objects composition/geometry and the composition of the atmosphere, it may experience enough "drag" that it burns/breaks apart on re-entry making a 'dynamic' terminal velocity I guess. I dunno I don't physics much anymore.


Drag force follows F<sub>drag</sub> = k v<sup>n</sup> with k a constant (depending on object shape) and n an exponent that itself depends on v, the velocity, IIRW.

It's not guaranteed the object will reach terminal velocity though: see also parachuting from too low an altitude (aka the 'splot effect' ;))

Such weapons have already been contemplated, BTW. I kid you not.

[Edited on 22-5-2015 by blogfast25]




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neptunium
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[*] posted on 22-5-2015 at 10:23


by throwing something from a space station i am assuming by hand? something smaller than the space station itself right?
since the ISS is already going at orbital speed when you throw that thing from there IT will also go at orbital speed.
Its vector is however pointed towards earth.
if the object is much smaller it might reach another orbital altitude and remain there for a time before the residual air slow it down over a time dificult to calculate without details on the object geometry size etc.. and eventually burn up on re entry.
if an astronauts let a tool escape in space (although they must be tattered or attached)
the tool generally remain in orbit and become another space junk ..




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smaerd
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[*] posted on 22-5-2015 at 10:31


You're definitely right I forgot about the case where an object may not reach terminal velocity.

This sounds like the worst idea for a weapon ever lol. I can't imagine anything worth dropping given the amount of fuel it takes to get things out that far. Then again the desire and monetary force behind hurting people has and likely will always be in high demand. I can't think of a single physical exploit is not without the ability or drive to inflict harm. I can imagine the conversation at some seedy gov't dept.
"Hey we can build sattelites now", "How can we kill people with them?", "Can we drop bombs from them?", "Let's call NASA."," Sir we cut all funding to NASA years ago", "Right, they weren't killing enough people."

Wait, maybe this is a trick question. Maybe the OP wanted us to consider that there is no philosophical "down" in low gravity locations. So if the object was dropped "down" as in away from the earth would it drift at the same speed? Well an object in orbit has a velocity of the escape velocity (I think?). So chucking an object toward the normal of the orbit path or centrifugal force direction vector should be the X &/or Y component of the velocity of the orbit plus the velocity of the object (1m/s)? nevermind the op said 400km seems way too low to avoid gravity, it'd probably travel outward for a little bit and then get sucked back to earth over time.




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[*] posted on 22-5-2015 at 10:39


If you're in orbit, which way is 'down' ?



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turd
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[*] posted on 22-5-2015 at 11:08


Down is where the acceleration vector (second derivative of the position) points.

Edit: This is probably the Wikipedia page that OP should try to understand: https://en.wikipedia.org/wiki/Delta-v

[Edited on 22-5-2015 by turd]
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aga
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[*] posted on 22-5-2015 at 11:14


Ah. Good. Mars then.



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[*] posted on 22-5-2015 at 11:41


Rods from the gods.

"The system most often described is "an orbiting tungsten telephone pole with small fins and a computer in the back for guidance". The system described in the 2003 United States Air Force report was that of 20-foot-long (6.1 m), 1-foot-diameter (0.30 m) tungsten rods, that are satellite controlled, and have global strike capability, with impact speeds of Mach 10."

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




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[*] posted on 22-5-2015 at 12:17


If you want to deorbit something, you don't throw it down. You throw it retrograde (opposite to your orbital velocity). 1 m/s wouldn't be enough to deorbit something from the ISS.



As below, so above.
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blogfast25
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[*] posted on 22-5-2015 at 13:42


Quote: Originally posted by Cheddite Cheese  
If you want to deorbit something, you don't throw it down. You throw it retrograde (opposite to your orbital velocity). 1 m/s wouldn't be enough to deorbit something from the ISS.


The ISS already has a velocity component pointing toward the Earth (it's in effect 'falling' towards Earth, in an orbit). There should be no problem releasing an object from it at 1 m/s.




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Metacelsus
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[*] posted on 22-5-2015 at 13:58


You might have already seen this, but it does a good job explaining stuff.
https://what-if.xkcd.com/58/

The ISS has an acceleration pointed towards (more or less) the center of the Earth, but its velocity is almost all horizontal. The ISS has apogee of 416 km and perigee of 409 km (on average), so its orbital eccentricity is quite low.




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SupFanat
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[*] posted on 22-5-2015 at 14:21


The question is whether such small speed as just 1 m/s towards Earth is enough for some physical body to leave the orbit.

Another question - could some object be thrown from geostationary orbit (or similar high orbit) down to Earth with small speed (it's about 36 million meters above Earth so it would take about million seconds or 11.6 days with speed 36 m/s).
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jock88
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[*] posted on 22-5-2015 at 14:25



I think 'down' in the case of the original question is towards the surface of the earth (at the time of throwing).

Assuming no air resistance, an object thrown towards the surface of the earth (or towards its center) at a velocity of 1m/sec will initially have an velocity of about 17000 miles per hour in the direction of a circle going around the earth (orbital path and speed of shuttle) and a velocity of 1m per second in the direction straight 'down'. The 17000 miles per second direction will stay the same and the downward vector (or speed if ya like) will get larger and larger. Assuming no burn up/no air resistance it will hit the earth at an angle. (so there you have it).
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blogfast25
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[*] posted on 22-5-2015 at 14:28


Quote: Originally posted by Cheddite Cheese  
You might have already seen this, but it does a good job explaining stuff.
https://what-if.xkcd.com/58/

The ISS has an acceleration pointed towards (more or less) the center of the Earth, but its velocity is almost all horizontal. The ISS has apogee of 416 km and perigee of 409 km (on average), so its orbital eccentricity is quite low.


Great stuff, Cheddite but we're throwing something down from the ISS. And it's not like we have to overcome some centrifugal velocity component. Even if the Earth pointing component was zero that would not matter.

It's true that if you throw something 'down' (toward Earth) from the ISS, its trajectory will not be linear but the radial velocity component would still be calculated the way I indicated. Potential energy would still be converted to kinetic energy, which is what causes the velocity increase.




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[*] posted on 22-5-2015 at 14:35


Quote: Originally posted by jock88  
Assuming no burn up/no air resistance it will hit the earth at an angle. (so there you have it).

Wrong. Cheddite Cheese gave the correct answer - thread can be closed.
(With air resistance you would not have to give it any impulse - ISS is in a decaying orbit).

Quote:
The ISS already has a velocity component pointing toward the Earth (it's in effect 'falling' towards Earth, in an orbit). There should be no problem releasing an object from it at 1 m/s.

Huh?
To me it's pretty obvious that OP was talking about 1 m/s towards the center of earth relative to the ISS.
Edit: which should only give it a small change in eccentricity (1 m/s = 3.6 km/h).

[Edited on 22-5-2015 by turd]
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SupFanat
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[*] posted on 22-5-2015 at 14:43


What about starting from geostationary orbit?
Imagine throwing some object from about 36 millions meters altitude with initial speed of 36 m/s towards Earth.
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blogfast25
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[*] posted on 22-5-2015 at 15:36


For a far away object falling in a central gravitational field, e.g. a comet 'falling' toward the Sun:

F = m a

F = G (mM)/R<sup>2</sup>

Ergo:

a = GM/R<sup>2</sup>

dv/dt = GM/R<sup>2</sup>

dv/dR dR/dt = GM/R<sup>2</sup>

v dv = [GM/R<sup>2</sup>]dR

Integrate: 1/2 v<sup>2</sup> = - GM/R + C (C is integration constant)

Apply boundary conditions and solve for v<sub>end</sub>.

v = speed (m/s)

[Edited on 22-5-2015 by blogfast25]




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[*] posted on 22-5-2015 at 15:47


Cheddite Cheese and turd have the answers closest to correct. All discussion of air resistance and terminal velocity is irrelevant.

Let's assume the ISS is in a circular orbit for simplicity. If you want to argue that its small eccentricity makes a difference, we can take that up later, but it doesn't affect the main conclusion.

Throwing the object directly downward gives gives it a radial component to its velocity, which it didn't have before, so it increases the kinetic energy and therefore the total energy (kinetic plus potential) of the object. But it doesn't change the angular momentum. The semimajor axis and eccentricity are functions of the total energy and angular momentum. Work out the formulas and you'll find that the eccentricity after the throw is v(radial)/v(angular), or about 1m/sec / 7 km/sec = about 1/7000 roughly. Thus the difference in apogee and perigee of the new orbit is about 7000 km (the radius of the orbit) x 2 x 1/7000, or about 2 km. This is not enough to bring the object into the atmosphere.

The object moves in a new (elliptical) orbit with a perigee about 1km below the circular orbit of the space station and apogee about 1 km higher.

There is also a slight change in the semimajor axis of the orbit, but the fractional change is (v(radial)/v(angular))^2, so it is negligable in this calculation.




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[*] posted on 22-5-2015 at 16:21


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



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[*] posted on 22-5-2015 at 18:28


Quote: Originally posted by annaandherdad  
Work out the formulas and you'll find that the eccentricity after the throw is v(radial)/v(angular), or about 1m/sec / 7 km/sec = about 1/7000 roughly.


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.




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[*] posted on 22-5-2015 at 18:36


quoting xkcd twice in the one thread.

https://xkcd.com/1356/

Its funny, 'cause its true.

Seriously - if you want a deep intuitive understanding of orbial mechanics, drop a weekend or two into the kerbal space program.

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.

Throwing something away from an orbiting object will put a very very little wobble in its orbit.

When you throw your thing at the earth from the ISS - think about the _whole_ orbit - at the moment you throw the object, it will still be travelling horizontally at the same speed it was before - but vertically, it will be travelling toward the earth at 1 meter per second faster.

Now 1/4 of an orbit later - the object still has that same 1 meter per second speed in the direction you threw it - but now, because its travelled 1/4 of the way round - that extra speed is now horizonal - your object is orbiting faster !

Another 1/4 of an orbit, and your object is now moving vertically away from the earth at 1 meter per second faster than it would have otherwise been - but the horizontal speed would be the same as just before you threw it.

..another 1/4 - and its orbiting slower.

Kerbal Space Program. Its grouse.
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blogfast25
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[*] posted on 22-5-2015 at 18:45


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. The energy you put in to get an object into orbit is the same as the energy you get back de-orbiting it: Conservation of Energy. You don't somehow put in that energy twice.




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smaerd
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[*] posted on 22-5-2015 at 19:44


I'm thinking about it more so from a perturbation standpoint. It might not instantly go down to earth sure, but over time the orbit will be perturbed. Maybe an ideal orbit an object can circle a perfect sphere for infinite but I don't think that's how it works with real world low earth orbit. I very well could be wrong but I am well aware that sattelites fall out of orbit.



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