Sciencemadness Discussion Board

Throwing objects down from satellites - would it work? Why not?

SupFanat - 22-5-2015 at 08:56

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?

blogfast25 - 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]

smaerd - 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.

blogfast25 - 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]

neptunium - 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 ..

smaerd - 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.

aga - 22-5-2015 at 10:39

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

turd - 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]

aga - 22-5-2015 at 11:14

Ah. Good. Mars then.

Praxichys - 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

Metacelsus - 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.

blogfast25 - 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.

Metacelsus - 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.

SupFanat - 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).

jock88 - 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).

blogfast25 - 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.

turd - 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]

SupFanat - 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.

blogfast25 - 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]

annaandherdad - 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.

smaerd - 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?

blogfast25 - 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.

ahill - 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.

blogfast25 - 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.

smaerd - 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.

annaandherdad - 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.

annaandherdad - 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.

turd - 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.

smaerd - 23-5-2015 at 05:14

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

annaandherdad - 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 201 times


blogfast25 - 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! :)

jock88 - 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).

turd - 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]

jock88 - 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?

Metacelsus - 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]

blogfast25 - 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]

jock88 - 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.

blogfast25 - 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).

SupFanat - 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. :(

blogfast25 - 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.

jock88 - 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.

aga - 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.

blogfast25 - 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]

phlogiston - 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]

blogfast25 - 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]

phlogiston - 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]

blogfast25 - 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]

SupFanat - 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?

froot - 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.

SupFanat - 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.

blogfast25 - 28-5-2015 at 05:50

Quote: Originally posted by SupFanat  
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.


That's basically word salad.

Geostationary orbits are orbits too. AAHD's derivation holds perfectly for such orbits too.

blogfast25 - 28-5-2015 at 05:58

Quote: Originally posted by SupFanat  
About re-entry... Adiabatic compression can produce high temperatures.
But what about adiabatic expansion which gives low temperatures? How low are the reachable temperatures?


I don't know. The braking force due to air-drag is given by (IIRW) F<sub>drag</sub> = k v<sup>n</sup>, as explained high up.

The braking force performs mechanical work acc.:

dW = F<sub>drag</sub> dx

That work is converted to heat energy, at least in part, causing both the air (around the object) and the object itself to heat up. Very hard to calculate, I think...

It's obvious that the heat shields of the Apollo capsules and space shuttles heated up very strongly upon re-entry. Shooting stars are meteorites burning up in the higher atmosphere, due to entry heat.

[Edited on 28-5-2015 by blogfast25]

SupFanat - 28-5-2015 at 06:33

Quote: Originally posted by blogfast25  
Quote: Originally posted by SupFanat  
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.


That's basically word salad.

Geostationary orbits are orbits too. AAHD's derivation holds perfectly for such orbits too.

If the atmosphere still exists on such altitude and rotates around the Earth with the same speed as the satellite then the satellite doesn't move relative to atmosphere.

blogfast25 - 28-5-2015 at 07:20

Quote: Originally posted by SupFanat  

If the atmosphere still exists on such altitude and rotates around the Earth with the same speed as the satellite then the satellite doesn't move relative to atmosphere.


True but that doesn't change anything, fundamentally. Air drag is ZERO at 35,786 kilometres above earth.

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

SupFanat - 28-5-2015 at 07:49

The zero speed relative to atmosphere means the air drag would be zero even if the atmosphere were dense at such altitude?

blogfast25 - 28-5-2015 at 07:59

Quote: Originally posted by SupFanat  
The zero speed relative to atmosphere means the air drag would be zero even if the atmosphere were dense at such altitude?


Yes but it's a nonsensical question: orbital period and orbital radius (and Earth's mass) are tied to each other by Kepler's Law. No low altitude geostationary orbits, for instance.

SupFanat - 28-5-2015 at 08:24

I hope they make better geostationary weather satellites which show as many color channels as they want for some scientific purposes but they don't forget about "true color" version as well.

blogfast25 - 28-5-2015 at 08:25

Quote: Originally posted by SupFanat  
I hope they make better geostationary weather satellites which show as many color channels as they want for some scientific purposes but they don't forget about "true color" version as well.


Colour is in your mind. Not important...

SupFanat - 28-5-2015 at 08:38

I hope it's thinkable. I want to view Earth from great altitude but I can't reach even low earth orbit, a fortiori such high orbit that the Earth appears as disc. So the only chance is...virtual trip.

turd - 28-5-2015 at 11:12

Quote: Originally posted by SupFanat  
I want to view Earth from great altitude


Source: https://en.wikipedia.org/wiki/Pale_Blue_Dot

You're welcome.

SupFanat - 28-5-2015 at 20:07

Thank you but billions of kilometers/miles is too tall. I prefer thousands of kilometers/miles.