Spaceship using Kepler's 2nd Law
Moderators: gmalivuk, Moderators General, Prelates

 Posts: 893
 Joined: Fri Feb 07, 2014 3:15 pm UTC
Spaceship using Kepler's 2nd Law
I want to start off by saying that this is a physics question; not an engineering question. Even if it works, this system would be completely impractical and require materials magnitudes stronger than those needed for a space elevator.
A spaceship is orbiting a planet. The spaceship fires a harpoon at the planet. The harpoon now connects the spaceship and the planet with a rope. As the spaceship orbits around the planet, the rope is wrapped around the planet. As more and more of the rope's length is wrapped around the planet, there is less available to connect the spaceship and the planet. Therefor, the spaceship is pulled closer to the planet. By Kepler's 2nd Law, the spaceship's speed increases as it moves closer to the planet. Eventually, the spaceship travels faster than the planet's escape velocity. The spaceship releases its hold on the rope. The spaceship goes off into interplanetary space.
Does this system work?
A spaceship is orbiting a planet. The spaceship fires a harpoon at the planet. The harpoon now connects the spaceship and the planet with a rope. As the spaceship orbits around the planet, the rope is wrapped around the planet. As more and more of the rope's length is wrapped around the planet, there is less available to connect the spaceship and the planet. Therefor, the spaceship is pulled closer to the planet. By Kepler's 2nd Law, the spaceship's speed increases as it moves closer to the planet. Eventually, the spaceship travels faster than the planet's escape velocity. The spaceship releases its hold on the rope. The spaceship goes off into interplanetary space.
Does this system work?
Re: Spaceship using Kepler's 2nd Law
The spaceship will indeed speed up, but it's never going to reach escape velocity so it will always still be in a bound orbit and not escape into deep space.
my pronouns are they
Magnanimous wrote:(fuck the macrons)
Re: Spaceship using Kepler's 2nd Law
Well,I had to put my brain through a few twists and turns, but I think that in principle this should be capable of working.
It is possible to have an orbit with a characteristic energy greater than that required for escape. To illustrate: it takes more deltav to get from LEO to GSO than it does to go from LEO to an escape trajectory  hence a mass in GSO possesses more than enough energy to escape, were its orbit in a different shape.
So the energy is present and backofthenapkin calculations show that an object orbiting at GSO, if "retracted" to LEO would achieve somewhere in the region of 20km/s, more than enough to escape.
It is not free energy, in fact, the energy of the system (of course we are assuming perfect conditions here: no drag or friction, perfect weightless, infinitestrength materials etc.) does not change at all. To succeed, your initial energy conditions must be sufficient, if not (for example, your starting orbit is too low) you will achieve a higher speed, but if energy is not sufficient, all you will do is change your orbit from a circular one, to an elliptical one.
It is possible to have an orbit with a characteristic energy greater than that required for escape. To illustrate: it takes more deltav to get from LEO to GSO than it does to go from LEO to an escape trajectory  hence a mass in GSO possesses more than enough energy to escape, were its orbit in a different shape.
So the energy is present and backofthenapkin calculations show that an object orbiting at GSO, if "retracted" to LEO would achieve somewhere in the region of 20km/s, more than enough to escape.
It is not free energy, in fact, the energy of the system (of course we are assuming perfect conditions here: no drag or friction, perfect weightless, infinitestrength materials etc.) does not change at all. To succeed, your initial energy conditions must be sufficient, if not (for example, your starting orbit is too low) you will achieve a higher speed, but if energy is not sufficient, all you will do is change your orbit from a circular one, to an elliptical one.
 gmalivuk
 GNU Terry Pratchett
 Posts: 26129
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: Spaceship using Kepler's 2nd Law
This is false. The deltav calculation tells us that a rocket capable of getting to GSO is also capable of escaping, but it doesn't tell us that the energy in GSO is enough to escape.p1t1o wrote:To illustrate: it takes more deltav to get from LEO to GSO than it does to go from LEO to an escape trajectory  hence a mass in GSO possesses more than enough energy to escape
The reason it requires more deltav is because of when the thrust is applied. Applying it at apogee adds less total energy than applying it at perigee, and to circularize an orbit with its apogee at the height of GSO requires additional thrust at that point. (A small change in orbital velocity from LEO changes kinetic energy by 2.5 times as much as the same change in velocity from GSO.) All orbits with the same semimajor axis have the same energy. GSO has the same energy as an elliptical orbit that goes almost into the atmosphere at one end and nearly twice as far out as GSO at the other, but this is not enough to escape.
Right. All it will do is change a circular orbit to an elliptical one with the same total energy (e.g. it would change GSO to the highly elliptical orbit I described above).eSOANEM wrote:The spaceship will indeed speed up, but it's never going to reach escape velocity so it will always still be in a bound orbit and not escape into deep space.
However, if we imagine that the windingrope process doesn't use any energy, then this could be used to save energy on certain types of maneuvers that are more efficient at apogee or perigee of an elliptical orbit than they are if you start from a circular one.
Re: Spaceship using Kepler's 2nd Law
gmalivuk wrote:This is false. The deltav calculation tells us that a rocket capable of getting to GSO is also capable of escaping, but it doesn't tell us that the energy in GSO is enough to escape.p1t1o wrote:To illustrate: it takes more deltav to get from LEO to GSO than it does to go from LEO to an escape trajectory  hence a mass in GSO possesses more than enough energy to escape
The reason it requires more deltav is because of when the thrust is applied. Applying it at apogee adds less total energy than applying it at perigee, and to circularize an orbit with its apogee at the height of GSO requires additional thrust at that point. (A small change in orbital velocity from LEO changes kinetic energy by 2.5 times as much as the same change in velocity from GSO.) All orbits with the same semimajor axis have the same energy. GSO has the same energy as an elliptical orbit that goes almost into the atmosphere at one end and nearly twice as far out as GSO at the other, but this is not enough to escape.
Ah. Nurtz. I hate being wrong.
My gut did at first tell me that it should be impossible, but when I looked at it from a perspective of angular momentum  my backofthenapkin said that winding in an object from GSO to LEO would give you a velocity of around 1820km/s, more than enough for escape from LEO (according to L=rmv).
Which is how I was convinced that there would be enough energy available.
Im sure you are right but where did I go wrong with the angular momentum?:
If L and m are constant, then v changes in proportion to r.
v @ GSO = ~3.07km/s
r @ GSO = ~6370+35780 = 42150km
r @ LEO = ~6370+200 = 6570km
(42150/6570)*3.07 = 19.69km/s
I presume I made a basic arithmetic error, or have missed some concept entirely. This is what happens when you stop playing KSP for a while.
**edit**
All orbits with the same semimajor axis have the same energy. GSO has the same energy as an elliptical orbit that goes almost into the atmosphere at one end and nearly twice as far out as GSO at the other, but this is not enough to escape.
Does this not mean that a starting circular orbit could be chosen, that is arbitrarily close to being nongravitationally bound, where this technique would achieve escape? (And hence showing that there do exist circular orbits that posses the energy required?)
Re: Spaceship using Kepler's 2nd Law
L and m aren't the only conservation laws though. You also have to conserve energy. Working it through, I find that, if we take the particle mass to be much smaller than the planet mass
[; \frac{\Delta L}{L} = \pm \frac{r'}{r}  1 ;]
for a particle starting at radius [;r;] with angular momentum [;L;] (assumed to be small wrt the planet's original angular momentum) and ending at radius [;r';] with angular momentum [;L+\Delta L;].
Now, it should be clear that the ve root is unphysical and so, letting [;r'=r+\Delta r;]
[;r\Delta L = L \Delta r = \frac{1}{2} \Delta (rL);]
And we've arrived at a situation that looks a bit thermoy which I suppose makes sense.
[; \frac{\Delta L}{L} = \pm \frac{r'}{r}  1 ;]
for a particle starting at radius [;r;] with angular momentum [;L;] (assumed to be small wrt the planet's original angular momentum) and ending at radius [;r';] with angular momentum [;L+\Delta L;].
Now, it should be clear that the ve root is unphysical and so, letting [;r'=r+\Delta r;]
[;r\Delta L = L \Delta r = \frac{1}{2} \Delta (rL);]
And we've arrived at a situation that looks a bit thermoy which I suppose makes sense.
my pronouns are they
Magnanimous wrote:(fuck the macrons)

 Posts: 893
 Joined: Fri Feb 07, 2014 3:15 pm UTC
Re: Spaceship using Kepler's 2nd Law
Can you please remake your post? I am having some trouble understanding the equations.
 gmalivuk
 GNU Terry Pratchett
 Posts: 26129
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: Spaceship using Kepler's 2nd Law
No circular orbit (however big it may be) on its own has enough energy to escape, because the kinetic energy must be less than the gravitational energy in any closed orbit, or else it wouldn't be a closed orbit.
However, *pulling* the ship in toward the planet does work on it, because it applies a force over a distance. The winding situation might be a bit complicated because it depends on the planet having a radius and the pull of the rope isn't directly toward the center but rather toward the planet's horizon behind the ship, but if we instead imagine the planet like a (spherical) yoyo, with the rope being actively reeled winding around an arbitrarily thin rod at the core, then it pulls directly to the center and our calculations need only worry about angular momentum. In which case p1t1o's equations work and it's going more than 19km/s in LEO, which is almost twice what it needs to escape.
But the OP isn't about someone actively reeling in the ship toward a fixed point. It's about the ship passively winding itself up on the planet. The fact that the rope pulls "backward" (toward the horizon) can't be ignored in this case, as that's what imparts some of the ship's angular momentum to the planet. No energy is added to the ship+planet mechanical arrangement in this situation, so in this case the ship won't have enough energy to escape. I'm not sure exactly how the balance works out,* but I am now pretty certain the ship couldn't escape in the scenario as described by the OP.
* It would depend on the winding radius, the speed at which the planet is already spinning, and maybe the mass of the ship. For example, if a ship did this to Earth from GSO, the harpoon wouldn't wrap at all and nothing would happen, while farther out from GSO and it would wrap in the other direction. (In this case, actually, the planet imparts some angular momentum and energy to the ship, so p1t1o's suspicion, that maybe it would work to escape from a large enough circular orbit, turns out to be true for different reasons.)
However, *pulling* the ship in toward the planet does work on it, because it applies a force over a distance. The winding situation might be a bit complicated because it depends on the planet having a radius and the pull of the rope isn't directly toward the center but rather toward the planet's horizon behind the ship, but if we instead imagine the planet like a (spherical) yoyo, with the rope being actively reeled winding around an arbitrarily thin rod at the core, then it pulls directly to the center and our calculations need only worry about angular momentum. In which case p1t1o's equations work and it's going more than 19km/s in LEO, which is almost twice what it needs to escape.
But the OP isn't about someone actively reeling in the ship toward a fixed point. It's about the ship passively winding itself up on the planet. The fact that the rope pulls "backward" (toward the horizon) can't be ignored in this case, as that's what imparts some of the ship's angular momentum to the planet. No energy is added to the ship+planet mechanical arrangement in this situation, so in this case the ship won't have enough energy to escape. I'm not sure exactly how the balance works out,* but I am now pretty certain the ship couldn't escape in the scenario as described by the OP.
* It would depend on the winding radius, the speed at which the planet is already spinning, and maybe the mass of the ship. For example, if a ship did this to Earth from GSO, the harpoon wouldn't wrap at all and nothing would happen, while farther out from GSO and it would wrap in the other direction. (In this case, actually, the planet imparts some angular momentum and energy to the ship, so p1t1o's suspicion, that maybe it would work to escape from a large enough circular orbit, turns out to be true for different reasons.)

 Posts: 7073
 Joined: Thu Jun 03, 2010 12:21 am UTC
Re: Spaceship using Kepler's 2nd Law
Perhaps this is a naive point of view. Since the tangential velocity the ISS is listed at 17,150 mph and the earths tangential velocity at the surface is 1037 mph, I'm missing how this works. For me conceptually, gravity is a taut string anchored at the earths center. The string has a constant angular velocity. GSO is the point where the surface and the object in orbit have the same angular velocity, not the same tangential velocity. Is my conceptual picture flawed? I was under the impression that I had read somewhere of a proposal to do something similar to this, backwards, that is to pull something into orbit by trailing a tether from orbit which could grab it.
 gmalivuk
 GNU Terry Pratchett
 Posts: 26129
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: Spaceship using Kepler's 2nd Law
It seems you're missing the fact that the rope is attached to the surface and reels the spacecraft in, where conservation of momentum means it speeds up.
Re: Spaceship using Kepler's 2nd Law
jewish_scientist wrote:Can you please remake your post? I am having some trouble understanding the equations.
I'm using TeX the world, take a look in the sticky for instructions on getting it to display properly.
my pronouns are they
Magnanimous wrote:(fuck the macrons)
Re: Spaceship using Kepler's 2nd Law
eSOANEM wrote:jewish_scientist wrote:Can you please remake your post? I am having some trouble understanding the equations.
I'm using TeX the world, take a look in the sticky for instructions on getting it to display properly.
that requires firefox or chrome, and probably a PC.

 Posts: 7073
 Joined: Thu Jun 03, 2010 12:21 am UTC
Re: Spaceship using Kepler's 2nd Law
Ok. But that doesn't help me conceptually. The attachment point on the spacecraft is moving faster than the attachment point on the surface. The orbital height of the spacecraft is a function of the energy level of the spacecraft as compared to the attraction of gravity at the center of the earth, not the surface. My conceptual problem is in the difference between a tether and gravity.gmalivuk wrote:It seems you're missing the fact that the rope is attached to the surface and reels the spacecraft in, where conservation of momentum means it speeds up.
Consider two bodies in orbit at different altitudes. If I tether the two I would expect that the final orbit would be an average of the two orbits with the final orbit being a function of the relative masses and their orbital energy levels. If this isn't true don't bother reading any more, it will be gibberish.
Spoiler:
Re: Spaceship using Kepler's 2nd Law
Tethered objects need not have the same velocity. In particular, the tether doesn't affect any velocity perpendicular to the tether; as you note, the tethered bodies will rotate about their centre of mass.
True, a less pretty version of the post is inside the spoiler below:
speising wrote:↶eSOANEM wrote:jewish_scientist wrote:Can you please remake your post? I am having some trouble understanding the equations.
I'm using TeX the world, take a look in the sticky for instructions on getting it to display properly.
that requires firefox or chrome, and probably a PC.
True, a less pretty version of the post is inside the spoiler below:
Spoiler:
my pronouns are they
Magnanimous wrote:(fuck the macrons)
 gmalivuk
 GNU Terry Pratchett
 Posts: 26129
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: Spaceship using Kepler's 2nd Law
Gravity always pulls toward the center of the planet. A tether to the surface pulls toward that point on the surface, and if the angular velocity of the surface and the spacecraft differ, eventually the tether will be pulling toward the edge of the planet, as it begins to wind around. It will also get shorter and shorter, as more of it wraps around the surface of the planet.morriswalters wrote:My conceptual problem is in the difference between a tether and gravity.
Gravity pulls toward C, while the tension on the rope points toward A. Also the rope shortens by the full circumference of the planet for every revolution the spacecraft makes around it.
Gravity slingshots don't steal any energy from a planet's rotation, they steal momentum from the planet's orbit. Rotation doesn't matter at all, except to the (tiny) extent that features on and within the planet can perturb the orbital path of something passing very near it.I'm assuming the OP is thinking in terms of gravity slingshots. But there is never a physical linkage. Just the mutual attraction of gravity. The craft which is already at escape velocity steals energy from the earth by slowing its rotation slightly through conservation of angular momentum.
If the OP were thinking of situations without a physical linkage, I suspect he wouldn't have posited a thought experiment with an explicitly physical linkage.

 Posts: 7073
 Joined: Thu Jun 03, 2010 12:21 am UTC
Re: Spaceship using Kepler's 2nd Law
Well this was fun for a second.
Yes.eSOANEM wrote:Tethered objects need not have the same velocity. In particular, the tether doesn't affect any velocity perpendicular to the tether; as you note, the tethered bodies will rotate about their centre of mass.
OK. I learned something. Something about the behavior of the tether bothers me though.gmalivuk wrote:Gravity always pulls toward the center of the planet. A tether to the surface pulls toward that point on the surface, and if the angular velocity of the surface and the spacecraft differ, eventually the tether will be pulling toward the edge of the planet, as it begins to wind around. It will also get shorter and shorter, as more of it wraps around the surface of the planet.
I'll add that to my conceptual picture.gmalivuk wrote:Gravity slingshots don't steal any energy from a planet's rotation, they steal momentum from the planet's orbit. Rotation doesn't matter at all, except to the (tiny) extent that features on and within the planet can perturb the orbital path of something passing very near it.
 gmalivuk
 GNU Terry Pratchett
 Posts: 26129
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: Spaceship using Kepler's 2nd Law
What bothers you about the tether?
It's like tetherball, but at the center is a planet instead of a pole.
It's like tetherball, but at the center is a planet instead of a pole.

 Posts: 7073
 Joined: Thu Jun 03, 2010 12:21 am UTC
Re: Spaceship using Kepler's 2nd Law
Ok. I found out what's twigging me. The anchor point isn't in orbit. Orbital velocity at the Earth's surface is just over 17,000 mph. In the diagram, the resultant vector of gravity's pull and the pull from the anchor would seem to aim at the earth no matter where you start. This setup has an additional gravity vector along line r. What I want to say naively is that there is energy available through the moment arm at AC but the the gravity vector causes the resultant vector to always point slightly below the horizon. I consider that questionable though. After giving it some thought I don't think I can't make any reliable statements about this problem.
@Gmalivuk
Never saw one of those when I could enjoy it.
@Gmalivuk
Never saw one of those when I could enjoy it.

 Posts: 893
 Joined: Fri Feb 07, 2014 3:15 pm UTC
Re: Spaceship using Kepler's 2nd Law
morriswalters wrote:I was under the impression that I had read somewhere of a proposal to do something similar to this, backwards, that is to pull something into orbit by trailing a tether from orbit which could grab it.
You probably read about a spacetether. It basically works as a 'momentum bank', taking or giving momentum to spacecrafts it connects with. When the spacecraft has a greater linear velocity than that of the end it connect to, basic collision physics tells us that the spaceship will slow down and the end will speed up. Rotational physics tells us that increasing the linear velocity of one end will increase the angular velocity of the tether as a whole. The same thing happens in reverse when the spaceship has a lower velocity than the end it connects to. What is particularly great about this idea is that if the tether is several times more massive than any single spaceship, is used to increase momentum as often as decrease it, and is not particularly long, then the tether's orbit and the ends velocities remain practically constant no matter how much time passes or how often it is used.

 Posts: 7073
 Joined: Thu Jun 03, 2010 12:21 am UTC
Re: Spaceship using Kepler's 2nd Law
Yes. The one I read about was called Skyhook.
Who is online
Users browsing this forum: No registered users and 10 guests