If you were in between two planets that each have a gravitational pull of let’s say 1000g, would the two sources of gravity cancel each other out rendering you weightless, or would you get torn in two?
There will be a point directly in between the two planets where the gravitatiojnal pull will be cancelled out. It’s one of the 5 Lagrange points in a two body system. A body placed at this point will stay there for a while, although since it’s not a stable point, eventually other forces will perturb the body and it will drift away and go into orbit about one of the two planets.
You would be weightless overall, but the tidal force would indeed rip you apart. Or at least it could in theory.
The tidal force is usually only significant for large objects. It’s equal to 2F[sub]G[/sub] · x/R, where x is your size and R is the distance from the planet. For us on earth, x = 2m and R = 6400000m (assuming you’re at the surface), so x/R = 3×10[sup]-7[/sup]. So we experience a tidal force of a measly 6×10[sup]-7[/sup]g.
Now, in your situation, you had a gravitational force of 1000g. In that case, the tidal force would be 6×10[sup]-4[/sup]g. Since there are two planets, you double it to 10[sup]-3[/sup]g. Still not a lot. If the gravity force were instead 1,000,000g, you’d experience about 1g of tidal force. This would probably be painful but not rip you apart.
100 million g’s should do it.
Are there any stable points like that? And in that case, can you explain why they are stable?
The Lagrange points are points of equilibrium in the orbiting two-body problem. The OP didn’t specify that the two planets are orbiting one another, but if we assume they are, then the Lagrange points appear.
There are five Lagrange points. Three (L1, L2, and L3) are unstable, and two (L4 and L5) are stable.
I could show mathematically why L4 and L5 are stable, but I can’t explain it. It’s not an intuitive result to me.
The L4 and L5 points of Jupiter in its orbit around the sun has some asteroids occupying it; they are called the Trojan asteroids, and are named after characters in the Iliad. These points have become known as trojan points.
Mars has six or seven trojan asteroids in its L5 point, and Neptune has at least one; Earth might have some, but they are consistently 150 thousand kilometers distant from the Earth so are difficult to see.
These asteroids might have been there for a relatively long time; what I cant understand is why they dont coalesce into a single body.
SF worldbuilding at
http://www.orionsarm.com/main.html
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For the simple case of one small body in a nearly circular orbit around one large body, i.e. Earth / Sun alone, where geometrically are the 5 points?
I’m looking for broad-brush qualitative descriptions, not necessarily mathematical quantitative precision.
Obviously there’s a point along the Earth-Sun line where the opposing gravities are equal, but after that I can’t see the others. And it’s not clear to me that a small 3rd body at that spot would follow (or even could follow) the curved trajectory needed to remain in position as the Earth drove around its orbit.
I recall something about two points along the Earth’s orbit leading and lagging the Earth’s position. But other than 180 degrees opposite I don’t see any of those as even remotely stable. And the 180 point will only work for a perfectly circular orbit due to Kepler’s law about orbital speed changing as the planet moves around an eccentric orbit.
Thanks to any/all responders.
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Calculating the five Lagrange Points. (WARNING: Lots of math, but there is a picture at the top of the page.)
This point is called L1, and the opposing gravities are not equal there in the stationary frame (otherwise it would stay still instead of orbiting along with the Earth).
If you know celestial mechanics, then you know that the closer you get to the Sun, the faster you orbit. There is no stable orbit around the Sun closer than the Earth which has a period of 1 Earth year. However, what if the Sun were a little less massive? Then the 1-year orbit would occur a little closer to the Sun than it does now. This is what the L1 point is; the pull of the Earth on the outside opposes the pull of the Sun and effectively makes the Sun a little less massive. Similarly, the L2 point (behind the Earth) effectively makes the Sun a little more massive, and moves the 1-year orbit out a little.
OK, take a top-down view of the Sun and the Earth. Three of the points are on the line between the Earth and the Sun. Since the Sun’s mass is so much greater than the Earth’s, these points are all close to the same distance from the Sun as the Earth is. Two are near the Earth, one closer to the Sun and one further, and one of them is 180 degrees from the Earth, and a little further from the Sun. These are the unstable points. They’re good places to put satellites: Since they’re unstable, assorted space junk doesn’t accumulate there, but a satellite can use onboard thrusters to maintain a balancing act and stay there.
Now, take the line from the Earth to the Sun as a base, and draw two equilateral triangles. The points of those triangles are the last two Lagrange points, and they’re stable. Because of this, junk tends to accumulate there: At least dust, and possibly asteroids.
Two equal and opposite forces yield no net force. Thus there is no net acceleration, and you would be motionless (or continue on your previous trajectory). However, you would be subject to internal stress.
Think of it like two people pulling on a rope. As long as they are pulling with equal force, the rope stays still. The tension in the rope, however, is not zero. If they pull hard enough, the rope would break.
I would just like to clarify an issue for the OP:
There is indeed a point where the forces cancel out. But real objects, such as people are not points. So one side of your body will be closer to one source than the other, and will feel a little bit of tugging. This difference is what tidal forces are all about.
As for asteroids at Lagrange points. The stable nature of the points means that an object can be a bit off from the point and not go wandering off. It in effect “orbits” that point. So you can also have multiple asteroids near such a point, their mutual attraction can even help (or sometimes hurt) their stability.
Interesting
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Achernar & Chronos, thank you for two very clear & nonoverlapping posts. I know just enough orbital mechanics to be dangerous. Thanks also to av8trmike for the wolfram ref. The math was more than I needed, but not more than I could follow.
The L1, 2, & 3 points are conceptually simple (after somebody explained it). I’m having a bit of conceptual trouble with L4 & L5. Since they’re geometrically equivalent, let’s just talk about L4, 60 degrees ahead of Earth in the same orbital path (ignoring Earth-orbit eccentricity).
Let’s assume a small satellite established at the L4 point. Now by the definition of the L4 point the satellite is at Earth’s orbital radius and is traveling at Earth’s orbital speed. So even in the absence of the Earth it’ll keep orbiting as it is. The Sun+satellite system is stable without the Earth in the equation.
Now we (conceptually) add Earth back to the Sun+satellite system and what happens?
I see a new attractive force between Earth and the satellite and due to the huge mass disparity, almost all the velocity change occurs to the satellite. It is accelerated towards Earth, which vector can decomposed into a tangential vector pointed orbitally-rearwards and a radial vector pointed Sunward.
Both those vectors result in the satellite “descending” to a lower smaller-radius orbit. The stable-orbit velocity requirement is less as it “descends”, so just considering the first dv/dt increment of delta-v, an equilibrium is eventually reached.
BUT, meanwhile the Earth continues to pull the satellite “rearwards” & “inwards”. So I see the process repeating, the delta-v compared to the pre-Earth situation accumulating, and the satellite getting drug a long way Sunwards off the point. I don’t see anything which counteracts the steady one-way accumulation.
That hardly seems even meta-stable, much less truly stable.
I now see that L5 is similar, but not identical, to L4. The Earth-attractive vector now decomposes into Sunward + orbittally forwards components. Those two velocity changes tend to cancel out in terms of which orbital radius they “prefer”, which sounds lots more stable than L4. But it’d be a heck of a coincidence for them to be in the exact balance required for L5 to be long-term stable.
Since the experts agree that both L4 & L5 actually work fine, obviously my qualitative arm-waving has distorted the true physics beyond the breaking point. So Help, Mr Wizard!!!
Thanks in advance.
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There was an L5 Society for years that was formed around the idea that the first logical places for permanent, large space stations would be the L4 or L5 points, since they’re so stable.
A link to their newsletters: http://www.l5news.org/. They’ve merged with the National Space Society/Association/Something, and I think they’ve modified their goals to be “let’s promote space travel.”
LSLGuy: I swear, 20-year-old groo could show you why L4 and L4 are so stable. The link (http://scienceworld.wolfram.com/physics/LagrangePoints.html) given by av8rmike actually describes this; here’s a little narrative of the treatment in the link:
You have to go into the calculus of variations and look for the incremental forces affecting the body at L4/L5 as it is perturbed from the actual L4/L5 point. Any perturbation results in forces drawing the object back to L4/L5, and the forces increase with distance (making the object locally stable*). You also have to do this in a rotating coordinate system, which is the only way (IIRC) to fool yourself into thinking there’s an analytic solution to the three-body problem. Again, mists of time and all that, but I think the “fool yourself” statement is because you have to posit that the two other bodies are much more massive than the third body and can be considered static (within your rotating coordinate system). You can see why 40-year-old groo would rather just say, “Ooh. Pretty planets.”
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- Again, IIRC, L1 is highly metastable – any perturbation and the object falls out of place. L2 and L3 are “sorta” metastable in that small perturbations will cause them to orbit the L2/L3 point, and larger ones will knock them out of orbit.
I think you’re making a common and tempting mistake here. In fact, it’s so tempting for me, that I’m doubting that it’s a mistake at all, but I’m going to point it out anyway.
Just because the Earth is pulling in a certain direction doesn’t mean the satellite will move in that direction. Such is the nature of acceleration vs. velocity.
I was explaining Lagrange points to someone and he asked why L2 was right alongside Earth in the orbit and not a little behind. I realized that he was expecting Earth to “tow” the satellite along, and I was unable to explain why this was wrong. I thought about this some more, and eventually came up with the picture I told you earlier, about effectively increasing or decreasing the mass of the Sun. I wonder if something similar might be useful to explain L4 and L5.
And, on a side note, an object inside a homogenous hollow sphere will feel no graviational effect from the sphere, because the forces exactly cancel each other at every point in the sphere. There, when a point is closer to one part of a sphere, the larger amount of sphere on the other side just exactly makes it up.
I’ve never been able to come up with an intuitive idea of why L4 and L5 are stable, but I think the problem in your argument is that when you add the earth back to the sun+satellite system, you change the center of rotation. When it’s just the sun+satellite, the satellite orbits the sun, but when you add in the earth, the whole system orbits the center of mass of the earth+sun. I guess that L4 and L5 are the spots where the “new” pull of the earth on the satellite matches up with the new center of rotation.
I doubt it would even be very uncomfortable. Grab a chinup bar and hang there. You’re now basically experiencing 1 g of force trying to pull you apart. Your shoulders and fingers will start to ache after a while, but only because all the force is concentrated in those areas. But the rest of your body feels no discomfort at all.
Not exactly. It depends on which direction you perturb your mass. Perturb it radially or north-south by a small amount, and it’ll drift back, but perturb it tangentially by any amount at all, and it’ll drift away (I think I got that right, but I might have gotten two directions mixed up). “Stable under small perturbations in any direction” isn’t “sort of stable” or “metastable”; it’s “stable”.