So I’ve been playing Star Wars: Knights of the Old Republic II - The Sith Lords again.
You start out on a mining colony called Peragus, which is orbitted by an asteroid field.
As you play further, you find that the asteroids are a result of trying to mine the planet itself, which resulted in a huge explosion.
But it got me to thinkin’, how big a hunk o’ the ol’ planet Earth could be blasted off of the planet before it changed the orbit enough to render it unliveable?
And how long would it take for the life on the planet to die after such an event?
When any object orbits another, its only acceleration is perpendicular to the direction in which it’s traveling.
If you could remove a chunk of the Earth without applying any force to the rest of it, what would change is that the gravitational force on the Earth would decrease, which in turn would lessen the radially inward acceleration exerted on the planet, causing the planet to change direction towards the sun less quickly than it currently does. However, the instantaneous tangential velocity of Earth would remain the same, and the radius of the orbit would therefore increase. So.
Call the current gravitational force on the earth F[sub]0[/sub]. Then reducing the mass of the earth, say dividing it by n, means that the new F = (1/n)F[sub]0[/sub]. But F[sub]0[/sub] = ma[sub]0[/sub], so F = ma[sub]0[/sub]/n.
From the mechanics of rotational motion, we know that a[sub]0[/sub] = (v[sub]T[/sub])[sup]2[/sup]/r[sub]0[/sub], where v[sub]T[/sub] is the instantaneous tangential velocity of the Earth (it doesn’t change at all in this scenario). So r[sub]0[/sub] = (v[sub]T[/sub])[sup]2[/sup]/a[sub]0[/sub], and the new radius r = n(v[sub]T[/sub])[sup]2[/sup]/a[sub]0[/sub]. But we defined n to be the ratio of old mass/new mass, so substituting that in and solving for the new mass means that M = m[sub]e/sub[sup]2[/sup]/r*a[sub]0[/sub], where we can plug in any radius we like for r and find the mass that the Earth has to have in order to be at that radius.
The source (PDF) I found says that the outer boundary of our solar system is at 1.2 AU. Converting that into SI units and plugging it in to the above equation, the mass of the earth would have to be reduced from 5.97x10[sup]24[/sup] kg down to 5.01x10[sup]24[/sup] kg. Converting the mass difference to volume, it turns out to be about 16% of the Earth’s current volume.
Now, considering that the outer boundary of the habitable zone is defined as the radius at which all water on the planet freezes, I’d say we’d all be dead shortly before we reached this point, because as water cools to 4 degrees Celsius it expands. Ever leave a glass jar full of water in the freezer overnight? We’d be the jar. Gruesome, eh?
Actually, the article you’ve linked to says that 1.2 AU is the outermost limit to the “habitable zone”, i.e. the farthest distance away from a sun-like star that an earth-like planet could support life. Habitable zone does not equal the boundary of our solar system.
I’m not at work yet, so I can’t double check my orbital mechanics book…but I’m pretty sure about this one. It’s centripetal acceleration wouldn’t change. As the mass decreases, so does the gravitational force. But the lower force acting on the newly lower mass produces the same acceleration. It’s the same reason all things on earth accelerate at the same speed when they fall (ignoring air resistance). So decreasing the mass won’t change the orbit.
Any object in the same position as the Earth moving with the same velocity will have the same orbit, regardless of mass. The only time this might change is when the mass gets large enough to not be negligible compared to the Sun. Then, the gravitational center of the object-Sun system moves appreciably far from the Sun’s center. When this happens, it’s much more like the object and the Sun orbit each other, rather than the object orbits a relatively stationary Sun. But for small masses, like the Earth, the Sun’s motion around the common center can be ignored, and the orbit of the object (whatever its mass) will look the same.
The short answer:
If you magically removed a big chunk of the earth’s mass without applying a force to the remainder of it, the orbit of the remaining part should stay the same.
If you blast a hunk off of the surface of the Earth, momentum and energy will be conserved, and the Earth’s orbit will be changed.
Technically the Earth’s orbit is very slightly altered every time a spacecraft acheives escape velocity and leaves the Earth. However, since the spacecraft’s mass is insignificant compared to the Earth’s, this is not a measurable effect.
If you blast some stuff on the Earth and it just stays in orbit around the Earth (or, say, you launch a spacecraft into orbit around Earth), then there will be no change to the orbit of the Earth-plus-debris system, and the Earth plus all the debris plus the Moon will all continue to orbit around the center of mass of the Earth-Moon system. This could introduce a measurable “wobble” to the Earth’s trajectory around the Sun, but it won’t change the size of the Earth’s orbit.
If the blast allows a significant portion of the debris to reach escape speed, then this will change the Earth’s orbit. The amount by which the orbit is changed, and how exactly it is changed, depends on what direction the debris flew off in. It could make the Earth’s orbit less round (change its eccentricity), change the Earth’s average distance from the Sun (the semimajor axis) or lead to a greater inclination (tilt) to the Earth’s orbit. (Yes, technically the Earth’s inclination is defined to be zero. Work with me here, people! Obviously I mean relative to the pre-blast ecliptic.)
As aerodave says, if you could some how magically change the Earth’s mass without exerting any other forces on the remaining chunk, this would have no effect on the Earth’s orbit.
1.) If you “blasted off” a chunk bymagically making one part split off from the other, withour any recoils, then the orbit doesn’t change at all. Jules Verne pointed this out in Off on a Comet (“Hector Servadac”)
2.) Even if you separated parts with a fairly hefty amount of momentum (large mass at small velocity or small mass at large velocity, or any other combination) the net effect would be to put both parts into elliptical orbits, with their center of mass still in the original orbit. There are a lot of ways to show this. Arthur C. Clarke used this in his short story “Jupiter 5”. Something shoved out the hatch on an orbiting spaceship basically came back into the hatch later on. You really want to get one part moving close to escape velocity of the remaining part if you want to separate them.
To avoid the recoil problem, think of it as having an Earth already split into two pieces (no necessarily equal) but loosely stuck together. Now, some big alien spacecraft puts its tractor beam on the smaller chunk and slows it down while the rest of Earth continues to move like it was before. (There will be some slight gravitational effects between the two planetary chunks, but nothing that will move the orbit by millions of kilometers)
Momentum is still conserved, because the loss in momemtum of the overall pieces-of-earth system is accounted for by the force applied by the tractor beam. Nowhere in the laws of physics does it say momentum has to remain constant…just that changes in momentum must correlate to forces.
Again, the above scenario is functrionally the same as the OP, and in the same spirit. It just avoids the unpredictable unnecessary effects of recoil associated with “blasting” a piece away.
Perhaps a physically realistic way to achieve the OP’s goal would be to blast two chunks from the Earth, in opposite directions, with equal momenta. Equal masses and equal velocities would do it, but all that really matters is that the net momentum sums to zero.
Assuming the middle chunk wasn’t annihilated in the explosion (though really it would be), and that the two departing chunks leave with sufficient speed never to return, the middle chunk should stay in the Earth’s original orbit, regardless of its mass.
Er, no, that’s not correct, I’m afraid. Gravity is what holds the Earth together in the first place.The Earth isn’t a big solid piece of rock. It’s mostly liquid on in the inside, with only a thin solid crust. Mechanical strength plays almost no role in holding the planet together. You couldn’t have two halves of the Earth “loosely” stuck together. Two masses of 3x10[sup]24[\sup] kg located 6400 km apart (I’m fudging; the true separation of their center of masses would be smaller) exert a heck of a gravitational pull on one another.
If you pulled gently on one half, the other half would just come along with it.
If you pulled hard enough to pull a chunk of the Earth away, the other side would get a good jerk too. Bytegeist’s scenario works better.
You read my post, and that’s the part you decide to gripe over? Perhaps you didn’t see the part about the tractor beam!!!
I’m sorry I didn’t add a “suppose”, “assume”, “let’s pretend”, and “for the sake of argument” to every single clause in my post. I thought it was implicit.
I know that the two halves would affect each other. But I was specifically trying to neglect that effect for the purpose of getting the important point across. As much as I love a good nit-pick, analyzing a completely implausible premise isn’t a good place to do it. It’s certainly OK to make implausible assumptions is a case like this. Surely you possess such powers of imagination…God knows I was able to make it through grad school without losing mine.
I admit, Bytegeist’s example is better, and I should have thought of it myself. :shrug:
aerodave, if you’d said that the aliens had some way to turn off gravity or compensate for it, that would be be fine, but you said: “There will be some slight gravitational effects between the two planetary chunks, but nothing that will move the orbit by millions of kilometers.” I took that to mean that in your scenario gravity would still be in effect, but would be negilgible, which is simply untrue—but perhaps I misunderstood what you meant.
Congratulations on making it through grad school, by the way. May I enquire what field? I defend my thesis in planetary astronomy this July, if all goes well.
Superdude, can you tell us where we could rephrase things to make them easier to understand? You can ignore ** Spatial Rift 47’s** post, since he made an error that negates his conclusion.
I’m afraid that there is no single answer to your question. It depends on how fast the mass is removed, and what direction it goes in. If you’d like a very oversimplified response, you’d need to blast off a significant amount of the planet’s mass. Say, at least something like 10%. But it wouldn’t be the fact that the Earth was getting less massive that would change the orbit, it would be the fact that you were blasting the material off. The material leaving the Earth would act like the gas leaving a rocket, and that’s what would change the Earth’s orbit.
Well, you got me there. I did say it was slight. To better express my point, I should have said, “For our purposes, we will ignore it, whether realistically negligible or not.”
That was intended to be the moral of my example.
Aerospce engineering. I don’t claim to be the board’s expert in all matters astrophysical, only having had two courses in orbital mechanics. But his question really just boils down to one of energy/momentum conservaton, so I was trying to tackle it from that vantage point.
Ah, okay, I see what you meant. And, duh, I should have guessed your field from your name.
My minor field is Theoretical and Applied Mechanics, so I have more orbital mechanics than average under my belt—five delightful semesters of epicylic approximations and higher order resonances, all told.
