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herewiss13So someone at Baen's Bar wants to reduce the eccentricity of Mars' orbit and move it closer to the Sun by throwing rocks at it. These rocks would be thrown by some Aliens she's created, who find rock-throwing on that scale to be trivial.
I suggested that this might shatter Mars.
She politely asked if this was a WAG or if I had numbers. I did not.
I got some. ;-)
Let me just say: Brute Force? Bad idea.
I was so impressed with my calculatory effort, an effort I haven't tried in 4-5 years, I didn't want it to go to waste in a single forum. So...
From: Julie
Eric: what's your background? Do you know on the "shatter it" or are you guessing?
Two years as a physics major, then a transition to General Science, which
gave me a little astronomy. Plus a life-long sci-fi reader (naturally). ;-)
[Note: the following is me typing out loud to myself, with frequent references to the calculator and Google. So it's a bit pedantic, and possibly condescending (as I explain even the most fundamental physics). Mainly, I'm out of practice, so I want to make sure I'm thinking about everything right and not leaving anything out. Even still, I expect at least one major conceptual correction from the Bar At Large, possibly more. ;-) ]
"Shatter" may be a slight hyperbole. The amount of energy needed lift the entire mass of Mars out of its own gravity well (i.e. send every piece away from every other piece) is, well, staggering in a way that makes the energy scales we're discussing seem like cap guns [Note: this opinion was formed pre-math. Now I'm not so sure we aren't in the Mars-shattering ballpark.]
Regardless of the technicalities of planet shattering, our Moon was formed when a Mars-or-larger sized body struck the Earth. So just because you don't technically shatter a planet doesn't mean you're going to leave it in one piece....and Mars is smaller than the Earth so the required size for a moon-splitting collider is also less...and stability-wise Mars can less afford to loose an equivalent moon-mass. ;-)
My main concern (despite what I just said) basically springs from the fact that Mars is really, really, really big. Incredibly big. More kilograms than any of us can readily imagine.
And you want to move those kilograms by throwing even more kilograms at them at some given velocity, either all at once or in sequence.
If you aren't careful, a single impact scenario is going to create several spheres of molten rock orbiting each other where Mars used to be. And while multiple smaller impacts may be spread out as the planet rotates beneath you, they're still big impacts. The area affected by each will almost certainly overlap...especially as collisions accumulate. (See: volcano equivalent impacts, below).
At best you'll have fractured the planet clear through the core and out the other side, the effects of which I can't even begin to imagine. At worst, you still end up with the multi-molten spheres scenario (you're still imputing the same amount of energy, regardless). Remember, nobody (to my knowledge) has mentioned orbit shifts following the impactors that cause Mass Extinction...and they wreaked a lot of havoc (not all of which was limited to the biosphere) Your potential Impactor(s) will need to be even larger or faster. A lot larger, or faster.
From previous discussions of nuclear explosions on the bar, I believe the larger the bang, the more gets lost in heat/noise/etc...in an ever increasing proportion. So scaling up the Impactor is going to scale up waste energy faster than transfered energy...and waste energy is bad. Heat's nice for bioforming, sure...but pulverizing rock (and/or throwing it into orbit) rather than simply altering its velocity is not. So that's yet another problem really big impacts will have.
Here is the math:
Mass of Mars: 6.4191x10^23 kg
Mean orbital velocity: 24130 m/s
Mars orbital momentum (mv) then is (rounding up slightly): 1.55x10^28(kg*m)/s
We're talking about an inelastic collision in that the masses stick together afterward. The equation for that is:
M1V1 + M2V2 = (M1+M2)V(final)
You can see where M2 and/or V2 is going to need to be _massive_ just to add anything significant to Mars' own momentum.
The next question is just how you're striking Mars. As you basically want to speed it up or slow it down in order to force it into a less eccentric orbit, we really only need to deal with head on (slowing) or tail-end (speeding up) collisions. This makes things simple. In the first case, one velocity is a positive and the other is negative (I'd make the impactor negative); in the second both velocities are positive. In the tail-end scenario, the impactor will also need to be traveling faster than 24130 m/s. If it doesn't, it won't catch up to the planet. The equation doesn't make that clear because it doesn't care. Mars could very well catch up to the impactor and speed it up...but that won't speed Mars up.
So what really needs to happen is, you need to figure out what final velocity (approximately) you want Mars to be traveling at. As your less eccentric orbit will be ideally closer to the sun (yes?) you want to speed Mars up. For comparative purposes, Earth's mean orbital velocity is 29780 m/s. For both Mars and Earth, the min and max velocities vary only by a few km/s. Mars' eccentricity is also .006 and Earth's is. 003.
So, just for the sake of argument (and I'm not sure how this would affect eccentricity) let's say we want to speed Mars up by 2 km/s to 26130 m/s. Thinking only in terms of ratios, I think this would put it into an orbit about half the distance from Earth it is now (a little further out, but still close to half).
Now for all the steps (there's good, mind-blowing stuff below it all):
MV(mars) + MV(impactor) = [M(mars) + M(imp)] * V(final)
____________________________________________________________________________________
1.55x10^28 kgm/s +MV(imp) = [6.42x10^23 + M(imp)] 26130 m/s
____________________________________________________________________________________
1.55x10^28 kgm/s + MV(imp) = 1.6775x10^28 kgm/s + [M(imp) * 26130 m/s]
That is = [M(mars)V(final)] + [M(imp)V(final)]
____________________________________________________________________________________
MV(imp) + -[M(imp)V(final)] = 1.6775x10^28kgm/s + -(1.55x10^28 kgm/s)
____________________________________________________________________________________
M(imp)[V(imp) - V(final)] = 1.275x10^27 kgm/s
____________________________________________________________________________________
And finally to put it into notation suitable to put in a graphing calculator (since it's a sliding scale of M vs. V)
Y = V(imp)
X= M(imp)
Y m/s= [1.275x10^27 kgm/s / X kg)] + 26130 m/s
Sadly, I don't _have_ a graphing calculator any more.
::consults the Almighty Google::
But a really nice Java Applet can be found here.
You've got to do some lovely scaling work just to find the line, since it's an extreme hyperbola, but the results are interesting.
If the Impactor is traveling at or near the speed of light, it needs to mass 4.8x10^18 kilograms.
That's 1/133731 of Mars' own mass...but still pretty big (it's actually smaller than I'd have thought though. All this scientific notation takes some re-adjusting to.).
Remember, that's if it's traveling as fast as physically possible! When you get to less relativistic speeds, the required mass increases dramatically. At half the size of Mars, it'll be traveling at 30000 m/s. Just a bit faster than the mean orbital velocity of the Earth. This is well within the moon-forming scenarios. The moon-forming object which struck the Earth was only about one tenth of Earth's size! I'm not sure how fast it was going, but I suspect it was at a comparable orbital velocity rather than anything
insanely faster. It was also a glancing blow. The tail-end collision we've just crunched the numbers for would be a much more severe impact.
If you want to distribute the momentum transfer into multiple impacts, look at how many you'd need!
The kinetic energy (.5MV^2) of these two scenarios are as follows:
At 'c', KE = 2.16 x10 35 joules
(I've ignored any relativistic effects for the sake of my sanity).
This is about twice the Sun's entire energy output for a single year! (10^34 joules)
At 30,000 m/s, KE = 1.44x10^32 joules
(1500 times less energy)
An H-Bomb only ranks at around 10^17 Joules and a volcanic explosion at 10^19 Joules.
So if you hit Mars with the slower of the two projectiles, it'd be the equivalent of around 1.4 trillion volcanoes!
Even if you somehow break it up into a four-day bombardment [one of her suggestions], that's a lot of wear-and-tear on the planet's surface (14.5 million volcanoes every hour...all in one general region!). If you break it up into 1 volcano equivalent impacts, one each hour, you're talking about around 160 million years...of continuous bombardment! Any attempt to speed it up ramps up the destruction tremendously, to the point where significant debris is reaching escape velocity. The first few thousand impacts might benefit a future biosphere, but the rest would just be...overkill, to say the least.
All this to speed Mars up by about 2 kilometers/second. Hardly seems worth it. ;-)
So I hope that gives you some perspective on what it takes to move a planet, the scale of the energies involved and why I find the brute force idea...doubtful. While god-like aliens might find the engineering aspects to be a cinch, I'm still not sure poor Mars could take it unless you scale it down to tiny nudges over a loooooooong period of time (how patient are your aliens? And how many generations of human perspective on the event do you want to cover?). ;-)
All figures and comparisons can be found with little difficulty using Google.
Hope this helped,
Eric
P.S. That was...surprisingly fun. I haven't done any serious physics number-crunching in 4-5 years. Here's hoping I haven't screwed it up too much. ;-)
P.S.S. Observation: Extremely hard to restrain the use of exclamation points when dealing with energy on that scale.
I suggested that this might shatter Mars.
She politely asked if this was a WAG or if I had numbers. I did not.
I got some. ;-)
Let me just say: Brute Force? Bad idea.
I was so impressed with my calculatory effort, an effort I haven't tried in 4-5 years, I didn't want it to go to waste in a single forum. So...
From: Julie
Eric: what's your background? Do you know on the "shatter it" or are you guessing?
Two years as a physics major, then a transition to General Science, which
gave me a little astronomy. Plus a life-long sci-fi reader (naturally). ;-)
[Note: the following is me typing out loud to myself, with frequent references to the calculator and Google. So it's a bit pedantic, and possibly condescending (as I explain even the most fundamental physics). Mainly, I'm out of practice, so I want to make sure I'm thinking about everything right and not leaving anything out. Even still, I expect at least one major conceptual correction from the Bar At Large, possibly more. ;-) ]
"Shatter" may be a slight hyperbole. The amount of energy needed lift the entire mass of Mars out of its own gravity well (i.e. send every piece away from every other piece) is, well, staggering in a way that makes the energy scales we're discussing seem like cap guns [Note: this opinion was formed pre-math. Now I'm not so sure we aren't in the Mars-shattering ballpark.]
Regardless of the technicalities of planet shattering, our Moon was formed when a Mars-or-larger sized body struck the Earth. So just because you don't technically shatter a planet doesn't mean you're going to leave it in one piece....and Mars is smaller than the Earth so the required size for a moon-splitting collider is also less...and stability-wise Mars can less afford to loose an equivalent moon-mass. ;-)
My main concern (despite what I just said) basically springs from the fact that Mars is really, really, really big. Incredibly big. More kilograms than any of us can readily imagine.
And you want to move those kilograms by throwing even more kilograms at them at some given velocity, either all at once or in sequence.
If you aren't careful, a single impact scenario is going to create several spheres of molten rock orbiting each other where Mars used to be. And while multiple smaller impacts may be spread out as the planet rotates beneath you, they're still big impacts. The area affected by each will almost certainly overlap...especially as collisions accumulate. (See: volcano equivalent impacts, below).
At best you'll have fractured the planet clear through the core and out the other side, the effects of which I can't even begin to imagine. At worst, you still end up with the multi-molten spheres scenario (you're still imputing the same amount of energy, regardless). Remember, nobody (to my knowledge) has mentioned orbit shifts following the impactors that cause Mass Extinction...and they wreaked a lot of havoc (not all of which was limited to the biosphere) Your potential Impactor(s) will need to be even larger or faster. A lot larger, or faster.
From previous discussions of nuclear explosions on the bar, I believe the larger the bang, the more gets lost in heat/noise/etc...in an ever increasing proportion. So scaling up the Impactor is going to scale up waste energy faster than transfered energy...and waste energy is bad. Heat's nice for bioforming, sure...but pulverizing rock (and/or throwing it into orbit) rather than simply altering its velocity is not. So that's yet another problem really big impacts will have.
Here is the math:
Mass of Mars: 6.4191x10^23 kg
Mean orbital velocity: 24130 m/s
Mars orbital momentum (mv) then is (rounding up slightly): 1.55x10^28(kg*m)/s
We're talking about an inelastic collision in that the masses stick together afterward. The equation for that is:
M1V1 + M2V2 = (M1+M2)V(final)
You can see where M2 and/or V2 is going to need to be _massive_ just to add anything significant to Mars' own momentum.
The next question is just how you're striking Mars. As you basically want to speed it up or slow it down in order to force it into a less eccentric orbit, we really only need to deal with head on (slowing) or tail-end (speeding up) collisions. This makes things simple. In the first case, one velocity is a positive and the other is negative (I'd make the impactor negative); in the second both velocities are positive. In the tail-end scenario, the impactor will also need to be traveling faster than 24130 m/s. If it doesn't, it won't catch up to the planet. The equation doesn't make that clear because it doesn't care. Mars could very well catch up to the impactor and speed it up...but that won't speed Mars up.
So what really needs to happen is, you need to figure out what final velocity (approximately) you want Mars to be traveling at. As your less eccentric orbit will be ideally closer to the sun (yes?) you want to speed Mars up. For comparative purposes, Earth's mean orbital velocity is 29780 m/s. For both Mars and Earth, the min and max velocities vary only by a few km/s. Mars' eccentricity is also .006 and Earth's is. 003.
So, just for the sake of argument (and I'm not sure how this would affect eccentricity) let's say we want to speed Mars up by 2 km/s to 26130 m/s. Thinking only in terms of ratios, I think this would put it into an orbit about half the distance from Earth it is now (a little further out, but still close to half).
Now for all the steps (there's good, mind-blowing stuff below it all):
MV(mars) + MV(impactor) = [M(mars) + M(imp)] * V(final)
____________________________________________________________________________________
1.55x10^28 kgm/s +MV(imp) = [6.42x10^23 + M(imp)] 26130 m/s
____________________________________________________________________________________
1.55x10^28 kgm/s + MV(imp) = 1.6775x10^28 kgm/s + [M(imp) * 26130 m/s]
That is = [M(mars)V(final)] + [M(imp)V(final)]
____________________________________________________________________________________
MV(imp) + -[M(imp)V(final)] = 1.6775x10^28kgm/s + -(1.55x10^28 kgm/s)
____________________________________________________________________________________
M(imp)[V(imp) - V(final)] = 1.275x10^27 kgm/s
____________________________________________________________________________________
And finally to put it into notation suitable to put in a graphing calculator (since it's a sliding scale of M vs. V)
Y = V(imp)
X= M(imp)
Y m/s= [1.275x10^27 kgm/s / X kg)] + 26130 m/s
Sadly, I don't _have_ a graphing calculator any more.
::consults the Almighty Google::
But a really nice Java Applet can be found here.
You've got to do some lovely scaling work just to find the line, since it's an extreme hyperbola, but the results are interesting.
If the Impactor is traveling at or near the speed of light, it needs to mass 4.8x10^18 kilograms.
That's 1/133731 of Mars' own mass...but still pretty big (it's actually smaller than I'd have thought though. All this scientific notation takes some re-adjusting to.).
Remember, that's if it's traveling as fast as physically possible! When you get to less relativistic speeds, the required mass increases dramatically. At half the size of Mars, it'll be traveling at 30000 m/s. Just a bit faster than the mean orbital velocity of the Earth. This is well within the moon-forming scenarios. The moon-forming object which struck the Earth was only about one tenth of Earth's size! I'm not sure how fast it was going, but I suspect it was at a comparable orbital velocity rather than anything
insanely faster. It was also a glancing blow. The tail-end collision we've just crunched the numbers for would be a much more severe impact.
If you want to distribute the momentum transfer into multiple impacts, look at how many you'd need!
The kinetic energy (.5MV^2) of these two scenarios are as follows:
At 'c', KE = 2.16 x10 35 joules
(I've ignored any relativistic effects for the sake of my sanity).
This is about twice the Sun's entire energy output for a single year! (10^34 joules)
At 30,000 m/s, KE = 1.44x10^32 joules
(1500 times less energy)
An H-Bomb only ranks at around 10^17 Joules and a volcanic explosion at 10^19 Joules.
So if you hit Mars with the slower of the two projectiles, it'd be the equivalent of around 1.4 trillion volcanoes!
Even if you somehow break it up into a four-day bombardment [one of her suggestions], that's a lot of wear-and-tear on the planet's surface (14.5 million volcanoes every hour...all in one general region!). If you break it up into 1 volcano equivalent impacts, one each hour, you're talking about around 160 million years...of continuous bombardment! Any attempt to speed it up ramps up the destruction tremendously, to the point where significant debris is reaching escape velocity. The first few thousand impacts might benefit a future biosphere, but the rest would just be...overkill, to say the least.
All this to speed Mars up by about 2 kilometers/second. Hardly seems worth it. ;-)
So I hope that gives you some perspective on what it takes to move a planet, the scale of the energies involved and why I find the brute force idea...doubtful. While god-like aliens might find the engineering aspects to be a cinch, I'm still not sure poor Mars could take it unless you scale it down to tiny nudges over a loooooooong period of time (how patient are your aliens? And how many generations of human perspective on the event do you want to cover?). ;-)
All figures and comparisons can be found with little difficulty using Google.
Hope this helped,
Eric
P.S. That was...surprisingly fun. I haven't done any serious physics number-crunching in 4-5 years. Here's hoping I haven't screwed it up too much. ;-)
P.S.S. Observation: Extremely hard to restrain the use of exclamation points when dealing with energy on that scale.
