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SRQHivemind
2010-Jul-20, 04:05 PM
Hi all,
So I'm writing a sci-fi story. The overarching details are not of particular importance to this query.
One of the principle macguffins the Heroes must deal with is an asteroid heading for Earth.
The details:
Designation: Far Kuiper Belt object
Diameter: ~30km (massive enough to cause an ELE)
Orbital Period: ~4900 years (plot point)
Perihelion: 1 AU (Earth orbit)
Aphelion: ~47AU
Orbital Inclination: 80 degrees
Next Perihelion: ~13 years from 'present'

Is is remotely plausible that such an object has remained undetected by present-day (~2000) astronomers?

Our heroes have 13 years and change to deflect it. The plan is to attach a VASIMR-like drive to the asteroid and nudge its orbit so that it no longer impacts Earth.

I am trying my best to stick to scientifically-grounded principles in the story. (I am a sysadmin scifi geek, not an astronomer) the physics have to be real enough to not catch the wrath of many a Bad Astronomer.

Will the above 'work' without stretching credibility?

pzkpfw
2010-Jul-20, 07:29 PM
Welcome. As a new member your posts are held up to make sure they are not advertising for canned meat. Please don't post twice - just wait a bit. Thanks.(Please check out the rules, if you have not already.)

Rhaedas
2010-Jul-20, 08:45 PM
Certainly plausible as a plot. If you want to avoid bad science, then don't do the usual Hollywood and wait until it's inside lunar orbit and conveniently blow it up. Not sure based on your numbers if 13 years is enough time or not, remember, to get some method of thrust on the asteroid, you have to first get out TO the asteroid with enough time to make a difference. Maybe you can stretch the tech side of the science a bit, with a much more advanced propulsion than anything we have at the moment. VASIMR generation II? :)

I think you can keep it grounded in science and still have a good book...decent characters, plenty of technical challenges, good visual descriptions. Just don't have the usual rogue spy character who for whatever reason is bent on ruining the mission. It's a bit cliche I think...Murphy can make plenty of things fail without that kind of subplot.

Grypd
2010-Jul-20, 10:06 PM
The Kuiper belt object Haumea was only discovered in 2004 it has a diameter of 1436 Kms and is what we class as a dwarf planet. The Kuiper belt may well have larger objects still to be discovered and to be honest we can only guess as to how many objects there are. An object of 30 kms is small.

As a story point an object of a diameter of 30kms would be very hard for us to see from here on the Earth and perhaps its discovery would be the result of a new space telescope. But to have 13 years to work out that such an objects path was now a direct threat may well be possible if you use that a collision is detected and that would give you the chance to see that due to orbital mechanics would pose a threat.

mugaliens
2010-Jul-20, 10:23 PM
Is is remotely plausible that such an object has remained undetected by present-day (~2000) astronomers?

All but the largest and brightest asteroids remained undetected until the last hundred years or so. If there's an object with a perihelion of 47 AU and it's still 13 years out, it's unlikely our astronomers will have spotted it.


Our heroes have 13 years and change to deflect it. The plan is to attach a VASIMR-like drive to the asteroid and nudge its orbit so that it no longer impacts Earth.

That's plausible, even if no large-scale VASIMR drive has been built and the US is without an HL lauch vehicle. It would be a global effort, I'm sure.


Will the above 'work' without stretching credibility?

I'm not an expert in any particular scientific discipline, but I've received undergraduate and postgraduate education in a variety of them, and none of what you propose seems beyond the scope of reality. It would require a herculenian effort well beyond your average space exploration budget, but yes, it's feasible. Perhaps a specialist in a particular field would disagree, but hopefully would tell you how and why so you can modify your story to fit reality.

You may wish to check your orbit using an online orbital calculator (http://inkido.indiana.edu/a100/a100_ellipse.html), and you're going to want to check your intercept orbit, as well. That's a bit more tricky, as you're probably not going to have enough time for a Hohmann-type orbit, but you'll be out of fuel for anything faster than a "just in time" approach. Hey - it's a plot element :) So you're going to have to figure out an orbit that's somewhat faster than a hohmann orbit and use that. Perhaps throw in a tricky/risky slingshot maneuver

Here's a collection of various calculators (http://www.educypedia.be/education/calculatorspace.htm)for space applications.

SRQHivemind
2010-Jul-21, 01:27 AM
Thanks for the quick responses :)

I'm trying to keep the principles as real-world as I can. Though I will say that the issues of heavy lifting are not a problem, nor is building a craft small enough and fast enough to get to the 'roid in a reasonable timespan.
A ship the size of the space shuttle is a different animal from a 30km+ asteroid after all.

I'm hesitant to post more details at the moment (quite protective of the story and ideas behind it at the moment), but would certainly appreciate input on the nitty-gritty science. I'm not sure if a fully public forum is appropriate. I've been looking for a space-science advisor type for a while now and figured the BA forums were the place for it.

jlhredshift
2010-Jul-21, 12:32 PM
A couple thoughts:

The highly inclined orbit, to some degree, minimizes the amount of deflection necessary; ie. force applied.

A 47 AU orbit suggests that the object has passed through the inner solar system many times and could be lacking in volatiles and therefore be a very dark object.

Its' mass density extremes could be anywhere from chrondritic rubble pile to a solid iron, but probably somewhere in between. As the density goes up the amount of deflection energy rises, but there are easier methods of deflecting an iron than a rubble pile, and my guess would be rubble pile.

jlhredshift
2010-Jul-21, 12:57 PM
Here is an outlandish sick thought, what if the object was found to be a solid uranium/lead object and the plan was to set off neutron bombs next to it. That out to provoke some discussions amongst the various factions on Earth.

IsaacKuo
2010-Jul-21, 02:10 PM
Designation: Far Kuiper Belt object
Don't call this thing an asteroid. Maybe a comet, or KBO.

Orbital Period: ~4900 years (plot point)
Perihelion: 1 AU (Earth orbit)
Aphelion: ~47AU
This is impossible.

I'm a little sad that no one applied basic orbital mechanics laws to these numbers. Kepler's third law is that the period squared is proportional to the semi-major axis cubed. Therefore, the period is equal to (semi-major axis/AU)3/2 in years.

Semi-major axis is equal to half of perihelion+aphelion, or in this case 24AU. This implies a period of only 120 years!

Alternatively, a period of 4900 years implies a semi-major axis of 289AU (so, for example, a perhelion of 1AU and an aphelion of 577AU).

Since you state that a 4900 year period is a plot point, I'll assume the latter. Unfortunately, it is entirely implausible for astronomers to calculate the period with any degree of certainty. When a comet comes in from this far out, it will enter the solar system at almost exactly solar escape velocity.

Worse still, it's is extremely unlikely for the 4900 year period to be stable. Since the perihelion is so close to the Sun, it will experience a lot of thrust from cometary effects. Small changes in its trajectory will lead to huge changes in its orbital period, since it's dancing so close to escape velocity.

Orbital Inclination: 80 degrees
KBOs don't have such high inclinations. I'd just call it a comet.

I'm trying to keep the principles as real-world as I can. Though I will say that the issues of heavy lifting are not a problem, nor is building a craft small enough and fast enough to get to the 'roid in a reasonable timespan.
A ship the size of the space shuttle is a different animal from a 30km+ asteroid after all.
How do you define a "reasonable timespan"? If the comet is 13 years away, and the orbital period is 4900 years, then it's going to be screaming inward at nearly solar escape velocity. I'll have to do more detailed calculations to figure it out, but I'd guesstimate it taking at least four or five years to reach it using VASIMR level technology.

antoniseb
2010-Jul-21, 04:29 PM
... Worse still, it's is extremely unlikely for the 4900 year period to be stable. Since the perihelion is so close to the Sun, it will experience a lot of thrust from cometary effects. ...
One additional thing for this factor is that any trip to the inner solar system is likely to result in some kind of gravitational perturbation from Jupiter, which also screws up numbers for you, but if you mean in the story to only have one previous visit, that's not an issue.

Grey
2010-Jul-21, 06:55 PM
Per Isaac Kuo, we can assume that the object is inbound at pretty much solar escape velocity, with a highly elliptical orbit. If we do a few more calculations, we can figure out that if you want it to collide in about 13 years, that puts it at about 29 AU (a little closer than Neptune) when it's discovered, moving at about 7.8 km/s. There have been a few asteroids discovered at that range in the last few years (the Neptune Trojans (http://en.wikipedia.org/wiki/Neptune_trojan)), and they appear to range in size from about 40 km to about 200 km (with large uncertainties in those estimates). So it would be entirely reasonable for something of comparable size to be found with current technology, with one caveat. An inbound object would be coming nearly straight for us (remember that high eccentricity?), rather than moving across our field of vision. I think there would still be enough lateral motion to notice it, though.

IsaacKuo
2010-Jul-21, 07:16 PM
If we do a few more calculations, we can figure out that if you want it to collide in about 13 years, that puts it at about 29 AU (a little closer than Neptune) when it's discovered, moving at about 7.8 km/s.
How do you do that calculation? I was trying to do it in my head, and couldn't think of how you'd calculate distance from the Sun vs time from perihelion for a parabolic trajectory.

An inbound object would be coming nearly straight for us (remember that high eccentricity?), rather than moving across our field of vision. I think there would still be enough lateral motion to notice it, though.
The lateral motion would be due to Earth's motion rather than the object's motion. Since this comet is on a collision course with Earth at perihelion, it must be incoming from roughly the opposite direction of the collision point.

So, let's say the comet is set to collide with Earth in the middle of winter at perihelion. That means that during Winter, the comet is hard to view because of the Sun. During Spring and Fall, there is very little lateral motion because Earth is moving "toward" or "away" from the comet. However, during Summer the comet is both viewable at night and there is lateral motion due to Earth's motion.

The lateral motion is about 30km/s at 29AU away (4.3 billion km away), or about 2 arcminutes per day.

SRQHivemind
2010-Jul-22, 12:44 AM
to clarify details:
The 47AU figure is random figure.
The inclination is subject to change.

The only points that matter in the long run are:
1: Orbital Period (4900 years)
2: Intercepts Earth in January, 2013.
3: 30km, big, dense rock.
4: Modern-day astronomy technology hasn't picked it up yet (given the replies, it's squarely in "plausible" territory)

The 4900 year period has been calculated meticulously. There was someone around in that day and age who observed the last close-approach up-close and personal who's still with us. (long story, remember this is a science fiction story, and yes, it all makes sense in context)

I'm just trying to get the orbital mechanics and ship performance details as close to plausible as possible (think 'B5', 'Firefly' and 'BSG' here, newtonian physics + handwavium)

The ship being sent to intercept the rock has four engines with total thrust of 35MN between them. (an inertial-dampening handwave is in effect so as to not impart severe G's on the payload and crew).

Assume a single engine to push the rock is the same design in use aboard ship:
Max output: 8.75MN, helium-fueled plasma rocket (fusion reactors, superconductors and exotic nanomaterials are in play so just run with the 8.75 MN figure)
pushing a rock that is mostly uranium. (someone have numbers on hos massive said rock might be?), how long is the burn time to nudge it out of Earth's path (close shaves are acceptable, as long as it doesn't actually hit the surface)

Peter B
2010-Jul-22, 10:31 AM
to clarify details:
The 47AU figure is random figure.
The inclination is subject to change.

The only points that matter in the long run are:
1: Orbital Period (4900 years)
2: Intercepts Earth in January, 2013.
3: 30km, big, dense rock.
4: Modern-day astronomy technology hasn't picked it up yet (given the replies, it's squarely in "plausible" territory)

The 4900 year period has been calculated meticulously. There was someone around in that day and age who observed the last close-approach up-close and personal who's still with us. (long story, remember this is a science fiction story, and yes, it all makes sense in context)

I'm just trying to get the orbital mechanics and ship performance details as close to plausible as possible (think 'B5', 'Firefly' and 'BSG' here, newtonian physics + handwavium)

The ship being sent to intercept the rock has four engines with total thrust of 35MN between them. (an inertial-dampening handwave is in effect so as to not impart severe G's on the payload and crew).
Well, the G-forces will depend on the mass of the spacecraft as well as its thrust. Do you know how massive the spacecraft is? You mightn't need your inertial dampers, and given the story is set only three years in the future, that might be something worth dropping from the story.


Assume a single engine to push the rock is the same design in use aboard ship:
Max output: 8.75MN, helium-fueled plasma rocket (fusion reactors, superconductors and exotic nanomaterials are in play so just run with the 8.75 MN figure)
pushing a rock that is mostly uranium. (someone have numbers on hos massive said rock might be?), how long is the burn time to nudge it out of Earth's path (close shaves are acceptable, as long as it doesn't actually hit the surface)
Well, uranium's density is about 19 grams per cubic centimetre - that is, about 19 times that of water. Assuming the object is roughly spherical, a diameter of 30 kilometres means its volume is about 14000 cubic kilometres. That's 14 trillion cubic metres. At 19 tonnes per cubic metre, that's about 268 trillion tonnes.

I don't know how to calculate accelerations using Newtons, so I'll leave that to others.

My other concern is why this object is mostly uranium. Iron perhaps, but how would a lump of uranium that large ever come into existence? Of course, if that's part of the story, I'll leave it alone.

Peter B
2010-Jul-22, 10:46 AM
Okay, I've just had a look at the Wikipedia page on the S-IC, the Saturn V's first stage. Apparently it produced about 33.4 meganewtons of thrust, which I know was about 3350 tonnes of thrust. So 8.75 meganewtons is about a quarter of this, or about 877 tonnes of thrust.

Now, at lift-off, because the Saturn V weighed about 3040 tonnes, meaning acceleration was 1.1 g.

If instead we divide 877 by 268,000,000,000,000, we get acceleration of 3.27 x 10^-12 g. Multiply that by 9.8 m/s/s, and I think it's about 3.21 x 10^-11 metres per second squared.

Plug that into the displacement formula of s = ut + 0.5at^2, and you can work out how far it'll move in a given period of seconds. In ten years, for example, I get the figure of 315,576,000 seconds, and therefore a displacement of 3,196,781 metres, or about 3200 kilometres. Now the Earth's diameter is about 12,700 kilometres, so that amount of shift is about half the maximum possible needed (that is, if the rock was aimed at the dead centre of the Earth).

Hopefully my sums are all correct, but if there are any errors, I'm sure someone will point them out.

Grey
2010-Jul-22, 01:45 PM
How do you do that calculation? I was trying to do it in my head, and couldn't think of how you'd calculate distance from the Sun vs time from perihelion for a parabolic trajectory.I confess that I didn't bother to worry about the parabolic trajectory. It's eccentric enough that I figured a straight line path would be a close enough approximation, and that I could assume it would be travelling at solar escape velocity throughout. And then I just used numerical analysis. I'm sure I could go back to the basic equations for the orbit and get a closed form solution, but I took the lazy path. :)


The lateral motion would be due to Earth's motion rather than the object's motion.Ah, of course. So no worries about detecting it at that range, and an object that size is right in line with the other things we're spotting in that region today.

Grey
2010-Jul-22, 02:06 PM
Hopefully my sums are all correct, but if there are any errors, I'm sure someone will point them out.Math looks about right to me. Looks like they'll need a bigger rocket, and an awful lot of fuel and reaction mass to run it for ten years. Of course, if the rock is mostly uranium, that gives you another option. Send equipment to enrich uranium and build yourself some fission bombs for your propulsion. Make sure it's all mostly automated, or you've got astronauts willing to go on a suicide mission, because I don't think they'll survive the radiation exposure of walking around on a giant hunk of mostly pure uranium.

SRQHivemind
2010-Jul-22, 02:35 PM
Thank you all for the responses. I think I have what I need to make this idea not trigger a "you fail astronomy forever" response amongst the Astronomy-savvy crowd.
I have other questions relating to orbital mechanics, but those are for other threads.

Thanks :)

IsaacKuo
2010-Jul-22, 03:41 PM
The only points that matter in the long run are:
1: Orbital Period (4900 years)
As noted before, this sort of orbital period would not be stable. A shorter orbital period would be more stable, and it would have the additional bonus of plausibly removing the volatiles out of the comet--turning it into a "extinct comet". Without any more outgassing of volatiles, the extinct comet's orbit will be hard to see.

I recommend using a dark comet with a shorter period...maybe between 200 and 300 years. Note that even though perihelion occurs once every 2XX years, the time period between close passes with Earth can be less frequent.

You could take inspiration from the comet109P/Swift-Tuttle (http://en.wikipedia.org/wiki/Comet_Swift-Tuttle). Note how the comet had indeed been sighted before, but connecting previous sightings is non-trivial.

2: Intercepts Earth in January, 2013.
Does this mean that the story takes place in the year 2000?

3: 30km, big, dense rock.
Conveniently, this is roughly the size of Swift-Tuttle. But your fictional extinct comet could be much harder to see--so it would take a very close pass to Earth to make observations (and the last time could have been 5000 years ago).

The 4900 year period has been calculated meticulously. There was someone around in that day and age who observed the last close-approach up-close and personal who's still with us. (long story, remember this is a science fiction story, and yes, it all makes sense in context)
Assuming a brilliant ancient astronomer were around back then, it's possible that he could have calculated the next time the extinct comet would pass near Earth (in the same way Swift-Tuttle is predicted to pass near Earth in 2469 years). This requires the extinct comet to be in a stable orbit which requires the orbital period to be much shorter.

Is it absolutely important that the object does not have an orbital period of, say, 272.76 years (1:23 resonance with Jupiter)? If the object is an extinct comet, the last time it would have been observable would have been in the year 922. The ancient astronomer could have gotten a good look at the object 17 orbits ago (2625BC, or 4637 years before impact in 2013).

I'm just trying to get the orbital mechanics and ship performance details as close to plausible as possible (think 'B5', 'Firefly' and 'BSG' here, newtonian physics + handwavium)
None of B5, Firefly, or BSG get orbital mechanics or ship performance anywhere close to plausible. You have already gone into more detail and thought about orbital mechanics than any of them (most terribly Firefly).

The ship being sent to intercept the rock has four engines with total thrust of 35MN between them. (an inertial-dampening handwave is in effect so as to not impart severe G's on the payload and crew).
There is no need for any intertial-dampening handwave. Severe G's are neither required nor desired for this spacecraft.

The question of thrust does not give much idea of the performance of the spacecraft. You would need to know the ship's mass to determine acceleration, although at this thrust level the acceleration will most likely be "enough". The limiting factor will be the specific impulse of the engines.

If we're talking about VASIMR type technology, then high thrust will mean low specific impulse. Which is bad. You will most likely prefer low thrust but high specific impulse. The basic idea is to exhaust the propellant with as high exhaust velocity as possible to get the most "bang for the buck". If you don't do this, then you will simply run out of propellant before you build up enough speed to reach the object in a reasonable time. You also need enough delta-v to kill off your velocity and reverse to land on the object.

I'll do calculations for the spacecraft later...

IsaacKuo
2010-Jul-22, 07:13 PM
Assuming a brilliant ancient astronomer were around back then, it's possible that he could have calculated the next time the extinct comet would pass near Earth (in the same way Swift-Tuttle is predicted to pass near Earth in 2469 years). This requires the extinct comet to be in a stable orbit which requires the orbital period to be much shorter.

Is it absolutely important that the object does not have an orbital period of, say, 272.76 years (1:23 resonance with Jupiter)? If the object is an extinct comet, the last time it would have been observable would have been in the year 922. The ancient astronomer could have gotten a good look at the object 17 orbits ago (2625BC, or 4637 years before impact in 2013).
Here are some more details for this suggested extinct comet, which I'll call "the dark comet":

Orbital period = 272.76 years (1:23 resonance with Jupiter)
Semi-major axis = 42AU
Perihelion = a bit less than 0.93AU
Aphelion = ~83AU

Inclination = 41 degrees (to make it even harder to see on the "off" years, as it travels around the Sun along with Earth)
Relative speed at impact is about 30km/s.

Year of impact (YOI) = 2013AD
Previous potential sighting = YOI-1091 = 922AD (4 orbits ago) --- closest approach of .3AU
Previous potential sighting = YOI-2182 = 170BC (8 orbits ago) --- closest approach of .6AU
Previous potential sighting = YOI-4637 = 2625BC (17 orbits ago) --- closest approach of .3AU

The ancient genius may have first done calculations on the object in 2625BC, and then refined them upon further observations in 170BC and 922AD.

SRQHivemind
2010-Jul-22, 07:51 PM
---
removed for being incomplete see below

SRQHivemind
2010-Jul-22, 07:55 PM
OK.
I've worked up more reasonable figures for the Big Rock based on what would be a world killer and what's easily deflected...
Diameter: 5km
Mass: 523,356,169,104 metric tons
Displacement required: 25,000 km (2 earth-diameters, for safety)
Engine output: 8.75MN
Acceleration rate: 1.6719 x 10^-5 m/s
Engine Burn time: 340 hours

Given Isaac's Orbital figures:
Orbital period = 272.76 years (1:23 resonance with Jupiter)
Semi-major axis = 42AU
Perihelion = a bit less than 0.93AU
Aphelion = ~83AU
Inclination = 41 degrees (to make it even harder to see on the "off" years, as it travels around the Sun along with Earth)
Relative speed at impact is about 30km/s.

That's a world-killer and easily deflectable given the engine specs available, there's even enough time to kill to adjust the orbit and have Earth's gravity capture it (or make it hit the moon)

Now my big question is, what would the dark comet's distance be at YOI - 9 (mission launch)

Is there an orbit plotting program I can plug these figures in to to get details?

SRQHivemind
2010-Jul-22, 08:07 PM
Edit my last (darn spam protection thing...). the orbital figures need to be adjusted based on the new mass.

Grey
2010-Jul-23, 03:31 PM
Edit my last (darn spam protection thing...). the orbital figures need to be adjusted based on the new mass.No worries. As long as the mass of the object is relatively small (i.e., significantly less than a planet), its own mass has very little effect on the orbital parameters. Unfortunately, I was wrong above when I said I thought there was a simple closed form solution of this problem; there isn't, so numerical analysis is the only way to go. For this much smaller orbit, my simplifying assumptions are probably too far off. Here (http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Position_as_a_ function_of_time) and here (http://spiff.rit.edu/classes/phys440/lectures/ellipse/ellipse.html) are the details of the calculation.

So, first we calculate the mean anomaly, M = 2 pi t / P. That one's easy, P is 272.76 years, and we want to know the distance at about 9 years from perihelion. M = 0.2073202.

Then we need to solve for the eccentric anomaly E in M = E - e sin(E), where e is the eccentricity, 0.9778387. Doing that numerically gives 1.06054.

We solve for the true anomaly theta, where tan(theta / 2) = sqrt((1 + e) / (1 - e)) tan(E / 2). That gives us theta (in radians) of 2.784.

Putting that into the basic orbit equation, r = p / (1 + e cos(theta)), gives us 21.9 AU for the distance from the Sun. Uranus is at an average of 19 AU, for reference.

My earlier simplification of straight line travel and moving at escape velocity would have given a distance of 23 AU when it's nine years out. So that wouldn't have been too far off, since this orbit is still quite eccentric.

IsaacKuo
2010-Jul-23, 10:48 PM
If I'm keeping up here, the goal is to get our spacecraft to the dark comet by YOI-9 (year of impact minus 9). The dark comet was detected on YOI-13, so the spacecraft needs to get there within at most 4 years.

The travel distance is about 22AU, so the average speed is 22AU/4years or 26km/s. It's more complicated than this, because the Sun's gravity will be slowing down the spacecraft on its way out.

Hmm...doing a simple numerical analysis assuming straight out motion, I get:

time:speed
4years: 46km/s
3years: 51km/s
2years: 63km/s
1.5yrs: 77km/s
1year: 107km/s

Subtracting Earth's own 30km/s, we get:

time:initial delta-v
4years: 16km/s
3years: 21km/s
2years: 33km/s
1.5yrs: 47km/s
1year: 77km/s

However, there's also the delta-v needed to match velocities and land on the object. The spacecraft needs to eliminate its own outward velocity plus the 9km/s inward velocity of the dark comet. So the total delta-v is:

4years: 16+22+9 = 47km/s
3years: 21+31+9 = 61km/s
2years: 33+48+9 = 90km/s
1.5yrs: 47+65+9 = 121km/s
1year: 77+99+9 = 185km/s

Assuming a VASIMR exhaust velocity of 50km/s, the mass ratio requirements would be:

4years: 2.6:1
3years: 3.4:1
2years: 6:1
1.5yrs: 11:1
1year: 41:1

The big issue is whether or not you can count on the dark comet to supply the fuel for deflection and to return back home. If so, then you could maybe afford one of the faster options, like 2years. If not, then...well, then you need to use a different strategy, like directly ramming the dark comet with kinetic impactors. With the direct impact strategy, you could use the 1 year strategy and slam the dark comet with impactors at 108km/s. At 108km/s, a 10 ton impactor hits with 60TJ of energy--the equivalent of the Hiroshima nuclear bomb.

SRQHivemind
2010-Jul-24, 08:03 AM
These figures are enough to work with.
The mass numbers for the rock were tweaked to make it movable given the tech at hand, yet big enough to cause major problems if it hits. A direct-impact is a good "plan B" :)

thanks.

Peter B
2010-Jul-24, 02:01 PM
One other thought occurs to me regarding the rocket used for the comet deflection...

If the comet is spinning (with respect to the Sun), then the rocket won't be able to fire all the time - it'll only consistently affect the comet's trajectory if the thrust is in a single direction (or at least a fairly narrow angle). Otherwise, if the rocket thrusts in all 360 degrees, the net effect will be zero.

Won't it?

IsaacKuo
2010-Jul-24, 02:32 PM
The rocket can thrust parallel to the spin axis, or it can de-spin the object if desired.

Peter B
2010-Jul-24, 03:24 PM
The rocket can thrust parallel to the spin axis, or it can de-spin the object if desired.

Ah. Point taken. But don't some smaller objects in the solar system have chaotic rotations? That is, they spin on multiple axes.

And how long would de-spinning take?

I suspect, though, the big showstopper is where the rocket's fuel is going to come from. That is, not just sourcing it, but refining it and putting it in the rocket's fuel tank. What are the tanking procedures for ion rockets these days? I assume, like car engines, refuelling while in operation is a Bad Idea. Could such a device be automated well enough that no further human intervention would be necessary? I mean, you'd feel like a bit of a dill if the device suffered some sort of failure - fairly likely I'd have thought given its operating environment.

Cougar
2010-Jul-24, 05:00 PM
newtonian physics + handwavium

Ha ha. Hey, that's good.

Cougar
2010-Jul-24, 05:41 PM
Diameter: 5km

I guess this would do some damage. Check out your experience if you were watching the impact from 500 km away, given the parameters shown:





Parameters:
Distance from Impact: 500.00 km ( = 311.00 miles )
Projectile diameter: 5.00 km ( = 3.11 miles )
Projectile Density: 8000 kg/m3 (iron, reportedly)
Impact Velocity: 60.00 km per second ( = 37.30 miles per second )
Impact Angle: 80 degrees
Target Density: 2750 kg/m3 (Crystalline Rock)

Energy:
Energy before atmospheric entry: 9.42 x 10^23 Joules = 2.25 x 108 MegaTons TNT
Average interval between Earth impacts of this size during the last 4 billion years = 3.0 x 108years

Major Global Changes:
The Earth is not strongly disturbed by the impact and loses negligible mass.
The impact does not make a noticeable change in the tilt of Earth's axis (< 5 hundreths of a degree).
The impact does not shift the Earth's orbit noticeably.

Crater Dimensions:
Transient Crater Diameter: 97 km ( = 60.2 miles )
Transient Crater Depth: 34.3 km ( = 21.3 miles )
Final Crater Diameter: 177 km ( = 110 miles )
Final Crater Depth: 1.41 km ( = 0.873 miles )
The crater formed is a complex crater.
The volume of the target melted or vaporized is 8260 km3 = 1980 miles3
Roughly half the melt remains in the crater, where its average thickness is 1.12 km ( = 0.694 miles ).

Thermal Radiation:
Time for maximum radiation: 3.27 seconds after impact
Visible fireball radius: 176 km ( = 110 miles )
The fireball appears 80.2 times larger than the sun
Thermal Exposure: 1.57 x 109 Joules/m2
Duration of Irradiation: 42.5 minutes
Radiant flux (relative to the sun): 616

Effects of Thermal Radiation:
Clothing ignites
Much of the body suffers third degree burns
Newspaper ignites
Plywood flames
Deciduous trees ignite
Grass ignites

Seismic Effects:
The major seismic shaking will arrive approximately 1.67 minutes after impact.
Richter Scale Magnitude: 10.2 (This is greater than any earthquake in recorded history)
Mercalli Scale Intensity at a distance of 500 km:
VII. Damage negligible in buildings of good design and construction; slight to moderate in well-built ordinary structures; considerable damage in poorly built or badly designed structures; some chimneys broken.
VIII. Damage slight in specially designed structures; considerable damage in ordinary substantial buildings with partial collapse. Damage great in poorly built structures. Fall of chimneys, factory stacks, columns, monuments, walls. Heavy furniture overturned.

Ejecta:
The ejecta will arrive approximately 5.57 minutes after the impact.
At your position there is a fine dusting of ejecta with occasional larger fragments
Average Ejecta Thickness: 6.32 meters ( = 20.7 feet )
Mean Fragment Diameter: 1.71 cm ( = 0.673 inches )

Air Blast:
The air blast will arrive approximately 25.3 minutes after impact.
Peak Overpressure: 1.09e+06 Pa = 10.9 bars = 154 psi
Max wind velocity: 798 m/s = 1790 mph
Sound Intensity: 121 dB (Dangerously Loud)
Damage Description:
Multistory wall-bearing buildings will collapse.
Wood frame buildings will almost completely collapse.
Multistory steel-framed office-type buildings will suffer extreme frame distortion, incipient collapse.
Highway truss bridges will collapse.
Highway girder bridges will collapse.
Glass windows will shatter.
Cars and trucks will be largely displaced and grossly distorted and will require rebuilding before use.
Up to 90 percent of trees blown down; remainder stripped of branches and leaves.


Input your own parameters here. (http://impact.ese.ic.ac.uk/ImpactEffects/?dist=20&diam=5&pdens=8000&pdens_select=0&vel=size%3D10&theta=30&tdens=2500&tdens_select=0) :cool:

SRQHivemind
2010-Jul-24, 09:11 PM
Nice way to ruin someone's afternoon...
The link is broken, by the way.

Cougar
2010-Jul-25, 12:13 PM
Nice way to ruin someone's afternoon...
The link is broken, by the way.

Hmm. Try this.

http://impact.ese.ic.ac.uk/ImpactEffects/

IsaacKuo
2010-Jul-25, 01:07 PM
I suspect, though, the big showstopper is where the rocket's fuel is going to come from.
Oh, forgot to answer this--the only plausible possibility is to get reaction mass from the target itself. If you need to bring the propellant with you, then it makes far more sense to use direct impactors instead. With an impact velocity of 40km/s or more (the three year plan), you get an order of magnitude more deflection from direct impacts rather than landing and thrusting. The difference is two or more orders of magnitude for the faster plans.

caveman1917
2010-Jul-25, 07:20 PM
Just a thought, depending on when (in the year) it will hit. If it is calculated to hit at a time when its inbound trajectory is ~tangential to earth's motion, could it be that it requires less work if we try to speed it up / slow it down to make it miss, instead of deflecting sideways? In the best scenario, with earth moving 30km/s, changing impact time by a mere 5mins would make it miss.