Mr Skeptic Posted March 24, 2009 Posted March 24, 2009 Well if units give you a hard time, just type the equation into google search. So long as you type it in a way it recognizes as a math problem, google calculator will give you the result and will take care of the units for you.
moth Posted March 24, 2009 Posted March 24, 2009 (edited) for a 50 meter solid iron meteor traveling at 12 km/sec the volume is [math] \frac{4}{3}*pi*2500^3 = 65449846949.787cm^3[/math] the mass is about [math]volume*7.5\frac{g}{cm^3}= 490873852123.4grams[/math] the kinetic energy is [math] \frac{mv^2}{2}=\frac{490873852.1234kilograms*(12000\frac{m}{s})^2}{2}=35,342,917,352,885,173.9joules[/math] i think:confused: Edited March 24, 2009 by moth divide by 2
Airbrush Posted March 24, 2009 Posted March 24, 2009 (edited) Thanks for the help folks. I like the looks of Moth's formulas which yield a number of over 35 quadrillion joules which convert to over 8 megatons. http://www.unitconversion.org/energy/joules-to-megatons-conversion.html Those numbers come from the Barringer Crater impact of 50,000 years ago. "The object that excavated the crater was a nickel-iron meteorite about 50 meters (54 yards) across, which impacted the plain at a speed of several kilometers per second. The speed of the impact has been a subject of some debate. Modelling initially suggested that the meteorite struck at a speed of up to 20 kilometers per second (45,000 mph), but more recent research suggests the impact was substantially slower, at 12.8 kilometers per second (28,600 mph). It is believed that about half of the impactor's 300,000 tonnes (330,000 short tons) bulk was vaporized during its descent, before it hit the ground. "The meteor hit the ground at an 80 degree angle from the north or northeast and it is theorized that the bulk of the remaining unvaporized 150,000 tons of the meteorite is under the crater's south rim which shows signs of uplift. The last major mining effort to recover the meteorite in that area was abandoned in 1929. "The impact produced a massive explosion equivalent to at least 2.5 megatons of TNT – equivalent to a large thermonuclear explosion and about 150 times the yield of the atomic bombs used at Hiroshima and Nagasaki. The explosion dug out 175 million tons of rock. The shock of impact propagated as a hemispherical shock wave that blasted the rock down and outward from the point of impact, forming the crater. Much more impact energy, equivalent to an estimated 6.5 megatons, was released into the atmosphere and generated a devastating above-ground shockwave..." http://en.wikipedia.org/wiki/Barringer_Crater Maybe because of the 6.5 megatons released into the atmosphere (whatever that means), plus the 2.5 megaton explosion, add up to 9 megatons, which is close to the number Moth calculated. "The blast and thermal energy released by the impact would certainly have been lethal to living creatures within a wide area. All life within a radius of three to four kilometers (1.9-2.5 miles) would have been killed immediately. The impact produced a fireball hot enough to cause severe flash burns at a range of up to 10 km (7 miles). A shock wave moving out at 2,000 km/h (1,200 mph) leveled everything within a radius of 14-22 km (8.5-13.5 miles), dissipating to hurricane-force winds that persisted to a radius of 40 km (25 miles)." Edited March 24, 2009 by Airbrush
moth Posted March 24, 2009 Posted March 24, 2009 that's encouraging, if you can look up a number that's close, i may not be too far off. the main thing (as mr.Skeptic said) is to keep all the units straight. the 6.5 megatons t.n.t. of energy to the atmosphere was probably given off as sound and light(heat) i guess?
Airbrush Posted March 24, 2009 Posted March 24, 2009 (edited) I don't know why they would specify different energy releases. Have you ever heard of two components to an explosion? That thing hit the ground, plain and simple. It was not an air burst. It burned off about half it's mass (165,000 tons of 330,000 tons) on the way down. I wonder how that data affects the calculation? Edited March 24, 2009 by Airbrush
moth Posted March 25, 2009 Posted March 25, 2009 if you imagine the energy being released over several seconds and smeared across the sky as a quick burst, it would add up to an explosion of that size. if it hit the seafloor, you could split the energy into three parts, air water, and ground since they all have different densities. the numbers from before are just the impact on the ground, ignoring the air.
Airbrush Posted March 26, 2009 Posted March 26, 2009 (edited) Refer to my earlier post. Kinetic energy, energy of motion, is converted to thermal energy and shock waves. Remember the equation is Energy = 0.5 x mass x Velocity ^2.The objects are travelling very fast - typically 20km/sec. Square that an d you get a very big number. At that speed, hitting the atmosphere for a small object is like hitting a brick wall. The object explodes violently. You say you can't imagine this. Don't. I'll assume you have Excel or some similar spreadsheet program. Create a small routine where you enter the diameter of the asteroid and its density. (Volume of a sphere = 4/ pi r^2). I used these values for density. Iron7.5 Stony-Iron4.5 Chondrite3.5 Carbonaceous Chondrite2.95 You can then calculate the mass of the asteroid. Offere a range of velocities, use the equation I gave above and there you have your energies. You no longer have to imagine, you can see what happens when one or other variable changes. You hear that folks? Since 70-80% of meteoroids are stony, when playing with the kinetic energy formula use the Stony-Iron density value of 4.5 to calculate energy released in the majority of impacts. The cool thing about the "Energy = 0.5 X Mass X Velocity-Squared" equation is that it looks a lot like Einstein's famous "E = mc^2". I never realized that until now. Why "times 0.5"? From wikipedia I get volume for a sphere as V = 4/3 X pi X r^3. Edited March 26, 2009 by Airbrush
lakmilis Posted March 27, 2009 Posted March 27, 2009 (edited) If the world was going to end in a year from now, would it be possible to supress that knowledge from the general public? It would result in total anarchy and social break down. In all honesty, If we were to be wiped out a year on, I'd personally love some mayhem and anarchy before going down ;p Nothing worse than finally quitting smoking, deciding to change your life in an exciting way, then when you start seeing the atmosphere turning red, the telly going: ok, this burning rock in the sky is about to destroy us.. sorry, we couldn't tell you because we didn't want anarchy before extinction..... DOH No, more to the point. ethically , it's insane that such things aren't informed. I till this day, can not fathom this idea that people panic lol. When sh** hits the fan, do people nto help out others? They act in panic, not shrivel up. If we had one year to try and survive as a species we'd need to frikkin know. So we could start digging racoon holes right away ^^ Merged post follows: Consecutive posts merged The main threat from the smaller ones would be tsunamis from an ocean impact. I propose that everyone living near the ocean should have an underground tsunamic shelter. You beachfront dwellers should start digging. I HAVE TO START LAUGHING... nice one.. one gets flooded with 12 meters of water above ya... ye.. get into an underground shelter... with a periscope to check when the water is gone.. PS... take a DEEEEP LOONG breath before going down Merged post follows: Consecutive posts mergedTrue. I wouldn't hesitate for a minute to take the kids to Yellowstone. A Yellowstone eruption would be a devistating event, but the threat from space COULD be an exstinction level event. Erm and so *would* a supervolcano eruption be, like Yellowstone or near Gibraltar ones. Merged post follows: Consecutive posts mergedlol ok airbrush.. I assume you are a high school pupil ye? Anyway.. now you know you yanks why the imperial system you use from the English just is useless. Why 0.5 ? Lol , it's called kinetic energy from Newtonian (classical mechanics kid): E = 1/2*m*v^2 . E = mc^2 is a measure of internal energy stored in mass (it is more a form of 'potential' energy than kinetic, loosely speaking). Anyway, about conversions. Stick to the base unit... So if something is given in g/cm^3 or so, to get it in kg/m^3 you would convert X g = X*E-03 kg / (10E-2 m)^3.. etc... when you finally get your unit in base values (m/kg^3) , you can then convert to other things... (when you get used to the units, you can save a lot of time by knowing which relationships algebraically are equivalent.. but it's a start). Edited March 27, 2009 by lakmilis Consecutive posts merged.
Airbrush Posted March 27, 2009 Posted March 27, 2009 Why 0.5 ? Lol , it's called kinetic energy from Newtonian (classical mechanics kid): E = 1/2*m*v^2 . E = mc^2 is a measure of internal energy stored in mass (it is more a form of 'potential' energy than kinetic, loosely speaking). Thanks for all your comments Lakmilis. Interesting subject. That is a beautiful comparison between Einstein's equation and the equation for kinetic energy. The energy stored in a given mass is exactly double the kinetic energy of that mass moving at light speed! Awesome revelation for me.
lakmilis Posted March 27, 2009 Posted March 27, 2009 (edited) No , it is not. First of all, no inertial mass (hmm in another topic someone states photons have some inertial mass but ok, conventional mass then) can not achieve the speed of light (infinite kinetic energy). Again, as stated, classical kinetic formula works for v << c (less than 0.1 c lets say as some arbitrary boundary). Merged post follows: Consecutive posts mergedThis by the way one is taught in last year of highschool (well, at least in Europe). However, all A-level books from around the world, use the wrong way of deriving the Swharzchild radius for example.. and uses classical escape velocity which wouldn't be the case when of relativistic escape velocities. It just happens to be that using classical kinetic energy , yields 2GM/c^2 (whereas special relativity predicts GM/c^2 so half the product). In GR however, through proper tensors, one gets the 2GM/c^2 again. This is something I pointed out (the half product from SR) when I did highschool physics but only to my teacher. No books have changed it or commented on it as I know of they still do it sigh. However, I am also an opponent of the S.R. as no black holes (bar perhaps SMBHs? although pfff) have no momentum, hence extreme Kerr holes I think are the real natural ones... i.e. GM/c^2). This however will have little consequence for centuries probably till we actually might be able to interact with them somehow. but anyway, this was off-topic. Merged post follows: Consecutive posts mergedPS. I don't know why I get doubleposted... really sorry..I clicked on that apostrophe once, don't know how ot use it and this has been happening since :/ Edited March 27, 2009 by lakmilis Consecutive posts merged.
Airbrush Posted March 28, 2009 Posted March 28, 2009 ...First of all, no inertial mass (hmm in another topic someone states photons have some inertial mass but ok, conventional mass then) can not achieve the speed of light (infinite kinetic energy).:/ This is not about conventional mass achieving the speed of light, this is about the striking similarity of E=mc^2 and Kinetic E=0.5mv^2. Energy contained in a mass is double its' own mass moving at light speed by the equation simply by substituting v with c. OK, matter cannot reach light speed, but the light energy equivalent of a given amount of matter does travel at light speed. So what's the problem?
Sisyphus Posted March 28, 2009 Posted March 28, 2009 (1/2)mv^2 is only applicable in classical systems, i.e. at insignificant fractions of light speed. Kinetic energy at relativistic speeds is [math]\frac{mc^2}{\sqrt{1-(\frac{v}{c})^2}}-mc^2[/math]
Airbrush Posted April 23, 2009 Posted April 23, 2009 (edited) To me it seems like there are 2 basic kinds of asteroid deflection methods. The direct method (kinetic impactor or nuclear explosion in close proximity), and the indirect method (lazer beam or particle cannons, solar sails, gravity tractors, etc). The direct method is short-range (or maybe even long-range) cheaper and quicker, a possible last resort to deal with short-notice asteroids. The indirect methods are preferable, but they cost more and take much, much more time, not only to develop but also to reach its' more distant destination. I like the simplicity of the kinetic impactor, especially if you can hit the object years before it comes near Earth. It wouldn't matter where you hit the object or even if you break it into pieces, because even a small change in course will cause it, and all the pieces of it, to miss Earth by a wide margin. With the direct method all you need is the mass and inertia of the rocket on an intercepting path at high closing speed, but the indirect method requires acceleration towards the asteroid, then a major course change to reverse direction to match the speed and path of the asteroid so as to fly alongside it, or gently "dock" with it to attach devices, or hover at a safe distance and do it's thing. I further propose that the indirect method spacecraft be multi-purpose. We need to test the object to find out its' composition and how solid it is, report back home, then we select from its' "quiver" from a number of on-board deflection techniques appropriate to the particular bolide. Edited April 23, 2009 by Airbrush
SH3RL0CK Posted April 23, 2009 Posted April 23, 2009 I don't necessarily see why the "indirect" method would be more expensive than the "direct" method. Consider a gravity tractor consisting of a small probe placed near the asteroid. The technology to do this already exists (unlike some of the other proposals), the math behind this is well understood (unlike some of the other proposals), and it is known for certain this will work (unlike some of the other proposals) without any unpleasant side-effects such as turning on asteroid into hundreds (some of which might possibly still be on a collision course with earth). The above makes me think an indirect gravity tractor wouldn't be terribly expensive (particularly since the research and development costs for the technology would be negligible). We currently have many interplanetary probes which are active, whats one more? Sure, this small probe may take years of time before it moves the asteroid sufficiently far to be safe...but if years of time are available that doesn't matter.
Airbrush Posted April 23, 2009 Posted April 23, 2009 (edited) You make a good point Sherlock about the gravity tractor, which will be effective on lower mass objects. Its' task is so simple that it won't require much coaching from home. It just senses were the object is and adjusts it's distance automatically. It would have to fire its' rockets in small bursts, pointed at angles away from the object, so as not to push the object away by the propulsion system, and also rockets on the opposite side to stop it at the correct distance. It may require a few bursts per hour, or day, or week? over a period of years. What kind of propulsion system can handle that long-term schedule? For giant ones we need a more aggressive method or a combination of different methods. Merged post follows: Consecutive posts mergedI don't know why they would specify different energy releases. Have you ever heard of two components to an explosion? That thing hit the ground, plain and simple. It was not an air burst. It burned off about half it's mass (165,000 tons of 330,000 tons) on the way down. I wonder how that data affects the calculation? I just saw a program about this impact on History Channel and they said the mass that vaporized upon impact was about 300,000 tons. Edited April 23, 2009 by Airbrush Consecutive posts merged.
SH3RL0CK Posted April 23, 2009 Posted April 23, 2009 It may require a few bursts per hour, or day, or week? over a period of years. What kind of propulsion system can handle that long-term schedule? For giant ones we need a more aggressive method or a combination of different methods. Those are good points as well. For the gravity tractor to work, it might require multiple probes, or some kind of refueling rocket in order to provide sufficient fuel to do the trick... Of course one key point in all this is that the more time available to address the situation, the less work has to be done overall. This is regardless of the actual method used to change the orbital trajectory of the object. Thus, it is critical to identify asteroids on a collision course with earth long before (I'm talking decades or even generations) before the collision will occur. This is where I would spend the research dollars first.
Mr Skeptic Posted April 23, 2009 Posted April 23, 2009 Another thing is you could have a kinetic impactor with a harpoon. As it flies by, it could harpoon the asteroid, and then, depending on how long the rope was, might decelerate slowly. Of course, then you have the danger of missing with the harpoon, the harpoon being loose, or harpooning a loose piece.
Airbrush Posted April 23, 2009 Posted April 23, 2009 (edited) For the gravity tractor to work, it might require multiple probes, or some kind of refueling rocket in order to provide sufficient fuel to do the trick... I may be wrong but I think there is a way to collect solar energy from solar cells to charge batteries that can send a pulse to the rockets of the gravity tractor. Kinetic impactors can be sent in numbers, so if one fails others can follow thru with the mission. There must be a way to blunt the impact so you don't break the asteroid into pieces, like a conventional or nuclear explosion just a second before impact, or the probe flattens out to splat against the asteroid. Or rather than a harpoon, a giant net is blasted around the front of the object and the probe swings around the other side attached by an elastic tether to pull on it. After the initial pull the probe will bounce back to the object which could be a problem, so the rebound impact could be avoided by cutting the probe loose after the pull. This would be tricky because the closing speed will be 10 to 20 miles per second. I'm trying to keep the technology as simple as possible. Of course, any of these methods should be tested in advance on non-threatening asteroids. Edited April 23, 2009 by Airbrush
Mr Skeptic Posted April 23, 2009 Posted April 23, 2009 No need for the tether to be elastic. Just wind it around a spool, and let it unwind slowly. Actually, if as you said you can get the impactor to bounce back, this would be like having extra mass (so long as you didn't hit the asteroid on the bounce). This could also be accomplished with the spool, if it had a motor to wind it and a thruster to miss on the way back.
SH3RL0CK Posted April 24, 2009 Posted April 24, 2009 I may be wrong but I think there is a way to collect solar energy from solar cells to charge batteries that can send a pulse to the rockets of the gravity tractor. Sure, you can gain energy from solar cells, but you need more than energy. Perhaps you are thinking of ion-thruster types of rockets which are far more efficient that chemical rockets? http://en.wikipedia.org/wiki/Ion_rocket As I understand it, the reason these are more efficient is because of the higher velocities of the ejected propellant and since E = 1/2mv2; high velocities for the propellant are highly desired in a rocket. But you still need to eject some kind of mass (which will eventually require refueling) or there is no net motion.
Klaynos Posted April 24, 2009 Posted April 24, 2009 It's laser not lazer. Light amplification by stimulated emission of radiation.
dragonstar57 Posted August 10, 2010 Posted August 10, 2010 (edited) I think the best way to defend against such NEOs would be this gravity is warping of space time right? so what if it could be warped in the opposite direction it would make a repulsion gravity ring/sphere around the earth yeah its a little off topic but i interpreted the topic as general neo detection/impact prevention Edited August 10, 2010 by cipher510
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