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A math question about acceleration over a distance
Let us say that you had a 62,000-mile-long Earth-anchored space
elevator and let us say it has an electro-magnetic repulsion accelerator along its entire length. If you were to accelerate a human so that he experienced only 2g (twice the force of gravity) during the entire length of the space elevator, what velocity would that human be expelled from the space elevator and how long would it take the human to travel the entire length? Where I'm having problems with the above question is the need for the human to be constantly increasing in speed during the entire length of the space elevator. As soon as the human goes at the same speed, the sensation of two gravities would stop. 32 feet (9.8 meters) per second squared = 1 gravity So 64 feet per second squared = 2 gravities? But that's for the first second. For the second second, the human would have to be going at 128 feet per second squared to continue to feel the force of two gravities, right? If the above question isn't hard enough, there are two other issues to deal with: 1) when the human is on the Earth's surface, there is also the Earth's gravity but as the human goes up the space elevator, that gravitational force decreases. This would mean a slower take-off speed so the human only experiences 2gs at all times. 2) once the human passes the 22,000 mile mark, centrifugal force begins to push the human away from the Earth so the further away from the 22,000-mile point, the more outward force is being felt by the human and his acceleration would have to be increased even more so that the human would feel 2gs in the opposite direction he is traveling. So without the impact of Earth's gravity and then centrifugal force past 22,000 miles, how fast would a human be traveling at the end of 62,000 miles and how long would he take to travel that 62,000 miles? Now if someone knows how Earth's gravity and post-22,000-mile-point centrifugal force would change the previous answer, I would love to know that too. Lastly, is there an equation for calculating the above? If so, I would appreciate knowing it since I could then enter any length and see what the final velocity would be. Thanks in advance! Scott Jensen |
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A math question about acceleration over a distance
On Sat, 31 Dec 2011 22:49:23 -0500, STJensen wrote:
Let us say that you had a 62,000-mile-long Earth-anchored space elevator and let us say it has an electro-magnetic repulsion accelerator along its entire length. If you were to accelerate a human so that he experienced only 2g (twice the force of gravity) during the entire length of the space elevator, what velocity would that human be expelled from the space elevator and how long would it take the human to travel the entire length? I've snipped some paragraphs where you confuse velocity and acceleration If the above question isn't hard enough, there are two other issues to deal with: [Snip re gravity as function of height and re centrifugal force. I think your take on centrifugal force treats it as a source of free energy, which so far as I know it isn't.] Lastly, is there an equation for calculating the above? If so, I would appreciate knowing it since I could then enter any length and see what the final velocity would be. To compute v(t) you need the value of the integral of a(t) dt. Ignoring centrifugal force, a(t) is 2*g0 - g0*(re/(re+h(t)))^2. [See 1] For h(t) you need the value of the integral of v(t) dt. In short, v(t) depends on a(t), which depends on h(t), which depends on v(t). I don't know whether the integrals have closed forms, but a simple program can give adequately-accurate results as shown by testing the following Python program with a range of values for dt. I tried several dt in the range 0.001 to 0.1 and for each got the same results within 4 decimal places. g0=9.80665 # surface gravity, m/s^2 re=6371000. # mean earth radius, m dt=0.05 # time increment, sec hmax=100000000. # elevator height, m v=0; h=0; t=0 while hhmax: .... h += v*dt .... p = re/(re+h) .... v += (2*g0 - g0*p*p)*dt .... t += dt .... print "At t=%f, v=%f m/s" % (t,v) At t=3548.900000, v=61686.788167 m/s Anyway, suppose this 100,000-km elevator goes straight up (which probably is a bad assumption, but no other is handy). Rotational speed of a stationary object at Earth surface is (2*pi*re)/(24*3600) m/s or 463 m/second. At the top, (2*pi*(re+h))/(24*3600) is about 7700 m/sec. This change in velocity over about an hour amounts to an 0.2 g acceleration the program doesn't account for, because it only looks at radial velocity. That may introduce about 10% error. The program also ignores centrifugal force. [1] The formula for gravity-versus-altitude is from wikipedia at http://en.wikipedia.org/wiki/Gravity_of_Earth#Altitude, g0 is surface gravity, and re = 6371 km = mean radius of Earth. -- jiw |
#3
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A math question about acceleration over a distance
On Sat, 31 Dec 2011, STJensen wrote:
Let us say that you had a 62,000-mile-long Earth-anchored space elevator and let us say it has an electro-magnetic repulsion accelerator along its entire length. If you were to accelerate a human so that he experienced only 2g (twice the force of gravity) during the entire length of the space elevator, what velocity would that human be expelled from the space elevator and how long would it take the human to travel the entire length? If you disregard angular momentum by assuming the Earth doesn't rotate, air resistance and weakening of gravity by distance from center of Earth, the problem becomes: If a person accelerates at one g for 62,000 miles, what is the terminal speed and how long does it take to reach terminal speed? d = distance, v = velocity, a = acceleration Set d(0) = 0 = v(0), a = g. v = integral(0,t) a dt = at d = integral(0,t) v dt = integral gt dt = gt^2 / 2 2d = gt^2; t = sqr 2d/g (= 76 min) v = at (= 28 miles/sec, if I've done the calculations correctly.) -- If you disregard angular momentum by assuming the Earth doesn't rotate, and air resistance, then the time will be shorter, the terminal speed larger and the calculation more complex. r = radius of Earth; s = distance above earth Set s(0) = 0 = v(0). Force of gravity at s is gr^2 /(r + d)^2. (Is there a physicist in the house?) Net acceleration at s is 2g - gr^2 /(r + d)^2. 2g - gr^2 /(r + s)^2 = dv/dt = d^2 s/dt^2 = s" 2g(r + s)^2 - gr^2 = (r + s)^2 s" gr^2 + 4grs + 2gs^2 = (r + s)^2 s" If I set it up right, that's a hard to solve differential equation. ---- |
#4
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A math question about acceleration over a distance
On Dec 31 2011, 10:49 pm, STJensen
wrote: Let us say that you had a 62,000-mile-long Earth-anchored space elevator and let us say it has an electro-magnetic repulsion accelerator along its entire length. If you were to accelerate a human so that he experienced only 2g (twice the force of gravity) during the entire length of the space elevator, what velocity would that human be expelled from the space elevator and how long would it take the human to travel the entire length? The boost force accelerates upwards, aways from earth. Gravity attracts downwards, in the amount mg . The net force is then boost - mg. F = ma, so a = F/m = (boost - mg)/m = boost/m - g. The mg , gravity, depends on elevation, so calling g g_local is better. Then acceleration = boost/m - g_local . A boost per kilogram of 9.8 Newtons/kg at earth's surface means acceleration = 0. I think this boost per kilo would be called '1g' ... A boost (rocket force) per kilo of 19.6 Newtons/kg would seem like 2g, but produce an acceleration of boost/m - g_local = 19.6 Newtons/kg - g_local, so at "inifinity" acceleration ~= 19.6 Newtons/kg = 19.6 m/sec/sec . With a space elevator that's anchored to Earth, things that don't move relative to earth woul be Ok. But the human in the space-elevator, being in motion relative to earth, would be subject to a Coriolis force: http://en.wikipedia.org/wiki/Coriolis_effect . 62,000 miles is about 7.5 earth diameters. Dave Where I'm having problems with the above question is the need for the human to be constantly increasing in speed during the entire length of the space elevator. As soon as the human goes at the same speed, the sensation of two gravities would stop. 32 feet (9.8 meters) per second squared = 1 gravity So 64 feet per second squared = 2 gravities? But that's for the first second. For the second second, the human would have to be going at 128 feet per second squared to continue to feel the force of two gravities, right? If the above question isn't hard enough, there are two other issues to deal with: 1) when the human is on the Earth's surface, there is also the Earth's gravity but as the human goes up the space elevator, that gravitational force decreases. This would mean a slower take-off speed so the human only experiences 2gs at all times. 2) once the human passes the 22,000 mile mark, centrifugal force begins to push the human away from the Earth so the further away from the 22,000-mile point, the more outward force is being felt by the human and his acceleration would have to be increased even more so that the human would feel 2gs in the opposite direction he is traveling. So without the impact of Earth's gravity and then centrifugal force past 22,000 miles, how fast would a human be traveling at the end of 62,000 miles and how long would he take to travel that 62,000 miles? Now if someone knows how Earth's gravity and post-22,000-mile-point centrifugal force would change the previous answer, I would love to know that too. Lastly, is there an equation for calculating the above? If so, I would appreciate knowing it since I could then enter any length and see what the final velocity would be. Thanks in advance! Scott Jensen |
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A math question about acceleration over a distance
On Sat, 31 Dec 2011 22:49:23 EST, STJensen
wrote: Let us say that you had a 62,000-mile-long Earth-anchored space elevator and let us say it has an electro-magnetic repulsion accelerator along its entire length. If you were to accelerate a human so that he experienced only 2g (twice the force of gravity) during the entire length of the space elevator, what velocity would that human be expelled from the space elevator and how long would it take the human to travel the entire length? Where I'm having problems with the above question is the need for the human to be constantly increasing in speed during the entire length of the space elevator. As soon as the human goes at the same speed, the sensation of two gravities would stop. 32 feet (9.8 meters) per second squared = 1 gravity So 64 feet per second squared = 2 gravities? You used the phrase "experienced only 2g". A person standing in an unmoving elevator is experiencing 1g but his acceleration is 0 ft/sec^2 as these problems are normally phrased. So experiencing 2g is accelerating upward at 32 ft/sec^2. You also need to decide how realistic you want to be (in other words, how much simplification are you willing to accept). Do you want the acceleration to be a constant (32 ft/sec^2 or 64 ft/sec^2 or any other fixed value)? Or do you want the acceleration to always be related to the current value of g (as you move away from the earth, g gets smaller). Acceleration is a vector. Is the total acceleration 2g or only the vertical component? If you include the Earth's rotation, then the path traveled is a spiral which means the total length is more that just the 62,000 mile length of the cable. But that's for the first second. For the second second, the human would have to be going at 128 feet per second squared to continue to feel the force of two gravities, right? You are mixing velocity and acceleration. And velocity is not a step function. At a fixed acceleration of 64 ft/sec^2, at the end of 2 seconds the velocity will be 128 ft/sec but only at that instant. Since the acceleration is continuous, the velocity is constantly changing. Just before 2 seconds have elapsed, the velocity would be slightly less than 128 ft/sec. Just after, it would be slightly more. If the above question isn't hard enough, there are two other issues to deal with: 1) when the human is on the Earth's surface, there is also the Earth's gravity but as the human goes up the space elevator, that gravitational force decreases. This would mean a slower take-off speed so the human only experiences 2gs at all times. 2) once the Speed does not determine how many g's you feel. Otherwise airplane rides would reduce you to a puddle. Change in speed (which is the definition of acceleration) determines what you feel. human passes the 22,000 mile mark, centrifugal force begins to push the human away from the Earth so the further away from the 22,000-mile point, the more outward force is being felt by the human and his acceleration would have to be increased even more so that the human would feel 2gs in the opposite direction he is traveling. Now you need to start being rigorous about what you mean by "feeling 2g". If you mean that he feels twice as heavy as he would if the elevator were to stop, then the acceleration would gradually decrease until he reached synchronous orbit. If you mean his legs always feel like they are holding up twice his weight, or equivalently, assume he is standing on a spring-type scale (not a balance one) and it constantly registers twice his "normal weight", then the acceleration must gradually increase to compensate for the lower value of g at altitude. Then the question becomes is the scale oriented vertically (as it is on the surface) or is it oriented to the direction of travel. So without the impact of Earth's gravity and then centrifugal force past 22,000 miles, how fast would a human be traveling at the end of 62,000 miles and how long would he take to travel that 62,000 miles? In classical mechanics, for uniform linear acceleration a, velocity v, distance d, and time t: v = at d =.5at^2 With a set to 64 ft/sec^2 and d set to 62,000 miles, you should be have no trouble answering both questions. (Watch the units.) Now if someone knows how Earth's gravity and post-22,000-mile-point centrifugal force would change the previous answer, I would love to know that too. The above formulas are simplified versions of the integration resulting from the differential equations v = dd/dt a = dv/dt You can derive the instantaneous value of g using Newton's equation mg = GMm/d^2 where G is the gravitational constant, M is the mass of the Earth, m is the mass of the man, and d is the distance from the center of the Earth. Lastly, is there an equation for calculating the above? If so, I would appreciate knowing it since I could then enter any length and see what the final velocity would be. How much integral calculus are you familiar with? -- Remove del for email |
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A math question about acceleration over a distance
Let us see if I get what William Elliot, James Waldby, and David
Bernier are saying. Time it would take to go the entire 62,000 miles: Elliot: 76 minutes Waldby: 3548.9 If the time is in seconds, that would equal 59.1 minutes. Bernier: Sorry, I couldn't find a time. Velocity human would be traveling at end of space elevator: Elliot: 28 miles per second Waldby: 61,686.8 meters per second. I'm assuming you mean meters and not miles so that would be ~ 38 miles per second Bernier: 19.6 miles per second If the above is correct, why so much difference between the numbers? Scott Jensen |
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A math question about acceleration over a distance
On Sun, 01 Jan 2012 22:42:56 -0500, STJensen wrote:
Let us see if I get what William Elliot, James Waldby, and David Bernier are saying. Time it would take to go the entire 62,000 miles: Elliot: 76 minutes Waldby: 3548.9 If the time is in seconds, that would equal 59.1 minutes. Bernier: Sorry, I couldn't find a time. Velocity human would be traveling at end of space elevator: Elliot: 28 miles per second Waldby: 61,686.8 meters per second. I'm assuming you mean meters and not miles so that would be ~ 38 miles per second Bernier: 19.6 miles per second If the above is correct, why so much difference between the numbers? First, "the above" is partly wrong: David Bernier said nothing about 19.6 miles per second; he wrote 19.6 m/sec/sec, where m is meters, and said that that is the acceleration equal to 2*g. Second, the numbers William Elliot supplied are for the case where a uniform 1 g acceleration is applied for a little over 75 minutes, causing one to arrive at the top of the elevator at a speed of about 27.5 miles per second. The numbers are from s = (1/2)*a*t^2 and v=a*t with s = 62000 miles, a = g. If you apply a uniform 2 g acceleration (as suggested in Barry Schwarz's post) and experience about 3G's on takeoff and about 2G's near the top, you arrive at the top in about 53 minutes going 39 miles per second. In my solution, with about 2G's total all the way up it takes about 59 minutes and top velocity is about 38.3 miles per second. -- jiw |
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A math question about acceleration over a distance
On 2012-01-01, STJensen wrote:
So 64 feet per second squared = 2 gravities? Yes, though in a coordinate system based at the Earth's surface, a stationary object feels one gravity already. But that's for the first second. For the second second, the human would have to be going at 128 feet per second squared to continue to feel the force of two gravities, right? No. Feet per second squared is an acceleration, and 128 ft/s^2 would be approximately 4 gees. So without the impact of Earth's gravity and then centrifugal force past 22,000 miles, how fast would a human be traveling at the end of 62,000 miles and how long would he take to travel that 62,000 miles? Now if someone knows how Earth's gravity and post-22,000-mile-point centrifugal force would change the previous answer, I would love to know that too. You can calculate it from a work-done point of view: delta-E = Integral F . dl This is easiest if you treat the person as being subject to thrust (real) force, and gravitational and centrifugal pseudo-forces. We may assume that the Coriolis forces have negligible net effect, due to the fact that the person's path stays along the beanstalk with its vastly greater mass preventing significant sideways motion. Then the forces at a given radius from the center of earth r a thrust = m a, with a = 2 gees ~20 m/s^2; gravity = -m g (R/r)^2, with R = Earth's radius ~6.4 Mm; centrifugal = m w r^2, with w = rate of Earth's rotation ~73 urad/s. The range of r is from R to R + h, with h = 60,000 miles ~100 Mm). Then delta-E / m = Int (a - g (R/r)^2 + w^2 r) dr = a h + g R (1/(1+h/R) - 1) + (1/2) w^2 ((R+h)^2 - R^2). E = (1/2) m v^2 and starts at zero, so you can just substitute the above formula into v = sqrt(2 E/m). -- Tim |
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A math question about acceleration over a distance
Thanks to all for the replies!
Scott Jensen |
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A math question about acceleration over a distance
In sci.math message
oglegroups.com, Sat, 31 Dec 2011 22:49:23, STJensen posted: Let us say that you had a 62,000-mile-long Earth-anchored space elevator and let us say it has an electro-magnetic repulsion accelerator along its entire length. If you were to accelerate a human so that he experienced only 2g (twice the force of gravity) during the entire length of the space elevator, what velocity would that human be expelled from the space elevator and how long would it take the human to travel the entire length? Since the elevator cable cannot be infinitely rigid, and will in practice be quite flexible, one will need to consider the effect of the sideways forces on its shape - unless the mass of a shortish length of the cable is large in comparison with that of your human and his accessories. -- (c) John Stockton, near London. Web http://www.merlyn.demon.co.uk/ - FAQish topics, acronyms, and links. Correct = 4-line sig. separator as above, a line precisely "-- " (RFC5536/7) Do not Mail News to me. Before a reply, quote with "" or " " (RFC5536/7) |
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