Escape velocity

Discussion in 'Physics & Math' started by plane, Mar 27, 2010.

  1. plane evolution Registered Senior Member

    Messages:
    115
    It’ll be stupid. Or perhaps you are. We are both human anyway.

    Your approach (to escape velocity) is to say potential energy is realized energy. It is only energy that will exist if a descent takes place. When you develop a theory that requires potential energy to be existing energy, just what are you doing?

    From the site that you say has it wrong.

    We say that the "Gravitational potential energy" of an apple is being converted to its kinetic energy as it falls.

    Are you disputing this.

    I have already (twice) defined potential energy in this way. It hasn’t drawn any comment or rebuke from you or anyone else.

    (Potential energy is an assessment of the energy that will be released if a descent is completed is what I posted when you questioned whether or not I believed in potential energy.)


    If a particle falls to the earth from as close to the earth as the moon, the earth’s gravity will accelerate it to over 15 km/sec by the time it reaches the earth. Well according to my calculations anyway doing it in 25,000 km stages.

    Which gives it a kinetic energy with a factor of 15 km/sec. Which is what it’s potential energy will have been before it descent began if its gravitational potential is being converted to kinetic energy as it falls.

    But according to escape velocity theory, once anything has a velocity of 11 km/sec going away from the earth, it’s potential energy is always less than it’s kinetic energy. Plainly it’s not if it can achieve a velocity of over 11 km/sec during a descent.

    Not sure if you will follow it or not but that’s all I’m saying to anyone.

    Integration assumes an end point. Infinite distance doesn’t have an end point. You will be adding up increments of F x S forever. Which will total an infinite amount of work done to get the particle from r to infinity. That’s indisputable unless you have found a way of making infinity sub infinite.
     
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  3. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    plane:

    I'm not sure what you mean by "realised energy". Energy is an accounting system. Viewed as a thing, it is like a substance that can be transformed from one form to another without being created or destroyed.

    Paying heed to the law of conservation of energy. Energy can't appear out of nowhere. It can only be converted from one form to another.

    It's a loose way of talking, but I'm not really disputing it. It is actually meaningless to talk about the "potential energy of an apple". The gravitational potential energy in question is a property of the system consisting of the apple and the Earth.

    When you're looking at the descent, if you released an object from rest an infinite distance from Earth (assuming there is nothing else in the universe), it would hit the ground with a speed of 11 km/s. If you fire it at the Earth with some speed, the 11 km/s will need to be added to that speed to determine the impact velocity.

    Do that in reverse and we find that an object launched from Earth with a speed of 11 km/s will reach infinity with zero speed. An object launched faster than that will reach infinity with speed left over. And an object launched with speed less than 11 km/s will be pulled back to Earth.

    Correct. It's a limiting process.

    Yes, but the further away from Earth you go, the smaller F becomes, so each increment is smaller and smaller and smaller as you go further away. Eventually, your increments are counting for practically nothing.

    No.

    Let me give you a simple mathematical example of why the sum of an infinite number of increments need not be infinite itself.

    Consider the following sum:

    1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...

    where is an infinite number of terms in the sum.

    I hope you will agree that the result of calculating the above sum is not infinity. It is, in fact, ONE.

    Do you agree?

    Your gravitational example is the same. Each term of the work/potential energy/kinetic energy sum gets smaller and smaller, and they add up to a finite total energy.
     
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  5. plane evolution Registered Senior Member

    Messages:
    115
    Good post JR. Enjoyed reading it. Good attitude.

    Could someone please put a link up to codes for posting x squared, stuff for showing inverse square law calculations. etc. Or just posts them. They are probably here and I can't see them for looking. (I dispute 11 km/sec from infinity to earth. See if I can do something about the matter tonight or tomorrow morning if armed with said codes).

    1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...


    Incidentally, the answer to this is infinity. 1 does not exist in this series. Never arrived at. 1 is in fact your theoretical infinity in this series.

    On the topic of escape velocity been trying to figure out why anybody would define potential energy as the work required to move a particle to infinity.

    As infinity is an indefinite, just seems a bizarre definition. Sure someone can give a reason.
     
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  7. BobG Registered Senior Member

    Messages:
    162
    Since the gravitational force has infinite range, a particle only escapes the gravitational force of another particle if it is an infinite distance away. And as James R shows this quantity is not infinite and has a finite value.

    Defining the potential energy this way is equivalent to defining the potential energy at infinity to be zero and for the potential energy at all finite distances to be negative. That way applying a force on the system to take it further away is carrying out positive work. And as the particle falls and increases it's velocity, the potential energy will be increasingly negative and the kinetic energy will be increasingly positive.

    You need to define some point as having a potential energy which you can compare the potential energy at all other points to so and in this case defining the zero potential to be at infinity is the most intuitive.
     
  8. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    Clearly you didn't take anything away from it, though.

    Just quote a previous post and look at the code. There's some in this thread.

    Everybody who's educated in this matter disagrees with you, but never mind.

    The answer is not infinity, and you asserting that it is infinity will never change that. You need to stop assuming you know some mathematics and actually learn some. Until then, there's little point in talking to you. You're arrogant and yet you know next to nothing. I'm happy to teach people who are willing and able to learn. My impression of you is that you're either incapable or unwilling, or both. If you don't improve rapidly, I'll write you off as another crank like Jack_ and not waste any more time on you.

    Also, 1 is not and never can be the same as infinity.
     
  9. CptBork Valued Senior Member

    Messages:
    6,460
    Alrighty then, I recommend you set about summing this series. I would say the result should always be less than one no matter how many terms you add, but because calculators only have finite precision, some weird effects might possibly come up and you might get 1.00000000 eventually because of this. Hence my recommendation is, if you think the sum is infinite, give us the number of terms it takes you to get past a total sum of 2, and we'll talk to you then.

    Because the amount of energy needed to move something an infinite distance away from Earth, in a universe where only the Earth's gravity matters, is virtually identical to the energy needed to carry objects a very large distance from Earth even in a universe where Earth isn't the only source of gravity. The limiting infinite case is also a bit easier and more convenient to calculate.
     
  10. 1100f Banned Registered Senior Member

    Messages:
    807
    According to your site mgh does not work for any distance.
     
  11. Fraggle Rocker Staff Member

    Messages:
    24,690
    This is not true. The upper limit of an integral can be plus infinity (I ain't messin' around with that damn symbol font!), the lower limit can be minus infinity, or both. There's nothing remarkable about a function which can be integrated from one infinity to the other and have a finite value.

    I don't think it will be algebraic, but setting up a trigonometric function with this property will probably be absurdly easy for one of you younger people whose math courses are fresher in mind.
     
    Last edited: Apr 6, 2010
  12. plane evolution Registered Senior Member

    Messages:
    115
    Bob G can not you see your contradiction in terms.

    Gravitational force has an infinite range.

    A particle only escapes the gravitational force of another particle if it is an infinite distance away.

    'shakes head'

    Now JR recommends (thanks for the quote advice JR) that a particle falling to the earth from an infinite distance away will hit the earth with a velocity of 11 km/sec.

    If a particle is about the moon distance away from the earth of 350,000 km, the rate of acceleration due to the earth's gravity is 0.0000033 km/s/s.

    6378\(^2\)/350,000\(^2\) x 9.8/1000 = 0.0000033

    As it falls the rate of acceleration increases. But lets say the rate is constant for the next 25,000 km of fall.

    The time taken to fall the 25,000 km is 123,953 seconds.

    S = 1/2at\(^2\)

    t = \( \sqrt{50,000/0.0000033}\)

    = 123091 seconds.

    After falling for this time at an acceleration rate of 0.0000033 km/s, the particle has a velocity of 0.41 km/sec

    v = at = 123091 x 0.0000033 = 0.41 km/sec.

    It will actually be faster than this after falling the 25,000 km because a constant rate of acceleration has been assumed.

    So the particle is now 325,000 km from the earth. At this distance from the earth, its rate of acceleration is 0.0000038 km/se/sec

    6378\(^2\)/325,000\(^2\) x 9.8/1000 = 0.0000038

    Assuming constant acceleration for the next 25,000 km fall, this time the 25,000 km takes 114708 seconds to fall

    S = 1/2at\(^2\)

    t = \( \sqrt{50,000/0.0000038}\)

    =114,707 seconds.

    Its velocity after 50,000 km of descent towards the earth is v = u + at = 0.41 + (0.0000038 x 114,707) = 0.85 km/sec

    Once again, its velocity will really be greater than this as constant acceleration is being presumed.

    *

    6378\(^2\)/300,000\(^2\) x 9.8/1000 = 0.0000044 km/s/s

    time to fall from 300,000 km to 275,000 km = t = \( \sqrt{50,000/0.0000044}\) = 106600 seconds

    Its velocity after 75,000 km of descent towards the earth is v = u + at = 0.85 + (0.0000044 x 106600) = 1.31 km/sec.

    *

    6378\(^2\)/275,000\(^2\) x 9.8/1000 = 0.0000053 km/s/s

    time to fall from 275,000 km to 250,000 km = t = \( \sqrt{50,000/0.0000053}\) = 97129 seconds

    Its velocity after 100,000 km of descent towards the earth is v = u + at = 1.31 + (0.0000053 x 97129) = 1.82 km/sec.

    *
    6378\(^2\)/250,000\(^2\) x 9.8/1000 = 0.0000064 km/s/s

    time to fall from 250,000 km to 225,000 km = t = \( \sqrt{50,000/0.0000064}\) = 88388 seconds

    Its velocity after 125,000 km of descent towards the earth is v = u + at = 1.82 + (0.0000064 x 88388) = 2.38 km/sec.

    *
    6378\(^2\)/225,000\(^2\) x 9.8/1000 = 0.0000079 km/s/s

    time to fall from 225,000 km to 200,000 km = t = \( \sqrt{50,000/0.0000079}\) = 79555 seconds

    Its velocity after 150,000 km of descent towards the earth is v = u + at = 2.38 + (0.0000079 x 79555) = 3.01 km/sec.

    *

    6378\(^2\)/200,000\(^2\) x 9.8/1000 = 0.0000100 km/s/s

    time to fall from 200,000 km to 175,000 km = t = \( \sqrt{50,000/0.00001}\) = 70710 seconds

    Its velocity after 175,000 km of descent towards the earth is v = u + at = 3.01 + (0.00001 x 70710) = 3.7 km/sec.

    *
    6378\(^2\)/175,000\(^2\) x 9.8/1000 = 0.000013 km/s/s

    time to fall from 175,00 km to 150,000 km = t = \( \sqrt{50,000/0.000013}\) = 62017 seconds

    Its velocity after 200,000 km of descent towards the earth is v = u + at = 3.7 + (0.000013 x 62017) = 4.5 km/sec.

    *

    6378\(^2\)/225,000\(^2\) x 9.8/1000 = 0.000017 km/s/s

    time to fall from 150,00 km to 125,000 km = t = \( \sqrt{50,000/0.000017}\) = 54232 seconds

    Its velocity after 225,000 km of descent towards the earth is v = u + at = 4.5 + (0.000017 x 54232) = 5.4 km/sec.

    *

    6378\(^2\)/125,000\(^2\) x 9.8/1000 = 0.000026 km/s/s

    time to fall from 125,00 km to 100,000 km = t = \( \sqrt{50,000/0.000026}\) = 43853 seconds

    Its velocity after 250,000 kmof descent towards the earth is v = u + at = 5.4 + (0.000026 x 43853) = 6.5 km/sec.

    *

    6378\(^2\)/100,000\(^2\) x 9.8/1000 = 0.000040 km/s/s

    time to fall from 100,00 km to 75,000 km = t = \( \sqrt{50,000/0.000040}\) = 35355 seconds

    Its velocity after 275,000 km of descent towards the earth is v = u + at = 6.5 + (0.00004 x 35355) = 7.9 km/sec.

    *
    6378\(^2\)/75,000\(^2\) x 9.8/1000 = 0.000071 km/s/s

    time to fall from 75,00 km to 50,000 km = t = \( \sqrt{50,000/0.000071}\) = 26537 seconds

    Its velocity after 300,000 km of descent towards the earth is v = u + at = 7.9 + (0.000071 x 26537) = 9.8 km/sec.

    *
    6378\(^2\)/50,000\(^2\) x 9.8/1000 = 0.00016 km/s/s

    time to fall from 50,000 km to 25,000 km = t = \( \sqrt{50,000/0.00016}\) = 17677 seconds

    Its velocity after 325,000 km of descent towards the earth is v = u + at = 9.8 + (0.00016 x 17677) = 12.6 km/sec.

    *

    So now we have a particle mathematically falling at 12.6 km/sec. In reality it will be faster than this because of the constant acceleration for each 25,000 of descent presumption.



    So what you have here JR is shown to mathematically wrong. If you can keep your pride out of the way you will begin to see how ridiculous your above quote is. The particle would be accelerating all the way from infinity to the earth.

    The last 25,000 km of descent.

    6378\(^2\)/25000\(^2\) x 9.8/1000 = 0.00064 km/s/s

    time to fall from 25,000 km to earth km = t = \( \sqrt{50,000/0.00064}\) = 17677 seconds

    Its velocity after 350,000 km is v = u + at = 12.6 + (0.00064 x 8838) = 18.3 km/sec.

    *

    So, just falling from the moon and using below value acceleration rates, we have a velocity of way over 11 km/sec.

    If a particle has a velocity of 18.3 km/sec at the end of its descent, it's potential energy at the beginning of the descent must have of a magnitude equal to its now kinetic energy.

    Escape velocity theory depends on a particle never having a potential energy magnitude above that of a particle travelling with a speed of 11 km/sec away from the earth.

    So the theory is shown to be wrong.

    CptBork, it was JR, not me who stated the answer to the addition of 1/2 + 1/4 etc is 1.
     
  13. CptBork Valued Senior Member

    Messages:
    6,460
    And I agree with JR's statement for that series. I said if you think the sum is infinite, please tell us how many terms you have to add together before you reach a sum of 2 or greater.
     
  14. BobG Registered Senior Member

    Messages:
    162
    Your answer is incorrect. Even if you don't want to directly use integration there is a straightforward way of calculating this.

    You agree that energy must be conserved? The decrease in the gravitational potential energy of a mass m falling towards a mass M (in this case the Earth) starting at r1 and ending at r2 is

    GMm/r2 - GMm/r1

    Now if energy is conserved this must be equal to the kinetic energy of the particle which is given by mv^2/2

    Therefore you get

    GM/r2 - GM/r1 = v^2/2

    And put in the numbers and you should get around 11.1km/s for v. This is actually not much less than the escape velocity and shows why it is a useful concept. Because the escape velocity might be the velocity necessary to for a particle to be projected to infinity but it also quite accurately describes the velocity to reach many intermediate distances.
     
  15. plane evolution Registered Senior Member

    Messages:
    115
    Perhaps I didn't explain it well enough.

    I said 1 does not exist in this series and the 1 JR is referring to is infinity. A number never arrived at.

    It doesn't add to 1 was the point. To say it does is to say the whole and the half are equal.
     
  16. plane evolution Registered Senior Member

    Messages:
    115
    So BobG you are saying this calculation is wrong for the rate of acceleration of a particle 350,000 km from the earth?

    6378\(^2\)/350,000\(^2\) x 9.8/1000 = 0.0000033 km/s/s

    The 6378 is the radius of the earth in kilometers which perhaps I should have mentioned. And the 9.8/1000 is the rate of acceleration due to the earth's gravity at a distance of 6378 km from the centre of the earth. Then it is just straight use of the inverse square law.

    If that is a correct calculation, the rest follows as a demonstration that JR has made an incorrect statement about a particle falling from an infinite distance to the earth having a terminal velocity of 11 km/sec.

    Before your formula could be taken seriously, you would first have to come up with something a little more coherent than this.

    Gravitational force having an infinite range jarringly contradicts a particle only escapes the gravitational force of another particle if it is an infinite distance away

    Whereas what I have done is quite coherent. It maybe inexact because I presume constant acceleration in 25,000 km blocks. But this only means the particle reaches a speed far greater than the calculated one.

    The problem you seem to be having is understanding what potential energy is.

    It is not energy that is in a system. It is energy in limbo.

    You are going about it back to front.

    If a body ascends in a gravity field it has a gain in potential energy and that gain is the product of the force required to lift the body and the upward displacement of the body in the gravity field. (F x S)

    That potential energy will become kinetic energy if the body falls. That is where the kinetci energy will come from. It is not active energy as you are presuming. (like a lump of wood has the potential to realease energy if it combusts. If it never combusts, that energy is always in limbo. That's just junior school physics)

    I have definitely shown that a body reaches a velocity of over 11 km/sec when descending in the earth's gravity field from a distance as great or greater than 350,000 km from the earth.
     
  17. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    plane:

    Yes, but it gets weaker and weaker and weaker with distance.

    The question is not whether the force stops acting. In fact you're right - it never stops acting (except at infinite distance, which is not reachable). The question you need to ask is: is the force strong enough to pull the object back to Earth? The answer to that is: if it was launched with a speed greater than about 11 km/s, Earth's gravity can't pull it back, even though there's always a gravitational force on it.

    Snip.

    An object falling to Earth from that distance will have a speed given by:

    \(\frac{1}{2}mv^2 = \frac{GMm}{6370~km} - \frac{GMm}{350 000~km}\)

    Putting in the mass of the Earth (\(6 \times 10^{24}~kg\)) and Newton's gravitational constant (\(6.67 \times 10^{-11}\)), we find the speed to be 11.107 km/s.

    Using the same values for the radius of the Earth etc., the escape velocity from the Earth is 11.209 km/s.

    So, we see that an object fired from Earth with a velocity of 11.107 km/s will barely make it to the Moon before falling back to Earth (ignoring the Moon's gravity, of course), while if it is fired at 11.209 km/s it will never fall back to Earth.

    Incidentally, plane, your calculations are approximate, which is why you get a value larger than 11.209 km/s.

    If you disagree, tell me how many terms I will need to add to get an answer greater than 2, say.

    By this I mean 1/2 is one term, 1/2 + 1/4 is two terms, 1/2 + 1/4 + 1/8 is three terms, and so on. How many terms will I need to add before this sum becomes greater than 2?

    Nonsense. Pick up any physics textbook and it will tell you what potential energy is. Or, look it up on the web.

    See if you can find any sites that define potential energy as "energy in limbo" that aren't crank sites.

    No. You've made mistakes in your calculations.
     
  18. BobG Registered Senior Member

    Messages:
    162
    FxS is the formula for a constant force. If the force varies you have to sum up the force at every point by an infinitesimal distance. You do this by integration.
    In this case the force is given by

    \(F = -\frac{GMm}{r^2}\)

    and the work done moving the particle from r1 to r2 is given by

    \(\int_{r1}^{r2} \frac{GMm}{r^2} \, dx = \frac{GMm}{r2} - \frac{GMm}{r1} \)

    This is the same as working out FxS but calculating the force at every point. You can also see it is the same result for the change in potential energy that I gave in my post.
     
  19. plane evolution Registered Senior Member

    Messages:
    115
    Bunkum. Take the approximations out and the velocity gets greater, not less. You can try forever and you won’t justify what you have posted here.




    I don’t say it gets to be 2 or greater. I say it never gets to 1. You say the answer is 1. This is fictitious. That is why I posted the 1 in your answer is the equivalent to infinity in this series.

    1,2,3,4……..

    No-one picked up what I meant. Probably still don’t.


    Stored energy seems to be the consensus. Not sure this is a variance to any defining I have done of it. Stored energy is not energy causing anything to happen.



    Afraid not, JR. The calculations are all about demonstrating that a body falling in the earth’s gravity field does not have a terminal velocity of 11 km/sec.

    If I was to do it at 10,000 km intervals (improving on the approximations), the answer would become greater than 18 km/sec.

    If I was to do this calculation from mars or the sun, obviously the starting velocity at a distance of 350,000 km from the earth would be above zero. Final answer again of greater again than 18 km/sec. Much greater in these cases.

    There is just no argument with the figures or what they demonstrate.

    Your basic problem is using zero potential energy at infinite distance. If it had zero potential energy, it would not descend.

    If it descends, its potential energy is above zero. You can argue whether or not potential energy is finite or infinite at an infinite distance. Either way it is not zero.

    Bob G I know what you are saying but you have the problem (as JR did in his calculation at a distance of 350,000 km from the earth) of saying potential energy at infinite distance is zero.

    Probably no point going any further. In this thread JR has had potential energy both zero and implied non zero at infinite distance. I’ve done the arithmetic to show you that the maximum speed of accelerating body does not have a terminal value. About all there is to it.

    If every particle in the universe is attracting every other particle in the universe, eventually every particle in the universe will collide. As I first posted with respect of the universe being a potential motion machine.
     
  20. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    plane:

    I haven't checked. What you say may be true, in which case you've made a calculational error. Either way, your conclusion is wrong, as I and others have simply shown by a full and correct calculation.

    I note that you have not addressed the direct calculation at all. Do you not understand it, or is there some other reason?

    What's the correct result then, if not 1? i.e. I'm asking you what is the answer to the sum:

    1/2 + 1/4 + 1/8 + 1/16 +...

    Note: the answer cannot be infinity, because you said it doesn't even get past 2 at any point. So, what's the answer?

    Ok. So what?

    But it does. I've given you the formula for escape velocity. IF you don't believe me, then look it up on the web. You'll find hundreds of sites all confirming that my formula is correct.

    IF you like, you can use the following formula instead:

    \(U = -\frac{GMm}{r} + 628~J\)

    According to this formula, the potential energy at infinity is 628 Joules. If you use this formula, it still won't change the escape velocity from the Earth.

    That's wrong. Since the potential energy is negative at the Earth's surface, and a negative number is less than zero, the object will fall, because doing do decreases its potential energy and things "want" to go to lower potential energy.

    You think you'll never understand? Ok then. Have a nice day.
     
  21. funkstar ratsknuf Valued Senior Member

    Messages:
    1,390
    Oh no, not that again. I suppose you're also one of those who do not "believe" that 0.999... = 1 ?
     
  22. BobG Registered Senior Member

    Messages:
    162
    It doesn't change the result. If you like you can say the potential energy at the surface of the Earth (r=6400km) is zero.

    In this case

    \(U = -\frac{GMm}{r} + \frac{GMm}{6400km}\)

    You will see in that formula that the potential energy is zero at r=6400km and is not zero at infinity. Again if you calculate the change in potential energy between r=6400km and r= 350000km you get

    \(U_2 - U_1 = -\frac{GMm}{350000km} + \frac{GMm}{6400km} - (-\frac{GMm}{6400km} + \frac{GMm}{6400km})\)

    \(U_2 - U_1 = \frac{GMm}{6400km} - \frac{GMm}{350000km}\)

    This is exactly the same answer and gives exactly the same velocity
     
  23. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    Hi plane,
    This equation for t is only correct if the particle's initial speed is zero. So, it works for the first iteration, but not for the second and later ones, when you will instead need:

    \(S = vt + \frac{1}{2} at^2\)

    \(t = \frac{-v + \sqrt{v^2 + 50000a}}{a}\)

    The corrected iterations look like this:
    Code:
       -----------------------------------------------   
      |    R     |       a	    |    t     |    v     |  
      |----------+--------------+----------+----------|  
      |  325000  |  0.00000325  |  123952  |  0.4034  |  
      |  300000  |  0.00000377  |   50191  |  0.5928  |  
      |  275000  |  0.00000443  |   37045  |  0.7569  |  
      |  250000  |  0.00000527  |   29913  |  0.9146  |  
      |  225000  |  0.00000638  |   25132  |  1.0749  |  
      |  200000  |  0.00000787  |   21556  |  1.2446  |  
      |  175000  |  0.00000997  |   18688  |  1.4309  |  
      |  150000  |  0.00001302  |   16268  |  1.6427  |  
      |  125000  |  0.00001772  |   14141  |  1.8932  |  
      |  100000  |  0.00002551  |   12202  |  2.2045  |  
      |   75000  |  0.00003987  |   10368  |  2.6179  |  
      |   50000  |  0.00007087  |    8558  |  3.2244  |  
      |   25000  |  0.00015946  |    6657  |  4.2860  |  
      |       0  |  0.00063785  |    4395  |  7.0896  |  
       ----------------------------------------------- 
    
    ...resulting in a final speed of 7km/s.

    As you say, this is an underestimate due to the iterative approximation. If you reduce the step size, you'll find that the result approaches a little over 11 km/s at a distance 6400km.
    A step size of 1000km, for example, gives 10.96km/s at 6000km, and a 100km step gives 11.02km/s at 6400km.

    Of course, it is better to use an integral to find an exact result, the limit you approach as you go to smaller and smaller steps. The integral is the continuous sum of the particle's acceleration over time. In this case, it's a bit ugly to do the integral directly for velocity (I can't do it myself), which is why potential energy is used as a stepping stone. It makes the maths much easier, for the same result.
     
    Last edited: Apr 9, 2010

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