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Vc - what is it?

  1. Jun 16, 2008 #1
    I'm pretty sure it's something to do with rocket science.... it's on a plaque at the Kennedy Space Center. It may be a dumb question here, but can anyone tell me what the formula

    Vc=Ro√(g/Ro-h)

    means?
     
  2. jcsd
  3. Jun 16, 2008 #2

    FredGarvin

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    Looks like some form of the equation for escape velocity. I have seen something similar, but not what you have here. The form I am used to seeing is

    [tex]u_e = R_o\sqrt{\frac{2g_o}{R_o+h}}[/tex]

    Ro is the radius of the Earth
    h is altitude
    go is the acceleration due to gravity at the surface
     
    Last edited: Jun 16, 2008
  4. Jun 16, 2008 #3
    That's great, thank you! The formula I entered was on a plaque on the launch site of Apollo I, I imagine it was a precursor to the formula you gave me. I'd be curious to look up the evolution of the science, which I should be able to do with some knowledge of what the variables stand for.

    Thanks again!

    Yours,
    Ben
     
  5. Jun 17, 2008 #4
    Formula's do not 'evolve' and they don't have 'pre-cursors'. Relations like the one you mentioned are not found by just randomly guessing or by experimenting but through mathematical derivation. (I'm sorry if that sounds a little naggy but it bugs me when people think that something like E=mc^2 comes from like randomly putting together variables and seeing what works). Now as for the meaning of the one you've posted your representation is a little ambigious. Was the formula:

    a) [tex]R_o \sqrt{\frac{g}{R} - h}[/tex]

    or was it

    b) [tex]R_o \sqrt{\frac{g}{R-h}[/tex]
     
  6. Jun 17, 2008 #5

    mgb_phys

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    Quite a few formulae are found by just guessing from dimensional analysis.

    There is a famous example (that I can't remember) about von Neumann coming up with a formula for some part of the Manhattan project that no-one else had been able to derive. When it was eventually derived the constant was somehting like 1.2 - he had assumed from Benford's Law that it would be one.
    Even more interestingly this was before Benford's Law was widely published.
     
  7. Jun 17, 2008 #6
    what you entered as b) Forgive me for being a little naive.... I'm not a physicist, just someone interested. Your tone came across as a little hostile, and while I appreciate that the people coming up with these things are very, very smart and a lot more knowledgeable about math than I am, surely there are attempts at finding the underlying math that aren't perfect the first time they're written down? Especially, it seems to me, when exploring brand new ideas in untestable environments.... such as space flight.

    I definitely don't thing a formula like like E=mc^2 was found by random putting together, but mustn't there have been SOME trial and error, some experimentation?

    I'm not trying to be rude, just trying to understand. Thanks!
     
  8. Jun 17, 2008 #7
    I apoligize for the hostile tone. It is, however, an important bit of understanding that most things in physics are derived mathematically without any reference to experiment and just follow naturally from a string of logic. Of course experment can certainly give one an idea of where to start (it is however considered quite bad form to do things the opposite way (try to create a theory to match experiment), although there are some time when this has occured). For the example of E=mc^2 I believe Einstein originally deriving it by considering the fact the a particle which emits two photons in opposite directions must have its momentum conserved regardless of the reference frame. However, that example may be a tad complicated (and unfortunately I'm afraid without more context I can't really see what the equation you wrote describes) but for a simple example I will look at orbital velocity. We'll derive how fast a satellite or the likes must be going to maintain a circular orbit at a height h above the earth: (continued in next post)
     
  9. Jun 17, 2008 #8
    Now when something is in a stable circular orbit the centripetal acceleration it feels will be exactly equal to the force of gravity pulling it inwards (i.e. If the amount it wants to fly away from the planet is exactly equal to the amount it wants to fall down towards it then the net result will be that it neither flys away or fall towards).

    Now the force of gravity is given by [tex]\frac{GMm}{(R+h)^2}[/tex] where M is the mass of the earth, m is the mass of the satellite, R is the radius of the earth and h is the height above the earth that the orbit is (therefore, R+h is the distance from the satellite to the center of the earths mass). This is Newton's Law of Gravitation and is basically just a mathematical way of saying that if you double the distance between two masses the gravitational force between them will be a quarter of what it was (the gravitationaly force, like the coulomb force decreases proportional to 1/r^2 where r is the difference between the masses). G is a 'proportionality constant' and this is determined from experiment. So we know the force of gravity and now the force of centripetal acceleration is [tex]\frac{mv^2}{R+h}[/tex] this is just a result from calculus. A derivation of it can be found here http://en.wikipedia.org/wiki/Centripetal_acceleration#Geometric_derivation . This is a purely mathematical result of the 'perceived' force that one will experienced if one assumes that a rotating frame of reference is not rotating. So finally we have our relations and we know that they must be equal for stable orbit so we have:

    [tex]\frac{mv^2}{(R+h)}=\frac{GMm}{(R+h)^2}[/tex]

    We multiply both sides by (R+h) to get:

    [tex]mv^2=\frac{GMm}{R+h}[/tex]

    Divide both sides by m:

    [tex]v^2=\frac{GM}{R+h}[/tex]

    and take the square root of both sides to finally get:

    [tex]v=\sqrt{\frac{GM}{R+h}}[/tex]

    Therefore, we see that a satellite which has a velocity (in the direction which is 'parallel' to the earths surface) [tex]\sqrt{\frac{GM}{R+h}}[/tex] will stay in geosynchronous orbit. Of course we didn't consider things like air resistance (most satellites aren't in 'outer space' but the exosphere so there is still some atmosphere) and radiation pressure from cosmic rays and such but all those are very very minor. Therefore we have a 'formula' for what the orbital velocity of a satellite needs to be without ever doing an experiment or fiddling or the likes. And also we can tell things like the mass of the satellite (which was m) doesn't matter at all for what the speed needs to be and such. This is how physics is done, experiment is used for two things, firstly to determine constants of proportionality (although this is very minor) and far more importantly to verify a theoretical assumption (i.e. if a particle which emits two photons of equal energy in opposite direction must have its momentum conserved then one must conclude that it must have had its mass decreased by x amount and then you check and you find that it has and then you rejoice).
     
  10. Jun 17, 2008 #9
    Fascinating.... Thank you for the background about how theories (and solid math) is developed. I wish I'd taken more physics and calculus when I had the easy opportunity, but my studies ended up going in different directions...
    Now I'm even more confused about this formula though.... It doesn't seem to take the gravitational constant into effect at all, or the masses of the two bodies involved (also you put the word formula in single quotes.... am I using the wrong word here?) I feel pretty safe assuming that [tex]V_c[/tex] is the velocity of a capsule (as shuttles were called at that point), that g is the gravity of the planet being rocketed from, that h is altitude and that [tex]R_o[/tex] is the radius of the planet being rocketed from. That would mean that

    [tex]V_c=R_o\sqrt{\frac{g}{R_o+h}}[/tex]

    is Velocity of capsule=the radius of the planet times the square root of gravity over the radius of the planet plus the altitude over the planet, yes? Unless at least one of those assumptions is wrong....

    That doesn't look like any of the escape or orbital velocity equations I've seen though. The search continues.... it must mean something!

    Ben

    p.s. At least I've learned how to code the notation right! Hopefully that will help people look at it at least.
     
  11. Jun 17, 2008 #10
    The experimentation usually comes after the equations...because the equations are in fact made with meaning. You are right, there are a few exceptions, such as Ohm's relation...but those are just few...Physics endorses both philosophy and empiricism, but when you lay eyes on an equation, what it represents philosophically is what is important...the equation you have there looks similar to one I would expect (the one mgb_phys wrote), but I don't think it was a precursor equation...you can derive the equation mgb_phys wrote using laws of gravitation and energy.
     
  12. Jun 18, 2008 #11
    Well BenTev the formula you wrote:

    [tex]v=R\sqrt{\frac{g}{R+h}}[/tex]

    IS the same equation for the necessary orbital velocity I derived. Was that what the plaque said? Because that's the speed necessary for geosynchronous orbit. See starting with the relation I derived:

    [tex]v=\sqrt{\frac{GM}{R+h}}[/tex]

    we're just going to put it in terms of another quantity. Say I don't KNOW the mass of the planet i'm on but I do know how fast an object accelerates on its surface ([tex]g[/tex]). Well let's think about what g is. If Force = mass x acceleration (F=ma) and the force of gravity on a mass m by a planet of mass M when standing a distance r from the center of mass (in this case r = R the radius of the planet since we're measureing this on the surface of the planet) is given by [tex]F=\frac{GMm}{R^2}[/tex]. So if the force on a ball is the gravitational force then Force = [tex]mg=\frac{GMm}{R^2}[/tex] (notice that i've replaced a with g since this is the symbol commonly used for the 'acceleration due to gravity'). So from that expression I solve for g to get [tex]g=\frac{GM}{R^2}[/tex], and this holds for any planet (and mass actually). So after all that if we go back to our original relation:

    [tex]v=\sqrt{\frac{GM}{R+h}}[/tex]

    and we rewrite it as

    [tex]v=\sqrt{(\frac{GM}{R^2})\frac{R^2}{R+h}}[/tex]

    (i.e. we introduce [tex]\frac{R^2}{R^2}[/tex] (we can do this because it's actually just equal to one. We now have our nice little expression [tex]\frac{GM}{R^2}[/tex] which we know is g. So if we replace that with g we get:

    [tex]v=\sqrt{(\frac{GM}{R^2})\frac{R^2}{R+h}} \rightarrow v=\sqrt{(g)\frac{R^2}{R+h}}[/tex]

    Now we simply take the R^2 in the numerator out of the square root to give the relation:

    [tex]v=R\sqrt{\frac{g}{R+h}}[/tex]

    and TADA! it's our same relation for the orbital velocity necessary for a satellite at an altitude h to maintain a geosynchronous orbit but this time it's written in terms of g instead of GM/R^2 (which may be more convenient to write because it may be easier to determine the acceleration due to gravity on the surface of a planet rather then determine its mass.
     
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