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Finding the formula for escape velocity

  1. Mar 15, 2009 #1
    first of all, we need to say that this question is not for homework.

    1. The problem statement, all variables and given/known data
    our doubt is in the process of finding the formula (not the value) of escape velocity in the earth; we want to use the principle of conservation of energy for that purpose.

    2. Relevant equations
    gravitational potential energy:
    we found that there are two formulas for gravitational potential energy:
    [tex]U=-\frac{GmM}{R}[/tex] or [tex]U=MgR[/tex], where [tex]G[/tex] is the gravitational constant, [tex]m[/tex] is the mass of the projectile, [tex]M[/tex] is the mass of the earth, [tex]R[/tex] is the radius of the earth and [tex]g[/tex] is the gravitational acceleration.
    kinetic energy:
    [tex]K=\frac{mv^{2}}{2}[/tex], where [tex]m[/tex] is the mass of the projectile and [tex]v[/tex] its velocity.
    principle of energy conservation:
    [tex]K_{0}+U_{0}=K_{1}+U_{1}[/tex]
    3. The attempt at a solution
    we found by searching in the internet that the escape velocity of the earth (or other object) is the minimum velocity needed for a projectile to escape from the earth's gravitational field (from the surface of the earth to infinity).
    therefore, by the principle of energy conservation:
    [tex]\frac{mv^{2}_{0}}{2}-\frac{GmM}{R} = \frac{mv^{2}_{1}}{2}-\frac{GmM}{r_{1}}[/tex], where [tex]r_{1}=\infty[/tex].
    if [tex]r_{1}=\infty[/tex] and the final velocity ([tex]v_{1}[/tex]) is zero:
    [tex]\frac{mv^{2}_{0}}{2}-\frac{GmM}{R} = 0[/tex]
    this derives:
    [tex]v_{0}=\sqrt{\frac{GM}{R}}[/tex]
    but it's not still our problem.
    our problem is the following:
    look what happens when we try to substitute the first formula for gravitational potential energy with the second one:
    [tex]\frac{mv^{2}_{0}}{2}+MgR = \frac{mv^{2}_{1}}{2}+Mgr_{1}[/tex]
    but [tex]r_{1}=\infty[/tex]; it leads to an indetermination:
    [tex]Mg\infty[/tex]
    if we consider that [tex]Mg\infty[/tex] equals zero, it leads to another impossibility:
    [tex]\frac{mv^{2}_{0}}{2}+GmR = 0[/tex]
    no more TeX is necessary here to perceive that it will lead to a negative square root.
    what are we doing wrong here? we think that the error is in the formulas for gravitational potential energy.
    another doubt: what does it mean to set [tex]r_{1}[/tex] as infinity?

    thank you in advance for your patience.
     
  2. jcsd
  3. Mar 15, 2009 #2

    Doc Al

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    That second equation for gravitational PE should be ΔU = mgΔy and is only valid near the earth's surface. You must use the more general formula, which measures distance from the earth's center.
     
  4. Mar 15, 2009 #3
    thank you for your reply.
    can we ask another question to see if we understood your point?
    here it is:
    so, we can affirm that, in the surface of the earth, [tex]U=-\frac{GmM}{R}[/tex] and [tex]\Delta U=mg{\Delta y}[/tex] (where [tex]\Delta y[/tex] is the displacement, not the distance), right? if that's right, can we affirm that, in the surface of the earth, [tex]U\neq 0}[/tex], but [tex]\Delta U = 0[/tex]?

    thank you very much.
     
  5. Mar 15, 2009 #4

    tiny-tim

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    Hi pc2-brazil! :smile:

    U = GmM/R means that if R = r0 + h, then U = GmM/(r0 + h)
    = (GmM/r0)1/(1 + h/r0)
    = gMr0/(1 + h/r0)
    where g = Gm/r02

    Subtracting the constant gMr0, that gives:
    -gMh/(1 + h/r0)

    For small h, obviously, that is approximately
    -gMh

    But that approximation only works for small h … for large h, you must use the original formula
    U = -gMh/(1 + h/r0)
    which at h = ∞ gives you U = -gMr0, = -GmM/r0, as expected :smile:
     
  6. Mar 15, 2009 #5

    D H

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    You have a sign error, tiny-tim. You should have had U=-GMm/r and U≈mgh, rather than the other way around.

    The reason both equations are valid for small values of h is because all that matters with potential energy is change in potential energy. Adding an arbitrary constant to a formulation for potential energy will yield exactly the same physics. One could just as well use U=-GMm/r + some constant or U≈mgh + some constant.
     
  7. Mar 15, 2009 #6

    Doc Al

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    The gravitational PE between two objects is U = -GMm/r, where r is the distance between (the centers of) the two objects. When the object is on the earth's surface, r will equal the radius of the earth.

    (Note that the expression for gravitational PE has U = 0 at r = ∞.)

    The formula ΔU = mgΔy is used to find changes in gravitational PE as long as you stay close to the earth's surface (Δy << radius of the earth). The two expressions are completely consistent, of course, but the second is only applicable near the earth's surface.
     
  8. Mar 15, 2009 #7

    D H

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    Which is key. The only reason for choosing this value is because it simplifies the notation. We could pick any arbitrary constant for the value of the potential energy at r=∞. Zero happens to be a very nice constant. Suppose instead we chose the value U=GMm/Re for the value of the potential energy at infinity. Then the gravitational potential would be U=GMm(1/Re-1/r). For r=Re+h, 1/r=1/(Re+h)=(1/Re)*(1/(1+h/Re)). If h<<<Re, 1/(1+h/Re))≈1-h/Re. Thus U=GMm(1/Re-1/r) becomes, for distances close to the radius of the Earth, U ≈ GMm/Re*(1-(1-h/Re))=GMm/Re2*h=mgh, where g=GM/Re2.
     
  9. Mar 15, 2009 #8
    thank you all for the answers.

    so, if U = -GmM/r calculates the gravitational potential energy, while ΔU = mgΔy calculates the change in the grav. PE:
    if we have an object moving from [tex]r_{0}[/tex] to [tex]r_{1} = r_{0} + h[/tex] (it will have a variation of grav. PE from U0 to U1):
    1. [tex]U_{0}=-\frac{GmM}{r_{0}}[/tex] and [tex]U_{1}=-\frac{GmM}{r_{0} + h}[/tex];
    2. [tex]\Delta U = U_{1}-U_{0}[/tex];
    3. for a very small h = (r1 - r0), [tex]\Delta U = mg(r_1 - r_0)[/tex].
    is this right?
    another doubt: in tiny-tim's deduction, why is gMr0 a constant? shouldn't g vary with the distance?
    we would also like to know why U has a negative sign. shouldn't it be positive? isn't it a modulus?

    thank you for your patience.
     
  10. Mar 15, 2009 #9

    Doc Al

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    Sure, as long as you realize that 3 applies only near the earth's surface.
    No, "g" represents the gravitational field strength near the earth's surface.
    No, it's not a modulus. Since work must be done to raise an object, and since we take U = 0 at r = ∞, U must be negative for all finite values of r.
     
  11. Mar 15, 2009 #10
    thank you for the answers. if you don't mind, as other doubts arise, we will ask them.
    just one more thing:
    is [tex]r_{1}[/tex] set as infinity because the gravitational field is theoretically infinite? if so, wouldn't a body thrown from Earth's surface with escape velocity have to go through an infinite distance until v = 0? (there is probably a misconception from our part here).
     
  12. Mar 15, 2009 #11

    Doc Al

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    Not sure what you mean.
    That's right. As it rises, it slows. Its speed will be zero at r = ∞.
     
  13. Mar 15, 2009 #12

    tiny-tim

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    r1 is set as ∞ because that's the way escape velocity is defined

    it's the speed needed to reach ∞ at 0 speed …

    (leave Earth any faster, and you reach ∞ at positive speed :wink:)

    we could of course define escape velocity to be something different, or not bother to define it at all :smile:
     
  14. Mar 15, 2009 #13
    in other words, the speed will never be zero, so, the Earth's gravitational field will never be able to slow it down to the point it gets negative?
    but it is still strange, because, then, what would be the maximum distance it can reach until g (gravitational acceleration) is 0? infinity? what does it mean to be at an infinite distance when v = 0?
    if escape velocity is to escape Earth's gravitational field, then the gravitational field only ends at "infinity"?

    thank you in advance.

    EDIT:
    we just thought of something a few minutes after sending this message.
    what if we propose the following statement:
    "Escape velocity is the velocity needed for an object to reach speed 0 only when the Earth's gravitational acceleration equals zero (g = 0) (then, it will be not able to bring the object back any longer)."
    well, g = GM/r². if G and M are constants, the only value of r (distance) that would make g = 0 is, mathematically, ∞.
    is our reasoning correct?

    thank you again.
     
    Last edited: Mar 15, 2009
  15. Mar 15, 2009 #14

    Doc Al

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    Think of escape velocity as being the speed you have to launch a projectile so that it never falls back down.

    And, yes, strictly speaking the gravitational force will never be zero so there will always be an acceleration towards the earth.
     
  16. Mar 15, 2009 #15
    OK, thank you for the clarification.
    is the g = GM/r² reasoning correct? if so, it concludes mathematically that g will never be zero, only at infinity (which means never).
     
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