1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Confused on escape velocity derivation

  1. Oct 18, 2014 #1
    I'm confused on how you get the formula v=root(2GM/R²). I know you can use work=forcexdisplacement (W=Fd) so F=W/d.
    W=½mv² and R=d. Hence, F=½mv²/R
    Equate that to F=GmM/R² and rearrange to get v=root(2GM/R²).

    That makes sense as it involves the kinetic energy needed/work done in reaching orbit (or is it for escaping orbit?).

    However, why can't you use the formula F=mv²/R and equate that to F=GMm/R² ? It gives you v=root(GM/R²). That's slightly different to the formula that involves kinetic energy.
     
  2. jcsd
  3. Oct 18, 2014 #2
    I would use the following formula were I to derive the escape velocity formula:
    [itex] \frac{1}{2}m{v_1}^2 - G\frac{mM_E}{r_1} = \frac{1}{2}m{v_1}^2 - G\frac{mM_E}{r_2} [/itex]
    The above equation can be derived using some integration and the W=fd equation (I wont post it because perhaps the class is not calculus based). If you would like to see the derivation, let me know.
    We can make some changes to the above equation for the case of escape velocity though:
    [itex] \frac{1}{2}m{v_{esc}}^2 - G\frac{mM_E}{r_E} = 0 + 0 [/itex]
    Rearrange the above equation and you should arrive at the equation for escape velocity
     
    Last edited: Oct 18, 2014
  4. Oct 18, 2014 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    "Escape velocity" is NOT the initial speed required to reach orbit. Reaching orbit is not "escaping". "Escaping" means escaping entirely from the planet's gravitational influence. And, since the gravitational field extends to infinity, that means having initial velocity sufficient to coast to infinity.
     
  5. Oct 18, 2014 #4
    For derivation, I've been taught to use the kinetic energy formula, calculate work done on moving R, then equate it to the universal law of gravitation formula and end with v=root(2GM/R). Why does using kinetic energy give you the right formula (I.e. why do you do that/what does it show?)

    However, using the formula F=mv²/R instead of the kinetic energy formula and equating that to F=GMm/R² gives you v=root(GM/R) instead. This is the wrong formula- what's the difference between using the F=mv²/r formula vs using the kinetic energy formula?
     
  6. Oct 18, 2014 #5
    I'm sorry, I'm not 100% certain what you're asking. However, here is the derivation I would use:
    [itex] \vec{F} = -G\frac{mM_E}{r^2} \hat{r} [/itex]
    The mass of earth is (Me) and r is a unit vector. The force on some object of mass m is directed toward the center of the earth (hence the minus sign). If, for instance, this object is a satellite, and said satellite moves from one point to another (it goes from r1 to r2) then work was done on this object. The work done is:
    [itex] W = \int_{1}^{2} \vec{F} \cdot d\vec{l} [/itex]
    [itex] W = -GmM_E\int_{1}^{2} \frac{\hat{r}\cdot d\vec{l}}{r^2} [/itex]
    However:
    [itex] \hat{r} \cdot d\vec{l} = dr [/itex]
    So our equation becomes:
    [itex] W = -GmM_E \int_{r_1}^{r_2} \frac{dr}{r^2} [/itex]
    [itex] W = \frac{GmM_E}{r_2} - \frac{GmM_E}{r_1} [/itex]
    It is wise here to remember that the gravitational force (which is the force acting on our object) is conservative. Also remember that change in potential energy is equal to the negative of the work by some force. Therefore:
    [itex] \Delta U = U_2 - U_1 [/itex]
    [itex] \Delta U = -\frac{GmM_E}{r_2} + \frac{GmM_E}{r_1} [/itex]
    So now we can calculate the potential energy at some distance r from the center of the earth:
    [itex] U(r) = - \frac{GmM_E}{r} + C [/itex]
    As usual, C is just some constant, for simplicity, set C=0 and the equation becomes:
    [itex] U(r) = - \frac{GmM_E}{r} [/itex]
    Now we know that:
    [itex] U=0 \hspace{2 mm} {at} \hspace{2 mm} r = \infty [/itex]
    For an object approaching earth, potential energy is decreasing and negative. If the object is near the surface of the earth, the above potential energy equation can be reduced to:
    [itex] \Delta U = mg(y_2-y_1) [/itex]
    Therefore, we now have the equation I posted earlier, which can be reduced to the escape velocity equation:
    [itex] \frac{1}{2}m{v_1}^2 - G\frac{mM_E}{r_1} = \frac{1}{2}m{v_2}^2 - G\frac{mM_E}{r_2} = c [/itex]
    I hope this clears up any problems you have understanding the derivation, and I'm sorry if this is information you already understand.
     
  7. Oct 18, 2014 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The gravitational force depends on distance. You cannot use the force at some point and then multiply it with an arbitrary distance to get the escape velocity. Sure, there is a specific distance value where this works just by coincidence, but that is not a derivation of the escape velocity.

    To find the escape velocity, using the potential energy is by far the easiest way.
     
  8. Oct 18, 2014 #7
    That's great thanks (I wasn't shown that in school) :)
    I thought you could literally equate ½mv²/R (work divided by displacement/radius) to GmM/R² and rearrange for v but it seems it was more involved than I initially thought.
     
  9. Oct 18, 2014 #8
    Yeah it requires some calculus, there is probably another way to do it. However, I believe the derivation I showed is probably the easiest (or most intuitive). I will try to post a drawing to go along with the above derivation so that it is easier to understand..
     
  10. Oct 18, 2014 #9
    Here is the drawing...
     

    Attached Files:

  11. Oct 19, 2014 #10

    gneill

    User Avatar

    Staff: Mentor

    A simple derivation is based on the total specific mechanical energy of the object. An object under the gravitational influence of another body of mass M has a specific mechanical energy:
    $$\xi = \frac{v^2}{2} - \frac{GM}{r}$$
    You can pick out the KE and PE contributions. This is just an expression of the conservation of energy for objects moving in the gravitational field.

    Now, ##\xi## characterizes the nature of the orbit of the body:
    • ##\xi < 0## Body is gravitationally bound and has an orbit that is a circle, ellipse, or straight line (a so-called “degenerate” orbit where the object just moves radially).
    • ##\xi = 0## Body is unbound and will just escape. The orbit is either parabolic or degenerate.
    • ##\xi > 0## Body is unbound and will have either a hyperbolic or degenerate shape.

    That middle case where ##\xi## is exactly zero defines the escape condition.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Confused on escape velocity derivation
  1. Escape Velocity (Replies: 1)

  2. Escape Velocity (Replies: 3)

Loading...