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Taylor series expansion for gravitational potential energy. GMm/r=mgh near the earth

  1. Oct 10, 2012 #1
    Using taylor series expansion to prove gravitational potential energy equation, GMm/r=mgh at distances close to the earth.
    R= radius of the earth h= height above surface of the earth m= mass of object M= Mass of the earth

    U = - GmM/(R + h)

    = - GmM/R(1+ h/R)

    = - (GmM/R)(1+ h/R)^-1

    do a binomial expansion or Taylor's series with (h/R) as the small variable;

    = -(GmM/R)[1 - (h/R) + ...].

    DOUBT: In this step why do we only use the FIRST order taylor expansion and no more?
    The function would not be complete would it?and nothing has been done to bound the error of the polynomial. Why would taking the first non-trivial term (first order term) of the expansion of 1+x (that is 1-x) account for the entire function 1+x or in this case 1+h/r.

    PLEASE REPLY ...

    Proof (continued): -(GMm/R)(1-(h/R)= -(GmM/R) + GmMh/R^2

    In the second term note that GM/R^2 =g

    = - (GmM/R) + mgh

    So the potential energy at "r" can be written as PE at "R" plus PE at "h"

    Now in planetary problems, where the distance h is not small and you use the general formula - GmM/r it is convienient to choose U = 0 at r = infinity
    However near the earth we can arbitrarily chose U=0 at U = - (Gmm/r)

    Therefore: U = mgh when h<<<R
     
  2. jcsd
  3. Oct 10, 2012 #2

    Doc Al

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    Staff: Mentor

    Re: Taylor series expansion for gravitational potential energy. GMm/r=mgh near the ea

    Succeeding terms would be higher order powers of h/R and thus increasing insignificant when h/R << 1.
     
  4. Oct 10, 2012 #3
    Re: Taylor series expansion for gravitational potential energy. GMm/r=mgh near the ea

    Wouldn't that mean that h/r would be off by all the succeeding terms all the way to infinity?
     
  5. Oct 10, 2012 #4
    Re: Taylor series expansion for gravitational potential energy. GMm/r=mgh near the ea

    If you're worried, you can write down the second order term, and then figure out what condition needs to hold in order for the second order term to be at most, say, 1% of the first order term.
     
  6. Oct 10, 2012 #5
    Re: Taylor series expansion for gravitational potential energy. GMm/r=mgh near the ea

    Okay... So your trying to imply that when h/r is way less than 1, higher order powers of h/r increase it insignificantly?
     
  7. Oct 10, 2012 #6

    Doc Al

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    Staff: Mentor

    Re: Taylor series expansion for gravitational potential energy. GMm/r=mgh near the ea

    The terms get smaller and smaller, so they can be safely ignored.

    Imagine if h/R = 1/1000

    Then (h/R)2 = 10-6, (h/R)3 = 10-9, and so on.

    Write out the next few terms of the expansion and see how small they are compared to the first.
     
  8. Oct 10, 2012 #7
    Re: Taylor series expansion for gravitational potential energy. GMm/r=mgh near the ea

    Thanks a lot! That explains it.
     
  9. Oct 10, 2012 #8
    Re: Taylor series expansion for gravitational potential energy. GMm/r=mgh near the ea

    Also that implies that U = mgh is just an incredibly good approximation right? Not exactly the potential energy.
     
  10. Oct 10, 2012 #9

    Nugatory

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    Staff: Mentor

    Re: Taylor series expansion for gravitational potential energy. GMm/r=mgh near the ea

    Yes. It's a good approximation anytime that h (which is really the difference between two values of the distance from the center) is small compared with the distance from the center. This will be the case for most reasonable heights from the surface of the earth.
     
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