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Gravitational Potential energy Problem

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data
    A projectile is fired vertically from Earth's surface with an initial speed of 3.1 km/s. Neglecting air drag, how far above the surface of Earth will it go?

    2. Relevant equations
    KF+UF=KI+UF
    U=-G(M1m2)/r2
    K=1/2mv2
    mass of earth = 5.972E24
    radius of earth = 6371km
    G=6.673E-11
    3. The attempt at a solution
    1/2mv2F+(-G(M1m2)/r2F=1/2mv20+(-G(M1m2)/r20

    I can cancel the m because they're in every term so and final kinetic energy since it should be zero at max height.

    (-GM1)/r2F=1/2v20+(-GM1)/r20

    Next I try and get the rF by itself

    (-GM1)=(1/2v20+(-GM1)/r20)*r2F

    (-GM1)/(1/2v20+(-GM1)/r20)=r2F

    √(-GM1)/(1/2v20+(-GM1)/r20)=rF

    then I subtract rF from the initial r0 to get my answer.
    Did I do this right?
     
  2. jcsd
  3. Apr 17, 2015 #2

    Dick

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    I can tell you one thing that's wrong. ##\frac{-G M m}{r}## is potential energy. Your exponent on the ##r## is wrong.
     
  4. Apr 17, 2015 #3
    Oh wow I guess I was looking the wrong equation in my notes thanks. I think I got it from here then thanks a lot!!
     
  5. Apr 17, 2015 #4

    Simon Bridge

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    Well done:
    Note: You could get it all in one go from: $$\frac{1}{2}v^2 -\frac{GM}{R} = -\frac{GM}{(R+r)}$$... and solve for r, where R and M the radius and mass values are for the Earth.

    Of course, GM/R^2 = g which is a nicer number to handle.
    So the relation becomes: $$\frac{1}{2}v^2 -gR = -\frac{gR^2}{(R+r)}$$
     
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