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Work and Potential Energy: To the Moon physics problem

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Part 1(complete and correct. I'm including this for context's sake) You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the earth with enough speed to make it to the moon.

    You leave the surface of the earth at v = 5534m/s. How far from the center of the earth will you get?


    2)Since that is not far enough, you consult a friend who calculates (correctly) the minimum speed needed as vmin=11068m/s. How fast will you be moving at the surface of the moon?

    Vmoon = ?
    Vmin = 11068m/s

    3) Which of the following would change the minimum velocity needed to make it to the moon? (I'm pretty sure this is correct, however let me know if I'm wrong)
    Mass of the earth *True*
    Radius of the earth *True*
    Mass of the spaceship *False*

    2. Relevant equations
    Ue = -(G*Me*m)/rem where rem is the radius of the earth plus the distance between earth and moon
    Kf = (1/2)mVf^2
    Ki = (1/2)mVi^2
    Um = -(G*Mm*m)/rm where rm is the radius of the moon

    Where Me is the mass of earth, Mm is the mass of the moon, m is the mass of the spaceship(which does not matter as it will be cancelled in all terms) and Vi is the minimum escape velocity needed.

    3. The attempt at a solution
    Part 2
    Ue + Ki = Um + Kf
    Ue - Um + Ki = Kf
    -G*Me/(rem) - -G*Me/(rm) + V^2/2 = Vf^2/2
    √2*(-G*Me/(rem) - -G*Me/(rm) + V^2/2) = Vf
    Calculating this gives me 21365.16m/s which is an incorrect answer.

    Thank you
  2. jcsd
  3. Oct 17, 2013 #2
    Your equation is incorrect. Do you know what the meaning of Ue is when you equate it with G*Mem/(rem) ?

    It is important to understand the physical meaning of every terms in order to write the equation correctly.
  4. Oct 17, 2013 #3
    This is the third time I've attempted to post this today... Not sure if it's getting through or not, but I was on a mobile device, maybe it didn't go through. Anyway, is Ue in fact the gravitational potential energy in terms of the radius of the earth only, and Um is the gravitational potential energy of the distance between the earth and the moon's surfaces? But then where does the gravitational potential energy of the moon come into play?
  5. Oct 18, 2013 #4


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    Staff: Mentor

    Sorry, for some reason your posts landed in moderation. Approved the last one only.
  6. Oct 18, 2013 #5
    Wow..it must be a hard time for you.

    Have you learnt about electric field generated by an electric charge? It is analogous to the gravitational field. For example, you will find the equation for electrostatic force ##F=k\frac{Qq}{r^2}## very similar to the equation ##F=G\frac{Mm}{r^2}##. I find it personally useful for me to understand the gravitational field when I view it as something similar to a electric field.

    Anyways, according to your equation, Ue is the gravitational potential energy at a certain point, and you should know where the point is. Do you?

    After you understand exactly what your Ue is, go back to look at the equation (Ue + Ki = Um + Kf) you set up and translate it to words. It should be something like "The gravitational potential energy at point X plus initial kinetic energy equal to ...." After doing so, you should clearly see that the equation doesn't make sense.

    How would you make adjustment to your equation then ?
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