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Moon orbiting Earth?

  1. May 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Calculate the potential energy and kinetic energy of the moon, as well as it's escape velocity. Give the moon's period and angular speed. Consider the Earth to have a mass of 6 x 10^24 kg, the moon to have a mass of 7 x 10^22 kg, and their center separation distance to be 4 x 10^8 m.

    2. Relevant equations
    I = mr^2
    F(g) = (GMm)/r^2
    KE = 1/2(I)(Omega)^2
    PE = ?
    Omega = ChangeinTheta/ChangeinTime

    3. The attempt at a solution
    So far, I have drawn my free body diagram, and I've calculated the force due to gravity (the centripetal force in this case) to be 1.750875 x 10^20. This is based off the the F(g) formula. From here, I don't know where to move, because to calculate the other items I need, I believe I need the velocity of the moon, which is not provided. Can anyone give me a hand with this? Thanks!
     
  2. jcsd
  3. May 9, 2017 #2

    TSny

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    How is centripetal force related to speed? (Review your lessons on uniform circular motion.)

    Also, you have a question mark for the formula for gravitational potential energy. This is a basic formula which you should be able to easily find in your textbook or online.
     
  4. May 9, 2017 #3

    phyzguy

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    Also, you know (or can look up) how long it takes the moon to make one orbit, and you know the diameter of the orbit, so it is a simple matter to calculate the moon's speed.
     
  5. May 9, 2017 #4
    I mean, for PE, the only thing that shows up is mgh when I search it online. I also know that Centripetal Force and speed are related in the fact that the object must be moving at a certain speed, or else it would get pulled to the center, or it would go flying away and leave orbit. I found an equation that stated Fc = (mv^2)/r. I solved for v in this case, and I got a velocity of approximately 1000 m/s. I plugged in this value for the v in the omega = v/r equation and got a value of omega as 2.500624923 x 10^-6. I then plugged the mass and the radius into the equation Inertia = mass(radius)^2, and got a value of 1.12 x 10^40 for my inertia. I plugged both of these values into the equation KE=(1/2)(inertia)(omega)^2 and got a kinetic energy of 3.921960003 x 10^68. Is all of that correct so far?
     
  6. May 9, 2017 #5

    CWatters

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    You know the radius so you know the circumference of the orbit and it's not hard to find out how long an orbit takes.
     
  7. May 9, 2017 #6

    CWatters

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    Actually I think your method is better as it's only based on data in the problem statement.
     
  8. May 9, 2017 #7

    mfb

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    You should work with units.
    That is only an approximation for small heights (e.g. in a building). There is a more general formula based on F = (GMm)/r^2.

    Okay so far, apart from the missing units.
    You multiply something with 1040 with numbers smaller than 1. How can the result be larger?

    You have the mass of the Moon and its speed, there is no need to use the moment of inertia as detour.
     
  9. May 9, 2017 #8

    TSny

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