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Trip to the moon

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data

    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. Some information that will help during this problem:
    mass of earth = 5.9742 x 10^24 kg
    radius of earth = 6.3781 x 10^6 m
    mass of moon = 7.36 x 10^22 kg
    radius of moon = 1.7374 x 10^6 m
    distance from earth to moon = 3.844 x 10^8 m (center to center)
    G = 6.67428 x 10^-11 N-m2/kg2
    1) On your first attempt you leave the surface of the earth at v = 5534.0 m/s. How far from the center of the earth will you get?

    2. Relevant equations

    Wnet = ΔKE
    (G * M * m)((1/rf) - (1/ro)) = (m * v^2)/2, M = mass of earth & m = mass of satellite
    (G * M)((1/rf) - (1/ro)) = (v^2)/2

    3. The attempt at a solution

    ro = radius of earth (?)

    (rf^-1 - ro^-1) = (.5v^2)/(G * M)
    rf^-1 = ((v^2)/(2G*M)) - (ro^-1)
    rf = (((v^2)/(2GM)) - (ro^-1))^-1

    I plugged in the corresponding values, but they seem to be wrong. I got 26022994.95 m.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 30, 2012 #2

    Simon Bridge

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    I'm not sure I follow the reasoning behind these manipulations ... did you try to use conservation of energy?

    Will you need to include the Moon's contribution?
     
  4. Oct 1, 2012 #3
    I tried using the work-energy theorem (W = ΔKE) & applied it to the problem. I'm not too sure about using the moon; I was hoping someone would know about whether to use it or not.
     
  5. Oct 1, 2012 #4
    The sum of the potential energy with respect to the Moon, the potential energy with respect to the Earth and the kinetic energy is constant.
     
  6. Oct 1, 2012 #5

    gneill

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    I would be tempted to treat it as an orbital problem. Take advantage of the Earth's daily rotation to choose an optimal launch location and direction (thus decreasing your required Earth-relative launch speed). I think a pretty good first approximation would be to plan a Hohmann style orbit for maximum apogee vs energy required.

    It might be worthwhile to determine the radial distance where the Moon's influence just equals the Earth's in terms of acceleration due to gravity -- if you can cross that line with the right Moon/Earth/spacecraft geometry, you can 'fall' the rest of the way to the Moon.
     
  7. Oct 1, 2012 #6

    Simon Bridge

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    What voko said - but I'd put it as "total kinetic energy + total potential energy = constant" ... you can work out the craft's potential energy due to the Earths gravity and due to the Moon's gravity and combine them for total potential energy. (The moon will help your spacecraft travel towards it.)

    the "orbital problem" treatment gneill mentioned is how you'd actually go about it IRL ... but may be more than your course wants from you. The question seems to be interested only in the radial contributions - so you'd have to aim really really carefully to hit the moon.

    gneill also talked about the radial distance where the Moon's influence is equal (and opposite) to the Earth's. This means you won't need to start out at the escape velocity from the surface of the Earth just to get to the Moon. If you arrange your initial kinetic energy to be just enough to get you there (IRL, you need just a bit more than that or it's 50:50) - the Moon will pull you the rest of the way in. I suspect this last bit is what the lesson is trying to get you to realize.
     
  8. Oct 2, 2012 #7
    Use conservation of energy.
     
  9. Oct 2, 2012 #8

    Simon Bridge

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    @Trolling: how did you get on so far?
     
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