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Planetary motion problem

  1. Mar 11, 2013 #1
    1. The problem statement, all variables and given/known data
    At a ertain point between earth and the moon the total gravitation force exted on an object by both planets is 0. The earth - moon distance is 3.84 x 10^5 and the moon has 1.2% of the mass of earth. Where is this point located.


    2. Relevant equations
    Fg=GmM/R^2


    3. The attempt at a solution

    So i drew a diagram and put the distance R between the earth and the moon and the object somewhere closer to the moon. I then wrote that the distance from the object from the earth is (R-X). I then said that the distance from the moon to the object was X.

    Sine it says the Gravitational force exerted on the object at a point is the same... i get

    Fgm=Fge
    (GmMm)/(X^2) = (GmMe)/(R-X)^2

    x<R

    i simplify this too
    X^2 = (R-X)^2
    0 = 0.988x^2 - 0.024RX + (0.012)*R^2

    if i put this into the quadratic formula i get values

    x= 0.5

    and x = 9327.44m


    Now if my answer are correct, does this mean that both are answers? so i would say?

    this point is located 9327.44 meters from the moon and 374.672.56 meters from earth.

    and this point is also located

    0.5 meters from the moon and 383999.5 meters from the earth?


    but at this second point it couldn't match my first point cause the force of gravity from the earth would decrease while the force of gravity on the moon would increase so how could the force exerted be the same??? Anyway i would appreciate any help i can get.
     
  2. jcsd
  3. Mar 11, 2013 #2
    Taking Me ti be the mass of the Earth, and Mm the mass of the Moon, (GmMm)/(X^2) = (GmMe)/(R-X)^2 is correct. The next step is not: X^2 = (R-X)^2 is incorrect. Me and Mm are different so cannot both disappear.
     
  4. Mar 11, 2013 #3
    If the two forces were equivalent, that would imply the two forces pull with the same magnitude and in the SAME DIRECTION. If the two forces pulled in the same direction, and were of the same magnitude, the net force wouldn't be zero. The key idea is that the net force is in fact zero; hence, the two forces have to balance each other out (or cancel).

    I hope this helps. If anyone finds fault, please correct me.
     
  5. Mar 11, 2013 #4

    SteamKing

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    You also need to get your units straight. Is the distance from the earth to the moon in feet, meters, miles, furlongs, etc? What about the units of G? What units is the mass of the earth measured in?
     
  6. Mar 11, 2013 #5
    Yea you're right it hsould have een X^2 =(0.012)*(R-X)^2 cause Mm = 0.012*Me but it was included in that step in calculation of the the next step
     
    Last edited: Mar 11, 2013
  7. Mar 11, 2013 #6

    So If i choose my + direction toward earth.

    i would get Fnet = Fge - Fgm = 0
    Fnet = (GmMe)/(r-x)^2 - (Gm(0.012)*Me)/x^2 = 0
    Fge = Fgm ?

    and solve for this?
     
  8. Mar 11, 2013 #7
    yea sorry i dont know why i didn't include it in the givens. Everything is in meters for distances. The radius R and X as well in meters. The mass of the earth is in kilograms.

    G is measured by (N*M^2)/Kg^2

    Sorry about not including that in my given statement.

    Me=5.98x10^24kg
    Re=6.38 x 10^6 meters
    G= 6.67x10^-11 (N*M^2)/kg^2
     
  9. Mar 11, 2013 #8
    Here is what happens. $$ \frac {M_m} {X^2} = \frac {M_e} {(R - X)^2}
    \\
    \frac {(R - X)^2} {X^2} = \frac {M_e} {M_m}
    \\
    (\frac {R - X} {X})^2 = \frac {M_e} {M_m}
    \\
    \frac R X - 1 = \pm \sqrt {\frac {M_e} {M_m}}
    $$

    You need to choose the right sign. Which one is right?
     
  10. Mar 15, 2013 #9
    The positive sign cause a distance can't be negative? So since this question asks for the Point at which the force of gravitational force exerted on the object is equal, R/X -1 is that point? Or do i solve for that X value, and express the point as a distance from the moon to the object and from the earth to the object?

    x=R/[√(1/0.012) +1]

    therefore the moon is "R/[√(1/0.012) +1]" from the object and the earth is "R - R/[√(1/0.012) +1]" from the object?

    I'm just having a hard time understanding how you consider the exact point.
     
  11. Mar 15, 2013 #10

    haruspex

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    Voko set X as the distance of the point from the moon. Did you understand voko's equations?
     
    Last edited: Mar 15, 2013
  12. Mar 15, 2013 #11
    Recall what R and X are - you defined them, so you should know!

    For R/X - 1 to be negative, R has to be less than X, or one of them has to be negative - is that physically sensible?
     
  13. Mar 15, 2013 #12
     
    Last edited: Mar 15, 2013
  14. Mar 15, 2013 #13


    R cannot be less than X because i defined X as the smaller portion, and no neither of them would be negative cause they are lengths

    Also thats a cool way of thinking about it, If R is LESS* than x the fraction is less than 1 so it will be negative... cooolio
     
    Last edited: Mar 15, 2013
  15. Mar 15, 2013 #14

    haruspex

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    I believe you have a sign wrong there.
     
  16. Mar 15, 2013 #15
    yessir i put a fix up top when i saw that.
     
  17. Mar 15, 2013 #16
    Also the Distance R is measured in kilometers... I dont know why i said meters, i must have missread the question.

    So when i solve the quadratic i get

    X1 = 37912.03 KM

    or x2 = -47239.97 KM
     
  18. Mar 15, 2013 #17
    The negative answer, as you have discovered, is meaningless physically.
     
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