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Homework Help: Gravitational Attraction

  1. Apr 18, 2008 #1
    1. The problem statement, all variables and given/known data
    The mass of the Moon is 7.35x10^22 kg. At some point between Earth and the Moon, the force of Earth's gravitational attraction on an object is cancelled by the Moon's force of gravitational attraction. If the distance between Earth and the Mon (centre to centre) is 3.84x10^5 km, calculate where this will occur, relative to Earth.

    2. Relevant equations
    Fg = (G)(m1)(m2) / r^2

    3. The attempt at a solution
    FgEARTH = (G)(m1)(mE) / r^2
    FgMOON = (G)(m1)(mM) / r^2
    dM + dE = 3.84x10^5 km = 3.84x10^8 m

    (G)(m1)(mE) / dE^2 = (G)(m1)(mM) / dM^2

    (5.98x10^24) / dE^2 = (7.35x10^22) / (3.84x10^8 - dE)^2

    (7.35x10^22) / (5.98x10^24) = [(3.84x10^8 - dE)^2](dE^2)

    0.01229 = (1.5x10^17 - 3.84x10^8dE + dE^2)(dE^2)

    I dont know where to go from here? can i move the 0.01229 to the other side and at the same time move the dE^2 to the other side (so that it gets cancelled out)??
  2. jcsd
  3. Apr 19, 2008 #2


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    I don't know what you are doing. If Fearth=Fmoon then G*m1*mE/rE^2=G*m1*mM/rM^2 and rM+rE=r, where rE is the distance from earth, rM is the distance from the moon and r is the total distance=3.84x10^5 km. That's two equations in two unknowns, rM and rE.
  4. Apr 19, 2008 #3
    thats exactly what i wrote in mine its dE instead of rE. and then you isolate for rE:
    dE = 3.84x10^8 m - dM

    and then you can solve by substitution and get what i have at the top, and now i still have the same question
  5. Apr 19, 2008 #4
    okay so i cross multiplied the ratios and then used the Quadratic Formula to solve for dE and i get a number of 5x10^43. Does that sound right?
  6. Apr 19, 2008 #5


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    Can you fix the error that occurs between these two steps?
  7. Apr 19, 2008 #6
    yea i multiplied wrong:
    so i cross multiplied like this instead:

    (5.98x10^24) / dE^2 = (7.35x10^22) / (3.84x10^8 - dE)^2

    (7.35x10^22)(dE^2) = (5.98x10^24)(1.5x10^17 - 3.84x10^8dE + dE^2)

    0 = 5.9x10^24dE^2 - 2.3x10^34dE + 8.97x10^42

    **using quadratic formula i then got an answer of: 5x10^43 m
  8. Apr 19, 2008 #7


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    You are on the right track, but here are more hints:

    The quadratic formula gives two solutions, not just one.
    You might double-check carefully the numbers you just posted in those equations.
    Perhaps it would help first to divide through by 10^22
  9. Apr 19, 2008 #8
    yea the other answer for the quadratic formula gave me a negative number, and since im looking for distance, negative numbers are meaningless, but ill try your suggestion of dividing by 10^22
  10. Apr 19, 2008 #9
    okay so i realized that some of my answers were off, but now i keep getting a negative under the square root sign??
  11. Apr 19, 2008 #10
    you made a mistake in multiplying out (3.84x10^8 - dE)^2
    I recommend finding the general formula for dE as a function of m_e m_m and R.
    you made additional errors in getting 2.3x10^34dE and 8.97x10^42 as well.

    you use the quadratic formula here with b^2 very close to 4*a*c so any mistake will result in
    an outcome that is many orders of magnitude off.
    Last edited: Apr 19, 2008
  12. Apr 19, 2008 #11
    okay so i restarted the question over (was making way too may mistakes lol) and i got as final answers dE = 7.78x10^10 and 1.7x10^8. Now because it is looking for the distance from the Earth do i just use 7.78x10^10m ??
  13. Apr 19, 2008 #12
    It's still not right. The point where the forces cancel must be between the earth and the moon, so 7.78x10^10m can't be right. the other point is closer to the earth than to the moon. This is really a case where it's better to find the general formula before you start pushing all those powers of 10 around.
  14. Apr 19, 2008 #13
    o okay i see what you mean, sorry i thought you meant there was a specific formula that i didnt know... so for what your saying then:

    (mE)(dM^2) = (mM)(dE^2)

    and then plug in? for:

    (5.98x10^24)(3.84x10^8 - dE)^2 = (7.35x10^22)(dE^2)

    and then id foil and multiple the left side and just distribute on the right side?
  15. Apr 19, 2008 #14


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    Don't be in such a hurry to plug in. Replace dM with R-dE, where R is the total distance from earth to moon. Now you have a quadratic equation in dE. Try solving it without putting numbers in, like kamerling suggests. It's easy to make mistakes with all of the numbers floating around. They all look alike.
  16. Apr 19, 2008 #15
    okay: no numbers at all would be:

    (mE)(r^2) - 2(mE)(R)(dE) + [(mE)(dE^2) - (mM)(dE^2)] = 0

    then from here there really is no way to solve except by quadratic formula is there?
  17. Apr 19, 2008 #16
    This is right if r is the same as R. There is quite a lot you can cancel after using the
    quadratic formula.
  18. Apr 19, 2008 #17
    yes r is the same as R i just forgot to capitalize...

    and i dont really see what can cancel:
    ** let the numbers inside {} be like under the square root sign
    = -b +/- {b^2 -4ac} / 2a

    = -[-2(mE)(R)(dE)] +/- { [-2(mE)(R)(dE)]^2 - 4[(mE)(dE^2) - (mM)(dE^2)](mE)(r^2)} / 2[(mE)(dE^2) - (mM)(dE^2)]
  19. Apr 19, 2008 #18
    you're solving for dE so you can't get that in the result of the quadratic formula.

    what are a, b and c in [tex]\frac { -b \pm \sqrt {b^2 - 4 a c}} { 2 a } [/tex]
  20. Apr 19, 2008 #19


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    Work on the terms inside the square root sign next. Something cancels.
  21. Apr 19, 2008 #20


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    I'd suggest going through some of your notes or a text book on algebra again. Most of the problems you're having will disappear and it will set you up for the rest of your studies.
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