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Gravitational Force Question

  1. Oct 31, 2006 #1
    The question i am stumped on is as follows
    The mass of the moon is 7.35 x 10^22 kg. At some point between Earth and the moon, the force of earth's gravitational attraction on an object is cancelled out b the moons force of gravitational attraction. If the distance between earth and the moon (center to center) is 3.84 x 10^5 kilometers, calculate where this will occur, relative to earth.

    Ive been fumbling with this question forever, can someone please help me get started. I dont even know where to begin.
  2. jcsd
  3. Oct 31, 2006 #2
    What is Newton's formula for gravitational force?
  4. Oct 31, 2006 #3
    first off.. let me say thank god someone is willing to help me..

    Fg = Gm1m2/r^2
  5. Oct 31, 2006 #4
    Ok, if M_E is the mass of the earth, M_m is the mass of the moon and m is the mass of some body in between the earth and the moon, what are the two forces on the body of mass m. Set them equal to each other. You will notice you have two variables (the moon-body and earth-body distances). But these are related (the sum of these distances is the earth-moon distance). Solve these two equations to find your answer. Post your work if you get stuck.
  6. Oct 31, 2006 #5
    what i get is... left side = Right side
    G(Me)(M)/r^2 = G(Mm)(M)/r^2

    but i dont know where to go from here.. how does the radiuses come into play? and how would i apply it
    man im so confused
  7. Oct 31, 2006 #6
    The r on the left is different from the r on the right.
  8. Oct 31, 2006 #7
    ok.. for purpose of this question lets call the first denote teh first r as (A) and the second r as (B)

    i know that A + B = 3.84 x 10^5
    so A = 3.84 x 10^5 - B
    can i use this to plug in somewhere?
  9. Oct 31, 2006 #8
    Take R = (distance from earth to moon) as given. Then A + B = R. You have
    [tex] \frac{G M_E m }{A^2} = \frac{GM_m m}{B^2}[/tex]
    [tex] \frac{M_E}{A^2} = \frac{M_m}{B^2}[/tex]

    Both A and B are unknown at this point. Eliminate A using B = R -A. Can you solve the remaining part?
  10. Nov 1, 2006 #9
    why isnt the mass of the earth given? am i just supposed to plug in the numbers now what i get is Me/3.84 x 10^5 - B = Mm/ B do i just plug in mass of moon, and mass of earth and use cross multiplying to solve?
  11. Nov 1, 2006 #10
    Thats exactly what you do. The mass of the Earth is a constant you'll probably be given in an exam, just look it up:

    Wikipedia says: [itex]5.9742 \times 10^{24} kg[/itex]

  12. Nov 1, 2006 #11
    here is what i have and it looks wrong after cross multiplying

    i have 2.82x10^28 - 7.35x10^22 B^2 = 5.98x10^24

    2.82x10^28 = 7.35x10^22 B^2 + 5.98x10^24

    2.82x10^28 = 6.10x10^24 B^2

    B^2 = 2.82x10^28 / 6.10x10^24

    B^2 = 4622.95

    B = 67.99 which seems to be way to small
  13. Nov 1, 2006 #12
    please someone correct my problem if there is one...
  14. Nov 1, 2006 #13
    OK, I'm not going to give you the answer, but I'll help you get there. You have:

    [tex]\frac{M_e}{A^2} = \frac{M_m}{B^2}[/tex]

    Let: [tex]B^2 = (R - A)^2[/tex], otherwise you're calculating the distance from the moon; you need relative to the Earth.

    [tex]M_e ( A^2 - 2AR + R^2) = M_m A^2[/tex]

    Rearranging gives:
    [tex]A^2(M_e-M_m) - A (2M_e R) + M_e R^2 = 0[/tex]

    And I'm sure you've solved lots of equations of this sort; let me know how you get on...

  15. Nov 1, 2006 #14
    i am now 10 times more confused than i was before..

    when i plug in numbers i get something that looks like a quadratic equation.. and im not good at solving those

    is there any other way? like could i have done it the way i did it before?
  16. Nov 1, 2006 #15
    A quadratic equation is exactly what you need to solve; and I can't see another way of doing it. I've 'bracketed' the things you need to plug into the quadratic formula.

    Hope that helps, Sam :smile:
  17. Nov 1, 2006 #16
    ok, first bracket (Me-Mm) equates to 5.9x10^24
    the second bracket (2MeR) equates to 4.6x10^33 *if its kilometers than its 4.6x10^30*
    the last part is (MeR^2) which i get equals 8.82x10^41

    I can not solve this question i have no clue what to do.. my calculator gives error, ive tried online quadratic solvers it comes up as imaginary
  18. Nov 2, 2006 #17
    OK, so lets use some new letters:
    [tex]Q = M_e - M_m[/tex]

    [tex]W = -2M_e R[/tex]

    [tex]E = M_e R^2[/tex]

    And the quadratic formula is:
    [tex]A = \frac{-W \pm \sqrt{ W^{2} - 4QE}}{2Q}[/tex]

    You take the positive solution as your answer. I'm sure that this will not give a complex answer (because [itex]W^2-4QE > 0[/itex]).

    Can you go from here?

    Last edited: Nov 2, 2006
  19. Nov 2, 2006 #18
    4.6x10^33 +/- sqrt (2.12x10^67 - (4)(5.9x10^24)(8.82x10^41)
    2 (5.9x10^24)

    =4.6x10 ^33 +/- sqrt (4.0x10^65)

    =4.6x10^33 +/- 6.32x10^32

    after doign that i get two answers.. how do i know which one is correct..
    both of them yeild positive answers..
    i get 4.43x10^8 as one answer
    and 3.36x10^8
  20. Nov 2, 2006 #19


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    One answer is a point between the earth and the moon. The other answer is a point farther away from the earth than the moon. If your calculations are correct, both answers are valid, i.e., the two forces are equal at two different places. Which one satisifies the statement of the problem?
  21. Nov 3, 2006 #20
    yes your right one of those answers is greater than the radius given..
    i guess its the one smaller.. thanks.. i believe this question is done now
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