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Gravitational Attraction Question

  1. Mar 15, 2009 #1
    1. 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 Moon (centre to centre) is 3.84x10^5 km, calculate where this will occur, relative to Earth.



    2. Relevant equations
    Fg = Gm1m2/r2


    3. The attempt at a solution
    I don't know where to begin.
     
  2. jcsd
  3. Mar 15, 2009 #2

    Doc Al

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    Imagine some mass (call it 'm') at a distance X from the center of the Earth. What force does the Earth exert on m? The Moon? What's the net force on m from both?
     
  4. Mar 15, 2009 #3
    So i have to find the gravitational attraction exerted on Earth and the moon first? and find the net force from those 2 forces?
     
  5. Mar 15, 2009 #4

    Doc Al

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    Yes. Do it symbolically first, before doing any arithmetic.
     
  6. Mar 15, 2009 #5
    Alright so what i did was i found the mass of the object first.

    I used Fg= Gm1m2/r2 and rearranged it to m1 = Fgr2/Gm1

    Then i put the information in and got

    m1 = (1.74x10^6m)2(1.67 N)/(6.67x10^-11 Nm2/kg2)(7.35x10^22 kg)
    m1 = 5.57x10^23 kg

    Now should i put that new mass into the Fg for the moon and the earth?
     
  7. Mar 15, 2009 #6

    Doc Al

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    The mass of the object is arbitrary. Don't try to "figure it out", just call it 'm'.

    In order to "figure out" the mass of the object, you'd have to know the gravitational force and its distance from the earth (or moon). You don't know any of that.

    Instead, just write expressions for the force on the object from the Earth and the force on it from the Moon. Call its mass 'm' and its distance from the center of the Earth 'X'.
     
  8. Mar 15, 2009 #7
    Alright hopefully this is the right step

    Gravitational Attraction of Earth

    Fg = Gm1m2/r2
    = (6.67x10^-11 Nm2/kg2)(m)(5.98x10^24 kg) / (6.38x10^6 m)2
    = 9.80 N

    Gravitational Attraction of the Moon

    Fg = Gm1m2/r2
    = (6.67x10^-11 Nm2/kg2)(m)(7.35x10^22 kg) / (1.74x10^6 m)2
    = 1.62 N

    Fnet = Fg of Earth - Fg of the Moon
    = 9.80 N - 1.62 N
    = 8.18 N

    now is this correct? lol
     
  9. Mar 15, 2009 #8

    Doc Al

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    Nope, not correct.

    In general, if someone asks you to find the force on an unspecified mass "m", do not respond with a number! You don't have the information to get a numerical answer.

    Instead, I just want you to write the formulas for the forces. The formulas will be in terms of 'm' and 'X'. Later, you'll solve for X.
     
  10. Mar 15, 2009 #9
    lol alright, this is harder than i thought....

    F = ma
    Fs = usFn
    Fk = ukFn
    F = mg

    Fg = Gm1m2 / r2
    F = mv2/r

    i think thats all of them
     
  11. Mar 15, 2009 #10

    Doc Al

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    I meant for you to write the force formula that you need, not every one that you can think of!
    That's the only one you need. What's m1, m2, and r for each case?
     
  12. Mar 15, 2009 #11
    ahah i knew that.. ah m1 is the mass of earth and m2 is the mass of the moon, and then r should be the distance between the centre of earth to the centre of the moon
     
  13. Mar 16, 2009 #12

    Doc Al

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    Earth's force on object:
    m1 = mass of earth
    m2 = m = mass of object
    r = distance of object to center of earth = X

    Moon's force on object:
    m1 = mass of moon
    m2 = m = mass of object
    r = distance of object to center of moon = D - X (where D is earth-moon distance)
     
  14. Mar 16, 2009 #13
    i think i finally got it....

    D = 3.84x10^5 km = 3.84x10^8 m

    Ge = Gm1m2/X
    Gm = Gm1m2/ D-X

    Ge = (6.67x10^-11 Nm2/kg2)(5.98x10^24 kg)m2/X
    = 3.99x10^14 Nm2/kg(m2)/ X

    Gm = (6.67x10^-11 Nm2/kg2)(7.35x10^22 kg) / (3.84x10^8m - X)2
    = 4.90x10^12 Nm2/kg m2 / (3.84x10^8 - X)2

    Now the X and m2's cancel and i am now left with

    (3.99x10^14 Nm2/kg) X (4.90x10^12 Nm2/kg) / (3.84x10^8 m)2
    = 1.33x10^10 m2
     
  15. Mar 16, 2009 #14

    Redbelly98

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    But X does not cancel, only m2 cancels.

    I have 3 suggestions for you:

    • Don't plug in numbers yet, wait until you have an equation in terms of X before doing that. Keep things in terms of symbols for now.
    • Let the Earth's mass be mE, and the moon's mass be mM, so that you're not using the same symbol (m1) for two different things.
    • If something is squared in your equations, write that as ^2. I.e., X^2 for "X squared". That way, you're not using m2 to mean both "mass 2" and "meters squared".

    Ah, one more thing. The expression for Ge should have /X^2 at the end of it.

    So, what equation results from setting Ge and Gm (using symbols, not numbers) equal to each other?
     
  16. Mar 16, 2009 #15
    GMe/X^2 = GMm/(X-D)^2
     
  17. Mar 17, 2009 #16

    Doc Al

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    Good. But since D > X, I'd use D-X (instead of X-D) to keep things positive. (That will matter when you take the square root.)

    Now cancel what you can, rearrange, and solve for X.
     
  18. Mar 17, 2009 #17
    alrght so i got square root(Me-D/X^2Mm) = X

    and if i rearrange the X^2 i get square root(Me-D/Mm) = X(X^2)
     
  19. Mar 17, 2009 #18

    Doc Al

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    Starting with this equation:
    Or better, this one:
    GMe/X^2 = GMm/(D-X)^2

    Cancel the G, invert, then take the square root of both sides.
     
  20. Mar 17, 2009 #19
    alright so then i get

    square root X/Me = square root(D-X)/Mm
     
  21. Mar 17, 2009 #20

    Doc Al

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    Almost. What you get is:

    square root {X^2/Me} = square root {(D-X)^2/Mm}

    What's the square root of X^2?
     
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