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Gravitational attraction question

  1. Mar 22, 2006 #1
    The mass of the Moon is 7.35 * 10^22kg. 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 3.84 * 10^5 km, calculate where this will occur, relative to Earth.

    This is what i've done so far..not sure if i'm on the right path. If I am..i'm not sure if that is my answer or if there is more to do.

    m1 = Mass of Moon
    m2 = Mass of Earth

    Fg = Gm1m2 / r^2
    = (6.67*10^-11)(7.35*10^22)(5.98*10^24) / (3.84 * 10^8)^2
    = 1.98 * 10^20

    Does that seem right?

    Also, in a different problem. I had to find the centripetal acceleration (which I did). Now if I have to express it in terms of g (acceleration due to Earth's gravity), would I use the formula mg = ma...then g = the answer I got for the centripetal acceleration? Not sure if that makes any sense...
  2. jcsd
  3. Mar 22, 2006 #2


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    It may be easier to do this by 'proportions' or field strength. If I say that at the equilibirum point, the gravitation field strength are equal (but opposite) thus;

    [tex]\frac{GM_{m}}{r^2} = \frac{GM_{e}}{R^2}[/tex]
    Where r is the distance from the moon and R is the distance from the earth

    Can you follow?:smile:
    Last edited: Mar 22, 2006
  4. Mar 22, 2006 #3

    Doc Al

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    No. You calculated the gravitational force that the moon and earth exert on each other, which is not what was asked. Instead, imagine an object of mass M a distance D from the earth's center. Find the gravitational force on the mass due to (1) the earth, and (2) the moon. Set those forces equal and solve for the distance.

    The acceleration due to gravity at the earth's surface is g = 9.8 m/s^2. If you have some acceleration measured in m/s^2 and you want to express it in terms of "g's", then just divide by 9.8. (In other words, an acceleration of 9.8 m/s^2 equals 1 "g", an acceleration of 19.6 m/s^2 equals 2 g's, etc.) Depending on the exact problem, you may be able to find your answer directly (algebraically) in terms of g without doing any arithmetic.
  5. Mar 22, 2006 #4
    What you have done is most likely wrong. If I understood correctly, the problem asks for a distance from earth (or it's centre) to a point at which the earth's and the moon's gravitational forces cancel out.
    The formula you used ( Gm1m2 / r^2 ) clearly gives you just a force, not any distance, so that is already wrong.

    What you should do is define a new variable, say r1, which is the distance to the point of equilibrium from earth's centre. Then the distance to that point from the moon's centre is r-r1.

    You know the forces are opposite by direction and equal by force at that point (r1).
    You also know how to calculate gravitational forces, so all you have to do is calculate both attraction forces from the Earth (using distance r1) and from the Moon (using distance r-r1). They are equal, so you can write an equation, and find r1 from it.

    Also note, that you will have to define some mass for the virtual object at the point (to use gravitation law). There is no problem with that, because it appears in both attraction forces and thus will cancel out in the equation.

    Edit: beaten by two posters :)
  6. Mar 22, 2006 #5
    Thanks for the replies. Let me know if I understood this correctly.

    First, I calculate due to the earth:
    Fge = (6.67*10^-11)(5.98*10^24)(1)/ (3.84 * 10^8)

    Then I calculate for the moon:
    Fgm = (6.67*10^-11)(7.35*10^22)(1) / (3.84 * 10^5 - r1)

    Then I set Fge = Fgm and solve for r1?

    Thanks for your patience!
  7. Mar 22, 2006 #6

    Doc Al

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    Almost there. In the expression for Fge, the distance should be r1.
    Last edited: Mar 22, 2006
  8. Mar 22, 2006 #7
    Ok, I did that and got r1 = 3.79 * 10^15m. That is hopefully right.

    Also, just for my information, if ever I need to find the gravitational force on the mass for a planet or some object, all I need to do is use 1kg as the second mass?
  9. Mar 22, 2006 #8


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    Gravitational potential is defined as the force per unit mass exerted on an object placed at point r in the field. It is given by

    [tex]g = -\frac{GM}{r^2}[/tex]

    and as units of [itex]N\cdot kg^{-1}[/itex] or [itex]m\cdot s^{-2}[/itex].

    Therefore, if you want to find the force exerted on an object, just multiply the gravitational field strength by the mass of that object resulting in the formula;

    [tex]F = -\frac{GMm}{r^2}[/tex]

    -Hope this helps :smile:
    Last edited: Mar 22, 2006
  10. Mar 22, 2006 #9

    Doc Al

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    Better check that. Since the entire earth-moon distance is only about 3.84 * 10^8m, how can that answer be right?

    Depends on the problem. In this problem, since the "test mass" is arbitrary, I would have used "M". It cancels from your final equation. (Using 1 kg is OK in this example, but it's better to realize that no actual mass is needed.)
  11. Mar 22, 2006 #10
    Thanks for all your help/patience guys! Really appreciated!
  12. Mar 22, 2006 #11

    Doc Al

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    It's better to call that gravitational field strength. Gravitational potential is usually defined as the gravitational potential energy per unit mass.
  13. Mar 22, 2006 #12


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    Indeed it is, I tend to use the too interchangably :frown: much to the frustration of my tutors:redface: . I use the correct formulae for the situation but tend to mix my terms:grumpy:
  14. Mar 23, 2006 #13
    If the point at which this happens is S metres from M
    If the earths and moons gravitational forces cancel eachother out, then:

    (Gm)/(r-S)^2 = (GM)/ r^2

    rearranging gives:

    S = r - (mr^2)/M

    where r is the distance between the moon and the earth
    M is the mass of the earth
    m is the mass of the moon
    G is the universal gravitational constant (6.67x10^-11)

    I plugged in the numbers (several times!) and ended up with 3.4x10^8m
    Does this sound right?
  15. Mar 23, 2006 #14


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    That looks good to me. :smile:
  16. Nov 10, 2008 #15
    maltesers, that does not look right. I know I am not a tutor, or even very good at physics, but I can see right away that that answer is impossible.

    The distance from the earth to the moon is 3.84 x 105. The answer you have is greater than that.

    If I am wrong could someone please explain where I made the mistake.
  17. Nov 10, 2008 #16


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    Mind your units.

    The distance to the moon varies and is in the range 3.6 to 4 times 108 meters or 105 in km.
  18. Nov 10, 2008 #17
    I see now where I went wrong. Thank you for the help.
  19. Feb 2, 2009 #18
    When I deduce a value for "r1", must I square root it?
  20. Feb 2, 2009 #19
    I got the same value but 10^5 m. Now that sounds right.
  21. Feb 3, 2009 #20
    Using the following values:

    Gravitational constant = 6.67e-11
    Earth mass = 5.98e+24 kg
    Earth’s GM = 3.98866e+14
    Earth radius = 6.4e+6 meters

    Moon’s mass (.0123 * Earth’s 5.98e+24 kg mass) = 7.3554e+22
    Moon’s GM = 4.9060518e+12
    Moon’s radius = 1.74e+6 meters

    Distance between Earth and Moon’s center of mass = 384,000,000 meters (238,606 miles)

    Earth’s gravitational attraction on Moon at distance of 345,664,007 meters (which is 54.01000109375 Earth radii) from Earth’s center of mass = .003338249 m/s^2

    verified by: a = GM / r^2

    (6.67e-11) (5.98e+24 kg) / (345,664,007 meters)^2 = .003338249 m/s^2

    Moon’s gravitational attraction on Earth at distance of 38,335,993 meters (which is 22.032179885 Moon radii) from the Moon’s center of mass = .003338249 m/s^2

    verified by: a = GM / r^2

    (6.67e-11) (7.3554e+22 kg) / (38,335,993 meters)^2 = .003338249 m/s^2

    So, at approximately 54.01 Earth radii (345,664,007 meters away from center of Earth) or 22.032 Moon radii (38,335,993 meters from center of Moon), the gravitational attraction on an object via the Earth and via the Moon is virtual unity. The object requires no orbit around the Moon though it does orbit the Earth in sync with the Moon’s cycle of approximately 27.3 days. The object remains in a straight line of sight between the Earth and Moon. For this reason, the object’s Earthly orbital velocity (approximately 921 m/s) will be a lesser velocity than the Moon’s orbital velocity (approximately 1023 m/s). Ordinarily, the closer an object is to the Earth, the greater will be its required orbital velocity, but this unity attraction between Earth and Moon allows the opposite to be true for the object in gravitational unity.
    Last edited: Feb 4, 2009
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