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Correcting for the Moon's gravity

  1. Dec 29, 2004 #1

    Andrew Mason

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    Here is an interesting question.

    When we make measurements of weight on earth using a spring scale should we correct for the moon's gravity? The correction would be:

    [tex]a_m = a_e\frac{M_m}{M_e}\frac{R_e^2}{d_m^2}[/tex]

    where:
    [itex]d_m[/itex] = 384,400 km (approximate distance of surface of earth to centre of moon)
    [itex]R_e[/itex] = 6400 km (radius of earth)
    [itex]M_m[/itex] = 5.98E24 kg (mass of moon)
    [itex]M_m[/itex] = 7.35E22 kg. (mass of earth)
    [itex]a_m[/itex] = acceleration due to moon gravity on surface of earth
    [itex]a_e[/itex] = acceleration due to earth gravity on surface of earth

    The correction (about g/293000 = 3.4 E-6g) would be subtracted when your location on the earth is facing away from the moon and added when your location is facing toward the moon. Or would it?

    AM
     
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  3. Dec 29, 2004 #2

    dextercioby

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    Well,Andrew it's basically the same problem with the one that explains tides.Lunar (and much less Solar,Jupiter,Mars and Venus) attraction dimineshes the gravitational force acting on a body on the surface of the earth,if the 2 forces have opposite sense (the same direction) which is realized when the body is facing the moon,and increases the gravitational force when the two forces have both the same direction and sense,which it's realized when the body is on the opposite side of the Earth wrt to the Moon.

    Daniel.

    PS.I'm afraid i would not agree with your computations though;lunar's gravity effects vary,since at one point the distance is 384400Km-6400Km (diminishing the effect of the Earth's attraction),and at the exactly opposite point is 384400Km+6400Km.I'm sure u know that the "d_{m}",which you've chosen as 384400Km,is measured actually between the centers of the 2 celestial bodies...
     
  4. Dec 29, 2004 #3

    Tide

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    Andrew,

    dextercioby is correct. Essentially, we are in freefall with respect to the Moon much like astronauts are in freefall with respect to the Earth and "feel" no gravitational force (i.e., they experience weightlessness). However, if the spacecraft were large then the freefall would be with respect to its center of mass and objects in the craft closer or further away from the Earth than the craft's center of mass would feel a force proportional to their distance from the center of mass. That would be approximately proportional to the inverse cube of the distance of the craft from the center of the Earth. The same argument will apply to weight corrections on Earth due to the Moon's gravitational pull.
     
  5. Dec 29, 2004 #4

    Andrew Mason

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    This is all true. But I am not talking about the difference in the magnitude of the gravity of the moon over the extra 12,800 km (earth diameter). I am talking about the direction of it relative to the up/down direction of a person on the earth surface. I don't think this is related to the tidal forces, which, as you quite correctly state, are due to the difference in magnitude of the gravitational force over that 12,800 km.

    AM
     
  6. Dec 29, 2004 #5

    Andrew Mason

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    I don't think this is a tidal issue. It has to do with the different direction of the moon's gravity relative to up/down at a particular location as the earth rotates. As you say, the magnitude of the moon's gravity would change as the earth turned, but I am really just concerned with its direction.

    Quite right. As the earth turns, there is a total difference of about 6%. But I am talking about the 200% difference due to the different direction.

    The key here, I think, is to recognize that the earth is actually in orbit around the centre of mass of the earth and moon system.

    AM
     
  7. Dec 29, 2004 #6

    dextercioby

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    The question is simple.At each moment of time a body at the earths's surface is acted upon by three forces:the gravity force of the earth,the gravity force of the moon and the centrifugal force due to the rotation of the earth.(4,if u include the normal reaction force of the Earth as a response to the body's pressure on the Earth).Taking into account different positions on the globe (meaning different latitudes and longitudes),composing these three vectors will have different results.The most dramatic change is observed when all these 3 forces have the same direction (support).
    I'm sure that u can draw some vector diagrams,use the correct figures and compute the total (resulting) acceleration.If u include the normal reaction force,the net acceleration should be zero...

    In that case,u need to make some corrections to the calculations made assuming the earth was fixed and rotating around its axis and the moon circling the Earth.I don't think the order of magnitude of these corrections will be significant wrt to the idealized case pictured above.

    Daniel.
     
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