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Thread: Why does the Moon-Earth Centre of Gravity not affect gravity experienced on Earth's surface

  1. #1 Why does the Moon-Earth Centre of Gravity not affect gravity experienced on Earth's surface 
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    Here is a diagram showing the earth-moon system with its centre of gravity which is nearer the surface than the centre.

    (I tried to post a link to a pic but the forum software says I don't have enough posts - but anyway the centre of gravity is far from the centre of the earth - it is nearer the surface than the centre)

    How come when you drop something at the north pole it does not fall at an angle towards the centre of gravity and how come you don't have to stand at a funny angle when you are at the north pole?


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    Mmmm OK.

    I guess it is the centrifugal force of the system's rotation that causes that.

    If the system was not in rotation then things would fall towards the centre of gravity!


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    Genius Duck Moderator Dywyddyr's Avatar
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    I'd say it's for much the same reason that we don't get attracted (i.e. stand at funny angles) to the centre of mass of the Solar system: we're on, and moving with, the Earth - not "involved" with Earth-Moon system except as things on the surface of one of the bodies of that system.
    (I know what I mean here, but the words don't come out right: perhaps someone else can explain it with more clarity).
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  5. #4  
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    Interesting question.
    So why don't plumb bobs point towards the barycenter instead of the center of the earth? Or do they not point at the center of the earth and nobody ever checked?

    Our Restless Tides - NOAA Tides & Currents

    And apparently we are not the only people who find it confusing.
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  6. #5  
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    Quote Originally Posted by mojorising View Post
    Mmmm OK.

    I guess it is the centrifugal force of the system's rotation that causes that.

    If the system was not in rotation then things would fall towards the centre of gravity!
    Kind of like this.

    NOAA National Ocean Service Education: Tides and Water Levels
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  7. #6  
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    Quote Originally Posted by mojorising View Post
    Here is a diagram showing the earth-moon system with its centre of gravity which is nearer the surface than the centre.

    (I tried to post a link to a pic but the forum software says I don't have enough posts - but anyway the centre of gravity is far from the centre of the earth - it is nearer the surface than the centre)

    How come when you drop something at the north pole it does not fall at an angle towards the centre of gravity and how come you don't have to stand at a funny angle when you are at the north pole?
    Obviously the center of gravity between the Earth -Moon system is not the center of gravity for someone on the surface of the Earth. Just like the astronauts that ventured to the Moon were still able to land on the Moon and weren't attracted back to the Earth.
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    The solar system barycenter is located within or just outside the surface of the sun, but we do not fly off the earth into the sun. The gravitational force follows an inverse square law and so the gravitational force we feel is dominated by the earth, which we are sitting on the surface of.
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    Quote Originally Posted by Harold14370 View Post
    The solar system barycenter is located within or just outside the surface of the sun, but we do not fly off the earth into the sun. The gravitational force follows an inverse square law and so the gravitational force we feel is dominated by the earth, which we are sitting on the surface of.
    So my plumb bob should not point at the sun either?
    But isn't that because we are actually in freefall as we orbit the sun and the centrifugal forces just happen to balance the gravitational forces?
    Last edited by dan hunter; August 15th, 2014 at 04:57 AM. Reason: correcting for wrong word.
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    ^No, it is not to do with centrifugal force as I earlier thought.

    I forgot that gravity decreases with distance! (Harold says it's inverse square and I believe him!)

    The moon's gravitational effect on tiny objects on earths surface is essentially zero. It does affect large objects on the earths surface (e.g. the ocean)

    The earth-moon barycenter is a property of the earth-moon system.

    Small objects on earths surface are entirely controlled by Earth's gravity.
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    Quote Originally Posted by mojorising View Post
    ^No, it is not to do with centrifugal force as I earlier thought.

    I forgot that gravity decreases with distance! (Harold says it's inverse square and I believe him!)

    The moon's gravitational effect on tiny objects on earths surface is essentially zero. It does affect large objects on the earths surface (e.g. the ocean)

    The earth-moon barycenter is a property of the earth-moon system.

    Small objects on earths surface are entirely controlled by Earth's gravity.
    The earth is 81.3 times the mass of the moon. The radius of the earth is 6371 kilometers and the distance to the moon is 385,000 kilometers. So the ratio of the gravitational force exerted on the plumb bob by the earth to that by the moon is
    81.3*(385000/6371)^2=296,891.
    The force exerted by the moon would be hard to notice.
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  12. #11  
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    Quote Originally Posted by mojorising View Post
    Here is a diagram showing the earth-moon system with its centre of gravity which is nearer the surface than the centre.

    (I tried to post a link to a pic but the forum software says I don't have enough posts - but anyway the centre of gravity is far from the centre of the earth - it is nearer the surface than the centre)

    How come when you drop something at the north pole it does not fall at an angle towards the centre of gravity and how come you don't have to stand at a funny angle when you are at the north pole?
    If you were several time the moon's orbital distance away from the Earth-Moon barycenter, you would indeed tend to fall towards this barycenter. However, once you started to get closer, things would change. This is because when you are dealing with non-spherical mass distributions, things aren't as straight forward as they are when they are with spherical one.

    Basically to get the direction in which the net gravitational force acts, you add up the sum of the gravitational forces of all its parts. If the total mass is spherical in shape, they add up so that they point to the center of the sphere. When you start to deviate from a sphere, this starts to change, and the direction of the net gravity can change depending on where you are.

    So for example, when you are standing at the North pole and you want to find the direction of gravity, including any effect due to the Moon, you sum up the gravity forces of mass involved, (Earth and Moon). In this case since both the Earth and Moon are themselves nearly spherical, we can treat each individually as if its gravity pull came from its center, and then sum up these two pulls to get the total net direction of pull.

    The Earth is 81 times more massive than the Moon and you are 60 times closer to the Earth's center than the moon's. Since the pull increases with mass and decreases by the square of the distance, and using the formula already given in an earlier post, the pull towards the center of the Earth is some 293615 times greater than the "sideways" pull of the Moon. In terms of how much this changes the net direction of the gravity pull, it works out to ~0.0002 degrees from pointing directly at the center of the Earth, at a point some 22 meters from the Earth's center . Obviously, this is nowhere where the Earth-Moon barycenter is located.
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  13. #12  
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    Quote Originally Posted by mojorising
    How come when you drop something at the north pole it does not fall at an angle towards the centre of gravity and how come you don't have to stand at a funny angle when you are at the north pole?
    It's quite easy to understand. The "funny angle" that you're thinking of is so small that you wouldn't notice it. Think about how fast the Moon is orbiting the Earth. While you might be thinking in terms the moon's orbital velocity of 1.023 km/s think about the velocity. Since it takes approximately 27.32 days for the moon to orbit the earth that means that the moon goes through an angle of about one degree per day. Since there are 86,400 seconds in a day then it goes through 1.157x10^(-5) seconds of arc in one second. One second of arc is defined as 60 arc seconds in one degree.

    Do you see how small of a deviation from straight down that is? It's so small as to be virtually zero.
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    Quote Originally Posted by dan hunter View Post
    Quote Originally Posted by Harold14370 View Post
    The solar system barycenter is located within or just outside the surface of the sun, but we do not fly off the earth into the sun. The gravitational force follows an inverse square law and so the gravitational force we feel is dominated by the earth, which we are sitting on the surface of.
    So my plumb bob should not point at the sun either?
    But isn't that because we are actually in freefall as we orbit the sun and the centrifugal forces just happen to balance the gravitational forces?
    I think that is the correct answer. If there was no orbital motion the Moon would be accelerated to the E-M barycenter and so would the any other mass in the Earth-Moon system.
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    sometimes I start to post something and then change my mind.
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    Quote Originally Posted by dan hunter View Post
    sometimes I start to post something and then change my mind.
    You are a wise person.
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    Quote Originally Posted by Robittybob1 View Post
    I think that is the correct answer. If there was no orbital motion the Moon would be accelerated to the E-M barycenter and so would the any other mass in the Earth-Moon system.
    No it would not.

    Harold is on the money.

    The gravitational effect of the earth is greater than that of the moon by many orders of magnitude for an object located on the earth's surface. The moon's gravitational effect is essentially zero.

    The barycenter location is linearly proportional to the relative masses of the 2 bodies.

    moon mass = 0.0123 earth masses.

    distance - 384000 Km * 0.0123 --> barycenter is 4700 Km from centre of earth (i.e. under, but close to, the earth's surface)

    But for a small object on earths surface the proportional gravitational pull is not 0.0123 it is 81.3*(385000/6371)^2=296,891

    so the effective gravitational centre is 384000km * 1/296,891 = 1.3 Km from earths centre

    i.e. moons gravitational effect is virtually zero
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    Quote Originally Posted by mojorising View Post
    Quote Originally Posted by Robittybob1 View Post
    I think that is the correct answer. If there was no orbital motion the Moon would be accelerated to the E-M barycenter and so would the any other mass in the Earth-Moon system.
    No it would not.

    Harold is on the money.

    The gravitational effect of the earth is greater than that of the moon by many orders of magnitude for an object located on the earth's surface. The moon's gravitational effect is essentially zero.

    The barycenter location is linearly proportional to the relative masses of the 2 bodies.

    moon mass = 0.0123 earth masses.

    distance - 384000 Km * 0.0123 --> barycenter is 4700 Km from centre of earth (i.e. under, but close to, the earth's surface)

    But for a small object on earths surface the proportional gravitational pull is not 0.0123 it is 81.3*(385000/6371)^2=296,891

    so the effective gravitational centre is 384000km * 1/296,891 = 1.3 Km from earths centre

    i.e. moons gravitational effect is virtually zero
    "If there was no orbital motion" .... that was my condition if that was the case the moon would crash into the Earth in a week or two. Everything else would be academic after that!
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