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Our Moon and Gravity

  1. Dec 2, 2009 #1
    Hi Forum,

    I am new and have just registered as I have a few quetions I was hoping you may know concerning our moon.

    At what distance must the moon be from earth for it to no longer have any gravitational effect on the earth?

    What is the strength of the moons gravitational effect on the earth?

    and

    How much weaker would this effect be over say an increased distance of say 10 miles and 100 miles?

    Please help, I'm no scientist or mathematician so responses in idiots english would be good :blushing:.

    (I am doing some research as I am writing my first Fantasy / Sci Fi Novel :biggrin:)

    Thanks forum!
     
  2. jcsd
  3. Dec 2, 2009 #2

    berkeman

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    Welcome to the PF. What you need is to use Newton's Law of Gravity:

    http://en.wikipedia.org/wiki/Newton's_law_of_gravity

    That article should get you started. You will need to use google or wikipedia to find the masses of the Earth and moon, and find out how far (on average) they are apart.

    Read through the article, look up those numbers, and see if that gives you what you need. Be sure to carefully convert all calculations into mks units (meters-kilograms-seconds), including the "miles" distances you mentioned.

    After doing that, if you still have questions, please post your work here, and ask away!
     
  4. Dec 2, 2009 #3
    In regards to your first question, the answer is that as long as the Moon exists it will always have a gravitational effect on the Earth. Why? The equation for the gravitational force between two bodies (F) is given by, F = GMm/r2. Where M and m are the masses of the two bodies, G is a constant, and r is the separation between the two objects. Using this equation you can see that regardless of how far away the Moon is from the Earth the gravitational force between them will never be 0. Presumably you probably want to know at which point this force will be negligible, but in that case it depends on what you mean by negligible. Since you are dealing with SF, you likely have some specific interaction in mind. If you can elaborate I would be glad to help. Though I understand if you are trying to safeguard your idea for this piece ;).

    Second question, the strength of the Moon's gravitational effect on the Earth. Again this question is a little ambiguous. I could give you a force by plugging in the math but something tells me this is not what you are looking for. Are you looking for something specific, like differences in tidal heights, etc.?

    Question three, now this one can be answered :). << solution deleted by berkeman >>

    Hopefully that helps, I'll cut it there since I'm running a little long, I'd hate to write a story on you, after all that's your job :). If you need anymore help or clarification you know where to ask. Good luck and I look forward to reading your story in the future. Cheers.
     
    Last edited by a moderator: Dec 2, 2009
  5. Dec 2, 2009 #4

    berkeman

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    Please check your PMs, pumpkin. Thank you.
     
  6. Dec 3, 2009 #5

    Matterwave

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    One little caveat to the "gravity has infinite range" rule.

    It is true that gravity's range is not limited (gravitons, if they exist, are thought to be stable and do not decay); however, according to the theory of general relativity, the effects of gravity should travel at the speed of light. Since our universe is finitely old; if the moon is sufficiently far away (beyond the horizon of the observable universe or about 47 billion light years taking into account co-moving coordinates), the effects of its gravity will not have had time to reach the Earth. The gravity of the moon, would thus only effect the Earth in the future, and not now.
     
  7. Dec 3, 2009 #6
    If we multiply the mass of the moon (7.3459*10^22 kg) by the gravitational constant (6.67*10^-11) and divide by the square of the average distance between the earth and the moon ((3.844*10^8m)^2), we get the gravitational potential due to the moon at this distance, approximately 3.3 times 10^-5 meters per second.

    http://www.wolframalpha.com/input/?i=7.3459*10^22*6.67*10^-11*kg*N*m^2*kg^-2*%283.844*10^8m%29^-2

    This means that, each second, the moon causes any object to increase the speed with which it falls towards the moon by about three hundredths of a millimeter per second. (Three hundredths of a millimeter is the width of a human hair.) At this rate, if we ignored the gravitational effect of the earth, it would take about four hours for an object, starting from rest, to reach a speed of one mile an hour (falling towards the moon)! So it's a very slight effect at the earth's surface compared to the earth's gravity, which causes objects to speed up towards the earth by 9.8 meters per second every second.

    The strength, at some point in space, of gravitational potential due to a mass is inversely proportional to the square of the point's distance from the center of mass. (Sorry about the jargon!) This just means that if you're twice as far away, you'll feel one fourth of the gravity you did.

    If you want to experiment with different distances, without doing calculations yourself, you can use Wolfram Alpha's gravity calculator:

    http://www.wolframalpha.com/input/?i=gravity

    Where it asks for "mass", fill in the mass of the body causing the gravitational effect, and where it asks for "radius" fill in the distance at which you want to know the strength of the effect. I'm sure there are lots of similar programs online.

    The distance between the earth and the moon varies as the moon orbits the earth. The difference between the biggest and the smallest distance is about 26 500 miles. This has an effect on tides, as does the alignment of moon and sun relative to the earth. The diameter of the earth is about 7 900 miles. Suppose you're standing on the earth's equator and can see the moon directly overhead. Unless my trigonometry is mistaken (fingers crossed...), traveling about 281 miles in any direction should put you ten miles further away from the moon, and traveling about 891 miles should put you 100 miles further away from the moon. Simplest way to travel: wait for the earth to rotate that distance. How much does the tide change in 16 minutes? That's how much difference it makes being 10 miles further away from the moon. How much does the tide change in 52 minutes? That's how much difference 100 miles makes.
     
  8. Dec 3, 2009 #7

    marcus

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    GuitarGirl, are you still around and still teaching yourself about gravity? I'll try to give a simplified approach. If you have already moved on, or if this doesn't work for you, that's fine, it might work for someone else.

    I'll assume that for some reason you want to talk miles---maybe it has to do with your Fantasy/SciFi story idea. We normally talk metric units, but let's do it in whatever units.

    Try typing this into the Google search window, don't ask why just type :biggrin:

    (G*mass of Earth)/(radius of Earth^2)

    Type in literally, word for word, and you can probably leave out the parentheses. You don't have to look up the mass of Earth in a book, you just put "mass of Earth" into the calculator. The google search window will act in certain cases like a calculator.
    It should say the strength of earth gravity at the surface of the earth. But maybe not in the UNITS you want. Suppose you want to be told the strength in "gee". Like the answer should be about "one gee"---that being earth surface normal gravity. Then you type this:

    (G*mass of Earth)/(radius of Earth^2) in g

    Putting that "in g" at the end makes the calculator tell it to you in the g unit.

    Basically that works for anything that the Google calculator knows about. You could say

    G*mass of anything/the square of any distance

    and it will tell you the influence of gravity of that thing, at that distance.


    So you could type in:

    G*mass of moon/(radius of moon)^2 in g
    and it will tell you the strength of gravity at moon surface, in units of earth normal gravity.
    It should come out around one sixth g, I think. Something like 0.16 g.
     
    Last edited: Dec 3, 2009
  9. Dec 4, 2009 #8

    marcus

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    Now you are asking about the influence of the moon at a very large distance, around 240 thousand miles. That is going to be a very very tiny number of gee. Because the formula uses the square of the distance and it is a long ways away. But you can still make the google calculator tell you. You just type this into the box:

    G*(mass of moon)/(240000 miles)^2 in g

    Just type the blue thing letter for letter into the box and press return
    it should tell you a tiny tiny number of gees. (looks like around 3 millionths of a gee, or more exactly 3.4)

    All it means is G times the mass of the moon divided by the square of the distance from the center of the moon will tell you the pull or the 'acceleration due to gravity' of a test mass at that distance. And you can get it for different distances, like 230 thousand miles or 250 thousand miles. Experiment and find out. You could, say, put in

    G*(mass of moon)/(245000 miles)^2 in g

    and that would tell you what adding 5000 miles to the distance would do. How much it would weaken the pull. I think it came to more like 3.2 millionths of a gee. (I hope you know that 10-6 is a millionth.)

    If this hands-on approach works for you, or anybody who happens to read it and try it, please get back with comment, and if you have any trouble---if something doesn't appear to work right---let us know.
     
    Last edited: Dec 4, 2009
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