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    Suicidebygasstationsushi's Avatar
    Suicidebygasstationsushi Posts: 2, Reputation: 1
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    #1

    Sep 22, 2007, 10:35 AM
    rate at which the moon orbits the earth.
    Okay, the question I'm having trouble with is this.

    Find an approximate value for the rate at which the moon orbits the earth. Assume that the moon's orbit is circular. To the side it says Hint: The average distance between the earth and the moon is 382,000,000 meters.

    I understand that I'd have to get the circumfrence of the circular path so it'd be:
    2 Pi * 382,000,000 m. which equals 2,400,180,000 m. (or at least that's what my calculator says), so that's my distance, but if distance = rate * time, then where exactly am I supposed to get the value of time to solve the problem?? It doesn't give me that in the question.
    CaptainRich's Avatar
    CaptainRich Posts: 4,488, Reputation: 537
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    #2

    Sep 22, 2007, 10:52 AM
    Quote Originally Posted by Suicidebygasstationsushi
    Okay, the question I'm having trouble with is this.

    Find an approximate value for the rate at which the moon orbits the earth. Assume that the moon's orbit is circular. To the side it says Hint: The average distance between the earth and the moon is 382,000,000 meters.

    I understand that I'd have to get the circumfrence of the circular path so it'd be:
    2 Pi * 382,000,000 m. which equals 2,400,180,000 m. (or atleast that's what my calculator says), so that's my distance, but if distance = rate * time, then where exactly am I supposed to get the value of time to solve the problem??? It doesn't give me that in the question.
    Quite simply, you don't have enough information to correctly calculate the answer.

    Pi - Wikipedia, the free encyclopedia
    s_cianci's Avatar
    s_cianci Posts: 5,472, Reputation: 760
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    #3

    Sep 22, 2007, 04:30 PM
    To follow up to CaptainRich's response, you need to know how long it takes the moon to orbit the earth.
    stoutworldinc's Avatar
    stoutworldinc Posts: 1, Reputation: 1
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    #4

    Jan 16, 2009, 01:37 PM

    OK well I am personally asking the question too lol I have it in my text book this week for homework but anyway from my understanding... if its d=r*t then it gives you d which is 382,000,000 meters which is distance apx from earth to moon. Then it gives you the moon orbiting the earth which is apx 30 days which is the cycle of the moon rotation to earth so that's time. So your finding rate which would be r=d/t. so 382,000,000 meters divided by 30 days = 12733333.33 but the thing I'm missing is what will that have to be converted to for the correct answer. Usually the final has to be changed to a multiple of days or meters. Like if you figure an equation finding distance from hours and mph then distance is in miles a multiple of hours and mph.
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #5

    Jan 16, 2009, 01:50 PM
    You found the distance the moon makes in one full revolution of the moon.

    Let's use km. That means 2,400,177 km around the Earth.

    Now, the moon does this in 27.3 days, not 30

    So, 2,400,177/27.3=87918 km per day. In hours, divide by 24

    3663 km or 3663000 m per hour. What rate units is it supposed to be in?

    Change it to whatever you need.
    Credendovidis's Avatar
    Credendovidis Posts: 1,593, Reputation: 66
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    #6

    Jan 16, 2009, 07:03 PM
    Just an add-on : there is more to this question.

    Both earth and moon are effected by the presence and gravity of the other. The trajectory of the earth - moon system is one of a twisted path around the sun, whereby the moon's path is much more profound than that of the earth.

    Here is a LINK to a website with an excellent animation that shows this effect.

    Note the wobble in the earth path due to the gravity influence of the moon.

    :)

    .
    Patsfan2019's Avatar
    Patsfan2019 Posts: 1, Reputation: 1
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    #7

    Jan 18, 2010, 08:03 PM
    Quote Originally Posted by galactus View Post
    You found the distance the moon makes in one full revolution of the moon.

    Let's use km. That means 2,400,177 km around the Earth.

    Now, the moon does this in 27.3 days, not 30

    So, 2,400,177/27.3=87918 km per day. In hours, divide by 24

    3663 km or 3663000 m per hour. What rate units is it supposed to be in?.

    Change it to whatever you need.
    Ok, I have this same question in my homework due tomorrow so I hope its not too late to get some clarification on this. Now the metric unit of time is seconds, correct ? So according to the way this was done above my answer would look like this right ?

    2,400,176,787 meters / 2,360,620.8 seconds (equivelant of 27.3 days) = 1016.8 m/sec
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #8

    Jan 21, 2010, 07:10 AM

    Well, it's approximately right.

    27.3 days = 2, 358, 720 seconds (well, only a small difference, but you put it to 8 significant figures... )

    And we say the SI unit of time is the second.
    ebaines's Avatar
    ebaines Posts: 12,132, Reputation: 1307
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    #9

    Jan 21, 2010, 10:36 AM
    Quote Originally Posted by Unknown008 View Post
    Well, it's approximately right.

    27.3 days = 2, 358, 720 seconds (well, only a small difference, but you put it to 8 significant figures...)
    Unknown008: tsk, tsk... never take a calculation to 8 significant places unless the data you're working with is good to 8 places. The moon's orbital period is given as 27.3 days, so good to only 3 decimal places. If you want more accuracy, you need to have more accurate input data. For example - you might use 27.322 days, and then you could go to 5 decimal places.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #10

    Jan 22, 2010, 09:20 AM

    ebaines, I think I didn't well formulate my post.

    If you convert 27.3 days into seconds, you get 2,358,720 seconds.
    The poster there put 2,360,620.8 seconds, which was way off, and even then, he posted it to 9 sf which is even worse!
    I didn't mention that I was going to use all the sf, but only showed the conversion, and where the poster went wrong, for the sake of saying double wrong.

    Sorry for the misunderstanding :o
    ebaines's Avatar
    ebaines Posts: 12,132, Reputation: 1307
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    #11

    Jan 22, 2010, 09:56 AM

    It seems that Patsfan actually used 27.322 days in his calculation (even though he wrote 27.3) to arrive at his 2,360,620.8 seconds, and then kept his fnal answer to 5 significant digits. But the fault here goes further back - the original data for the moon's orbit was given to only 3 digits. I think it was Galactus who then used that data to calculate the circumference of the orbit to 10 significant digits. And then that 10-digit figure got used in the subsequent posts.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #12

    Jan 22, 2010, 10:10 AM

    Well, anyway, I don't think the poster will get here again. Better get this thread closed... before some other newbie revive it!
    InfoJunkie4Life's Avatar
    InfoJunkie4Life Posts: 1,349, Reputation: 81
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    #13

    Jan 24, 2010, 12:38 AM

    Just to comment, another important misinformation. You guys have calculated this distance as the moon from surface of the earth not the center of its orbit. The earth has around a 6.3 million meter radius. Technically, that doesn't matter either, because the actual radius of the moon's orbit is somewhere around 385 million meters. This number is really an average. The pull of other planets, the elliptical nature of its shape, and a few other things actually make the moon's orbit quite irregular.
    InfoJunkie4Life's Avatar
    InfoJunkie4Life Posts: 1,349, Reputation: 81
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    #14

    Jan 24, 2010, 12:39 AM

    I know Unknown, I just had this awful feeling to put my two cents in...
    Fr_Chuck's Avatar
    Fr_Chuck Posts: 81,304, Reputation: 7692
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    #15

    Jan 24, 2010, 05:25 AM

    Ok, why are we opening a 3 year old post. Closed

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