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Post by fireballxl5 on Nov 9, 2010 8:48:09 GMT -4
I was having a discussion with someone and they said that we can work out the speed of the Moons rotation by looking at it through telescopes. This one has bent my head a bit as i know we only see one side of the Moon. Could anyone help me out and say if this is possible or not.
Thanks guys.
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Post by drewid on Nov 9, 2010 8:59:06 GMT -4
One rotation per day more or less, errrmm, That's it isn't it?
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Post by fireballxl5 on Nov 9, 2010 9:07:47 GMT -4
Sorry, what i meant was its said at the Moons equator the rotation speed reaches its maximum of 10.3 mi/hr. Can this be shown to be true by looking at it through a telescope? Its just im having a discussion with some HB and hes fine with the rotation speed quoted but seems reluctant to take on other stuff. So i asked how he knows that speed is true and he said you can check using a telescope.
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Post by echnaton on Nov 9, 2010 11:01:36 GMT -4
I don't know how a telescope would help, the rotation can be confirmed by sight. The moon rotates once in approximately 27.3 days and is about 2160 miles in diameter at the equator. Through a little math on that and you get 10.3 mph.
I guess you must ask him how to check with a telescope.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Nov 9, 2010 11:37:48 GMT -4
I guess you must ask him how to check with a telescope. That's what I was going to say. If he says it can be done, have him explain it to you.
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Post by fireballxl5 on Nov 12, 2010 0:33:56 GMT -4
I asked him how to check with a telescope and his responce was:
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Bob B.
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Posts: 3,072
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Post by Bob B. on Nov 12, 2010 2:02:36 GMT -4
I asked him how to check with a telescope and his responce was: (1) Why does one need a telescope for that, and (2) how does one determine the Moon's speed of rotation from that? I might know what they guy is driving at, but he's given you a BS answer that needs clarification. I'd like him to give a detailed answer of what observations must be made and how one calculates the rotational speed from those observations.
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Post by echnaton on Nov 12, 2010 11:18:41 GMT -4
I asked him how to check with a telescope and his responce was: There is still quite a lack of understanding between what he is saying and what we know of what he really means that I am not sure we can make a meaningful answer. Why don't you invite him here to ask the question directly?
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Nov 12, 2010 12:37:29 GMT -4
There is still quite a lack of understanding between what he is saying and what we know of what he really means that I am not sure we can make a meaningful answer. Agreed. In fact, I don't know what is meant by "speed of rotation". Does he mean the period of rotation? Or does he mean the speed that the surface is moving, and if so, at what latitude? Furthermore, observing the phases gives the synodic orbital period, not its rotational period. Could the person be confusing the terms rotation and revolution and actually be talking about the orbital period and/or its orbital speed? The bottom line is I don't know what the person is talking about because he's not being clear. All that being said, it is possible to determine the rotation period from observing the phases, but I don't know if my method is the same as this person has in mind. Observing the time it takes to go from one phase to the next (say from new moon to new moon) gives the Moon's synodic period, which is about 29.5 days. In 29.5 days Earth moves about 29 degrees around the Sun. Therefore, for the Moon to return to a previously observed phase it must complete one full revolution around Earth plus 29 degrees, or 360 + 29 = 389 degrees. Since it takes the Moon 29.5 days to revolve 389 degrees around Earth, the time is takes to complete a 360-degree revolution is 29.5 x 360 / 389 = 27.3 days, which is the Moon's sidereal period. Since the Moon always keeps one face toward Earth, its period of rotation must be equal to its sidereal period, or 27.3 days.
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Post by echnaton on Nov 12, 2010 13:39:00 GMT -4
Agreed. In fact, I don't know what is meant by "speed of rotation". Does he mean the period of rotation? Or does he mean the speed that the surface is moving, and if so, at what latitude? In reply 2, fireballxl5 did clarify that it was the rotational speed at the equator and gave 10.3 MPH as the speed. In my brief research that seems to be the correct speed of the travel for the solar zenith over the equator. Or how fast you would have to travel across the equator to keep the sun at roughly the same place in the sky. Figuring out what people really believe about space physics is tough enough for me, doing it second hand in next to impossible.
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Bob B.
Bob the Excel Guru?
Posts: 3,072
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Post by Bob B. on Nov 12, 2010 14:06:27 GMT -4
In reply 2, fireballxl5 did clarify that it was the rotational speed at the equator and gave 10.3 MPH as the speed. In my brief research that seems to be the correct speed of the travel for the solar zenith over the equator. Or how fast you would have to travel across the equator to keep the sun at roughly the same place in the sky. Now that you've reminded me, I do remember reading that. When I came back to the thread a second time I only re-read the first post and forgot about the subsequent clarification.
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Post by fireballxl5 on Nov 22, 2010 7:49:50 GMT -4
Ok, i asked him you're questions and his responce was:
This is wrong yeah?
I dont see what this has to do with his first claim that you can check with a telescope by:
As far as i can see he dropped his first claim then went onto another which is not answering the questions bob asked.
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Post by gwiz on Nov 22, 2010 8:37:47 GMT -4
Ok, i asked him you're questions and his responce was: This is wrong yeah? Yeah. You need to divide the circumference by the period to get the speed.
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Post by fireballxl5 on Nov 22, 2010 8:56:15 GMT -4
v=d/t = Moon's circumference / period = 2*pi*1737km/(27.32*24) = 16.645 km/hr, same as 10.3 mi/hr or 4.6 m/s.
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Post by fireballxl5 on Nov 27, 2010 6:25:50 GMT -4
The guy says he cant do the calculation v=d/t = Moon's circumference / period = 2*pi*1737km/(27.32*24) = 16.645 km/hr, same as 10.3 mi/hr or 4.6 m/s so hes come up with his own way lol.
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