PDA

View Full Version : The Mass of the Moon



Shawn
2010-Jan-27, 06:18 PM
HI. I'm new to astronomy, and I'm having a hard time figuring out how we can determine the mass of the moon. The figure cited is usually as being 7.36 × 10 to the 22nd power kilograms, but I have no idea how that number is arrived at. Explanation would be appreciated.

Shawn
2010-Jan-27, 06:24 PM
Hi. I'm new to astronomy, and I'm having a hard time figuring out how we can determine the mass of the moon. If someone could explain that to me, I'd greatly appreciate it.

PetersCreek
2010-Jan-27, 06:37 PM
Shawn,

Welcome to the BAUT forums. Please be patient when making new posts. Because you're a newcomer, your posts may be held in a queue for moderator approval, espcially is they contain links, images, or certain keywords. Don't worry, it's just a spambot countermeasure. Once you have a few more posts under your belt, your posts will appear immediately. If you have already done so, please take a little time to familiarize yourself with our rules, linked in my signature line below. You might also check out the other informational links.

Again, welcome to BAUT.

ETA: Duplicate posts merged.

AndreasJ
2010-Jan-27, 08:11 PM
It's ultimately derived from Newton's law of gravitation:

http://www.codecogs.com/eq.latex?F=-G\frac{m_1m_2}{r^2}

If we know the force F which the Moon attracts something, the mass m2 of that something, and the distance r between it and the Moon, the Moon's mass m1 falls out:

http://www.codecogs.com/eq.latex?m_1=-\frac{Fr^2}{Gm_2}

The "something" could be a space probe passing by the Moon, of known mass (we built it), and position and acceleration followed by radio. Acceleration gives the force via F = m2a (Newton's 2nd Law).

Or it could be the Earth itself. The arithmetic gets slightly complex, but with the help of Newton's Laws, the orbital radius (known from radar, parallax, etc) and orbital period one can determine the total mass of the Earth-Moon system - if the mass of the Earth is known, that of the Moon follows. Earth's mass follows from how fast rocks fall and the value of the gravitational constant G.

chornedsnorkack
2010-Jan-27, 08:27 PM
Earth's mass follows from how fast rocks fall and the value of the gravitational constant G.

That is the hardest part.

Because it involves trying to measure gravitational attraction to small objects, whose mass can be measured.

How precisely is G known?

Shawn
2010-Jan-27, 08:32 PM
@AnreasJ

I'm familiar with that equation (somewhat), but here is what I'm still puzzling over. F is the moon's gravitation force, right? So... how is that determined without knowing the moon's mass? From what I can gather, in order to determine the gravitation, you need to know the mass, and vice-versa. So... how do you determine the first of the pair? In your case, how did you determine F, without knowing m1?

Shawn
2010-Jan-27, 08:43 PM
@AndreasJ

I think what you're saying is that the only real way for us to figure it out (at this point) is through empirical testing (i.e. we need to send a satellite up there to orbit the moon). Is that correct (there's not really a theoretical way we can figure it out from Earth)?

ugordan
2010-Jan-27, 08:53 PM
How precisely is G known?
Not terribly precisely, in fact the product of a body's mass and G, GM can be known more precisely for celestial objects and is equally as useful for doing calculations.

Shawn
2010-Jan-27, 08:59 PM
Or it could be the Earth itself. The arithmetic gets slightly complex, but with the help of Newton's Laws, the orbital radius (known from radar, parallax, etc) and orbital period one can determine the total mass of the Earth-Moon system - if the mass of the Earth is known, that of the Moon follows. Earth's mass follows from how fast rocks fall and the value of the gravitational constant G.

And we can determine the mass of the Earth because we have man-made satellites that orbit the Earth?

One little thing that bothers me about Newton's Law of Universal Gravitation is the fact that it was intended for "point masses" (which I think means masses which have a uniform density). However, if we're assuming something about the density of the Earth or Moon, then there's a little bit of circular logic going on if we're trying to determine mass, isn't there (since density is a component of mass)?

AndreasJ
2010-Jan-27, 09:01 PM
@AndreasJ

I think what you're saying is that the only real way for us to figure it out (at this point) is through empirical testing (i.e. we need to send a satellite up there to orbit the moon). Is that correct (there's not really a theoretical way we can figure it out from Earth)?
Not quite.

Empirical testing is always necessary, but we need not put a probe in orbit around the Moon. A fly-by probe would suffice, or no artificial probe at all, as a natural one happens to be present already - the Earth. We can determine the Moon's mass purely from observations on Earth.

Shawn
2010-Jan-27, 09:17 PM
Not quite.

Empirical testing is always necessary, but we need not put a probe in orbit around the Moon. A fly-by probe would suffice, or no artificial probe at all, as a natural one happens to be present already - the Earth. We can determine the Moon's mass purely from observations on Earth.

In order to determine the mass of the Earth, don't we need to know the mass of the object orbiting it? If that's the case, then wouldn't we need to use an artificial satellite of human construction to first determine the mass of the Earth before we could determine the mass of the moon?

Hopefully these aren't stupid questions. I'm just trying to wrap my mind around this.

AndreasJ
2010-Jan-27, 10:09 PM
In order to determine the mass of the Earth, don't we need to know the mass of the object orbiting it?
No.

To determine the mass of the Earth, we first study how rocks fall. This gives us (via the gravitational law quoted above) the product GmEarth:

http://www.codecogs.com/eq.latex?Gm_{Earth}=-\frac{Fr^2}{m_{rock}}=-a_{rock}r^2

Then, we do very sensitive measurements of how much two test masses attract one another in the lab (using the same law). This is horribly difficult in practice, as the forces concerned are so small, albeit simple in concept. This gives us G.

Presto! and we've got the mass of the Earth.


(And no, I don't think your questions are stupid.)

Centaur
2010-Jan-28, 05:54 AM
HI. I'm new to astronomy, and I'm having a hard time figuring out how we can determine the mass of the moon. The figure cited is usually as being 7.36 × 10 to the 22nd power kilograms, but I have no idea how that number is arrived at. Explanation would be appreciated.

Welcome to the discussion group, Shawn. Excellent question!

Other responders are correct that by measuring the positions and velocities of spacecraft, and applying Newton’s gravitational formula, we can accurately determine the Moon’s mass. However, the Moon’s mass was fairly well known before the space age. In a sense the Moon does have a satellite. It is the Earth, although they both revolve around a common barycenter (center of mass) that is beneath the Earth’s surface. The monthly rotation of the Earth around the Earth-Moon barycenter could be measured by variations from the expected positions the Earth would have in its orbit around the Sun without the influence of the Moon. It was not hard to determine that the Moon’s center averaged 81 times further from the barycenter than was the Earth’s center. From this it could be deduced that the Moon had 1/81 the mass of the Earth. The Earth’s mass could be determined not only by the distance and orbital period of the Moon, but by measuring the acceleration of any object falling in a vacuum on Earth and applying Newton’s gravitational formula.

Centaur
2010-Jan-28, 06:06 AM
One little thing that bothers me about Newton's Law of Universal Gravitation is the fact that it was intended for "point masses" (which I think means masses which have a uniform density). However, if we're assuming something about the density of the Earth or Moon, then there's a little bit of circular logic going on if we're trying to determine mass, isn't there (since density is a component of mass)?

One of Newton’s motivations to invent integral calculus, was to prove that a body with a certain type of uniformity in its density could be considered a point object for calculations involving celestial motions. This type of uniformity was related to a spherical object which had a uniform density within each of its spherical shells from surface to center, but not necessarily the same density among different shells. This is actually a fairly good approximation of most large celestial bodies whose densities increase relatively uniformly from surface to center.

AndreasJ
2010-Jan-28, 11:28 AM
Other responders are correct that by measuring the positions and velocities of spacecraft, and applying Newton’s gravitational formula, we can accurately determine the Moon’s mass. However, the Moon’s mass was fairly well known before the space age.
I believe I've repeatedly said that an artificial probe isn't necessary. :)

Stroller
2010-Feb-07, 01:52 PM
Newton got it wrong by a 100% in this case
http://www.dioi.org/kn/newtonmoonerror.htm

Stone recalulated it fom Le Verriers tables, and spotted a numerical error
http://articles.adsabs.harvard.edu/full/seri/MNRAS/0027//0000241.000.html

One problem is that the Moon orbits slightly higher due to the sun than it's mass would suggest in a two body problem.
http://adsabs.harvard.edu/full/1928AJ.....38..181L

I'd like to find out the actual observation values of parallax from the Earth due to it's motion about the Earth - Moon barycenter if anyone knows where I can find some.

Hornblower
2010-Feb-07, 02:45 PM
Newton got it wrong by a 100% in this case
http://www.dioi.org/kn/newtonmoonerror.htm
Much ado about what is now a non-issue. Newton simply had no means of measuring the Earth's barycentric motion with any accuracy.

Stone recalulated it fom Le Verriers tables, and spotted a numerical error
http://articles.adsabs.harvard.edu/full/seri/MNRAS/0027//0000241.000.html

One problem is that the Moon orbits slightly higher due to the sun than it's mass would suggest in a two body problem.
http://adsabs.harvard.edu/full/1928AJ.....38..181L

I'd like to find out the actual observation values of parallax from the Earth due to it's motion about the Earth - Moon barycenter if anyone knows where I can find some.

hhEb09'1
2010-Feb-07, 03:45 PM
Newton got it wrong by a 100% in this case
http://www.dioi.org/kn/newtonmoonerror.htmAccording to that link, he'd calculated (fairly accurately) the ratio of the density of the earth to the density of the sun:
Newton arrived at an accurate estimate of the Sun’s mean relative density as one quarter that of Earth (modern value 0.27). This he found using his inverse square law of gravity; he compared the rate at which planets accelerate towards the Sun in their orbits with the rate at which the Moon accelerates towards the Earth in its orbit. However, as nothing is observ*able accelerating towards the Moon, he had no such means of obtaining an estimate of its relative mass.
Then he used the syzygy/quadrature tidal ratios at Bristol (45 feet/25 feet) and Plymouth (41 feet/23 feet), because they more or less agreed but they're huge tides, probably amplified by coastal resonances. That ratio is 45/25, or 9/5 (misprinted as 9/55 at link), but if it had been 8/3 Newton would have been right on track: if (L+S)/(L-S) = 8/3, then L = 2.2S--supposedly the modern value. 'Course, the article says Newton massaged the data further, so, who knows.

Stroller
2010-Feb-08, 12:12 AM
Much ado about what is now a non-issue. Newton simply had no means of measuring the Earth's barycentric motion with any accuracy.

Well this is what I'm wondering about. Didn't anyone try to observe the parralax caused by the monthly motion of Earth about the Earth-Moon barycentre against, say, Mars WRT to some backgound stars? Or was that beyond the margin of error of the equipment and the star catalogues of the time?

antoniseb
2010-Feb-08, 12:35 AM
Well this is what I'm wondering about. Didn't anyone try to observe the parralax caused by the monthly motion of Earth about the Earth-Moon barycentre against, say, Mars WRT to some backgound stars? Or was that beyond the margin of error of the equipment and the star catalogues of the time?

What you're asking is a more complex measurement than is needed AND it won't give a good answer because the Earth-Moon system is not a pure two-body system. The Sun, Jupiter, Venus, Mars, and lesser influences all contrive to make the Moon's orbit not very close to elliptical, and the whole barycenter concept only a vague approximation.

Stroller
2010-Feb-08, 12:08 PM
What you're asking is a more complex measurement than is needed AND it won't give a good answer because the Earth-Moon system is not a pure two-body system. The Sun, Jupiter, Venus, Mars, and lesser influences all contrive to make the Moon's orbit not very close to elliptical

Hi Antoniseb and thanks for your reply. Just trying to break that down a bit, purely for the sake of the history of science perspective. The effect of the sun on the moons orbit would be fairly constant, as the Earth moon system orbits the sun in an elliptical orbit with a fairly small amount of eccentricity and oblateness. Certainly in measurements taken 14 days apart as the Earth wobbled from one side of the Earth-Moon barycentre to the other, I wouldn't expect the sun to make a huge difference, especially as the moon would be at pretty much the same distance from the sun at the moments when parallax would be most apparent; i.e. WRT to (say) Mars closest approach. Jupiter wouldn't have moved much in 14 days in angular terms either (1.16 degrees), Though of course the Earth Moon system would move 15 degrees or so WRT Jupiter in two weeks. Venus is too small and too far away to make much difference in 14 days too.

So these other perturbations would be small compared to the Earth's motion WRT the Earth Moon Barycentre over two weeks.

So my question is, what sort of amount of parallax would we expect to see?

Tycho Brahe had several goes at observing Mars' diurnal Parallax, which would be of the same order of magnitude as the barycentric parallax. He was thwarted by refraction and poor instrumentation.
http://aas.org/archives/BAAS/v29n5/aas191/abs/S001003.html

David Gill got some observations in 1877:
http://tallbloke.files.wordpress.com/2010/02/gill-mars1.jpg
But what were the magnitudes of the arcs he observed? And at what time of the morning and evening?


and the whole barycenter concept only a vague approximation.

Only as vague as the concept of gravity one would hope. ;)

Of more direct relevance to this thread, and also to the barycentre question is the last paragraph of the piece quoted:
http://tallbloke.files.wordpress.com/2010/02/gill-moon.jpg

It turns out then, that Gill's methods of the measurement of parallax using the minor planets enabled a more accurate determination of the Moon's mass than had previously been managed by Le Verrier. As a bonus, he determined the barycentric motion in order to get that more accurate mass. It's going to be interesting to compare his results with the theoretical value.

So the hunt for the Cape Annals begins...

....And it turns out my home town university has the relevant volume in the special collection. Nice one, I'll call by tomorrow.