How do we measure the mass of the moon?

[Cardinal]

If the mass is immaterial to the orbit/speed, then how did we find the mass of the moon in the first place?

[dracoi]

A Google search using the question in your title turns up this article: http://curious.astro.cornell.edu/question.php?number=207

Much of it talks about determining the mass of the Earth, but it also talks about the moon. In short, there are three ways:

1. Estimates. We know its size and we can make assumptions about density. This apparently lead to a somewhat high estimate for mass.
2. Joint System Measurements: The moon’s gravity tugs on the Earth. This produces tides (not sure if these are useful for measuring the moon’s gravity), but it also means the Earth and moon are really orbiting a sort of invisible center of gravity between the two. We can measure that.
3. By putting people and spacecraft up on/around the moon, we got direct measurements of gravity that lets us go back to the techniques used in 1 with actual numbers rather than estimates.

[Leaffan]

I’m guessing Luna 10 provided the some pretty accurate numbers.

It was the first artificial satellite of the Moon (or another heavenly body, for that matter). The spacecraft entered lunar orbit on April 3, 1966 and completed its first orbit 3 hours later (on April 4, Moscow time). Scientific instruments included a gamma-ray spectrometer for energies between 0.3—3 MeV (50–500 pJ), a triaxial magnetometer, a meteorite detector, instruments for solar-plasma studies, and devices for measuring infrared emissions from the Moon and radiation conditions of the lunar environment. Gravitational studies were also conducted.

[K364]

@Cardinal

If the mass is immaterial to the orbit/speed, then how did we find the mass of the moon in the first place?

Are you sure that’s true?

The first orbiters of the moon must have given a very accurate reading based upon their orbital period.

[Chronos]

He means, you need to have something else in orbit around the Moon, not just have the Moon’s orbits around something else.

[Superfluous_Parentheses]

Can’t we just weigh a standard mass (on earth) with the moon above it and below it? Sure you’d need a pretty accurate scale, and the exact distance to the moon, but other than that it seems a fairly straight forward and relatively cheap experiment.

ETA: I’m not sure if the mass of the earth factors in with this experiment and it’s much too late to work it out myself.

[Strangelove]

Mass is not immaterial to the orbit. The orbit is all about the mass of the two objects.Kepler’s Third Law of Motion let’s us find the moon’s mass when we know how long it takes it to orbit the Earth. Mass doesn’t matter so much when you’re looking at a planet’s orbit around the sun because the difference between the two masses is so great.

[Beware_of_Doug]

1. Weigh self on standard bathroom scale. Note reading.
2. Take moon out of orbit. Hold over head.
3. Step back onto scale. Note reading.
4. Subtract first reading from second. The difference is the moon.
5. Please place moon back in orbit when finished. Thank You.

[Cardinal]

@Strangelove

Mass is not immaterial to the orbit. The orbit is all about the mass of the two objects.Kepler’s Third Law of Motion let’s us find the moon’s mass when we know how long it takes it to orbit the Earth. Mass doesn’t matter so much when you’re looking at a planet’s orbit around the sun because the difference between the two masses is so great.

I’m willing to believe you in general, but the physics textbook says the mass of the satellite doesn’t determine the orbit, and my dad points out that if the mass mattered, the astronauts in the space shuttle would instantly try to achieve different orbits when they let go of the Space Shuttle, and slam into the wall. Please explain.
I’ve been trying to figure the pull of the moon on a person, and I’m unsure of my figures. Please comment.

F=mmG/d^2

F = 7…5 E27kg100kg6.67 E -11 / 350600000m^2

F = mass of moon * mass of 100kg person * universal grav. constant / distance to moon in meters at closest approach^2

Convert Newtons to pounds by 1/.02242 ratio

I get the effect of the moon directly overhead to be .00894 pounds at most.

[Cardinal]

I’ve looked at your link to Kepler’s Third Law, but could you try to explain it? It seems to say that actually you weigh the object that is being orbited this way, not the satellite. Isn’t that correct? If so, that would mean this doesn’t apply to the moon, and that the mass of the satellite is still immaterial.

@/mechanical-physics.suite101.com

Kepler’s third law is extremely important to astronomers. Because it involves the mass it allows astronomers to find the mass of any astronomical object with something orbiting it. Astronomers find the masses of all astronomical objects by applying Kepler’s third law to orbits. They measure the mass of the Sun by studying the orbits of the planets.

Also, I note that my figures above seem to say that the difference should be noticeable, being 4.047 grams, but I’ve never heard of anyone taking the moon’s position into account. Am I just wrong?

[Francis_Vaughan]

@Cardinal

I’m willing to believe you in general, but the physics textbook says the mass of the satellite doesn’t determine the orbit, and my dad points out that if the mass mattered, the astronauts in the space shuttle would instantly try to achieve different orbits when they let go of the Space Shuttle, and slam into the wall. Please explain.

The constant of proportionality in Kepler’s third law is G(M+m), where M is the sun’s mass, and m is the planent’s mass. Or, in the case of astronauts in orbit, the Earth’s mass and the astronaut’s mass. So indeed, the orbits are determined by the sum of the mass of the Earth and the mass of the orbiting object, and different mass objects will orbit at different rates. However, the difference in rate is, well, somewhat on the insignificant side. 610^24kg versus 110^2 kg means the difference in orbital speed for doubling the weight of an astronaut would be about one part in sqrt(610^22) = 2.410^12, and for low earth orbit, with a speed of 7800m/s, that would mean a difference in speed between a thin, and a rather portly astronaut, of about 3*10^-9m/s. About the width of a hundred hydrogen atoms every second. For all useful intents, when we consider astronauts, satelites, small rocks, the mass of the orbiting object does not matter. Jupiter about the Sun, it makes about 0.1% difference. The Moon around the Earth, it matters.

[Chronos]

Oh, and someone asked about tides, too. In principle, you could use the tides to measure the mass of the Moon, but in practice, things like local geography have a huge and very complicated effect on the tides. The best bet would probably be to use the relative strengths of spring and neap tides to get the components of tides from the Sun and Moon separately, and use those together with the distances to both to determine the ratio of the Moon’s mass to the Sun (the Sun’s mass, in turn, can be determined from the many things orbiting it).

[Cardinal]

@Francis_Vaughan

The constant of proportionality in Kepler’s third law is G(M+m), where M is the sun’s mass, and m is the planent’s mass.

Ah. Could you reference me the whole equation?

Also, do you mean that Jupiter is much less of a fraction of the sun’s mass than the moon is to the earth’s? But even if it is, how does this let us use the orbit to determine the mass of the moon? If it makes only, say, a .6% difference that’s a small number, and without the mass, we wouldn’t know what that percentage was. What am I missing?
Does anyone have the inclination to check my math or explain how I’m down the wrong road entirely?

[ZenBeam]

The term you need to Google is “Reduced mass”. Here’s the Wikipedia page on it.

In principle, given the Earth’s mass, you could calculate the period expected for a small mass at the Moon’s distance, and from the difference between that and the Moon’s actual period, get the reduced mass, hence the Moon’s mass. The error bars might be kind of big, but before lunar satellites, I’m not sure how you do better (apart from just assuming a density for the Moon).

[Malacandra]

@Cardinal

Ah. Could you reference me the whole equation?

Also, do you mean that Jupiter is much less of a fraction of the sun’s mass than the moon is to the earth’s?

I’ll just remark on this one. The Earth is about four and a bit times the Moon’s diameter, about 50 times its volume, about 80 times its mass (the Moon’s less dense overall). The Sun is about ten times Jupiter’s diameter, about a thousand times its volume… so you’re looking at about three orders of magnitude for the masses. (2 x 10[sup]24[/sup] tonnes vs 2 x 10[sup]27[/sup] tonnes, to one significant figure.) By Solar System norms, the Moon is a huge fraction of its primary’s size.

[Dead_Cat]

@Beware_of_Doug

1. Weigh self on standard bathroom scale. Note reading.
2. Take moon out of orbit. Hold over head.
3. Step back onto scale. Note reading.
4. Subtract first reading from second. The difference is the moon.
5. Please place moon back in orbit when finished. Thank You.

That’s a silly way to do it - bathroom scales can be quite inaccurate (and think of how hard it would be to count the number of times the needle spins round!). As Archimedes could have told you, if you’re going to be in the bathroom anyway, just fill the bath with water and put the moon in the bath. Then measure the mass of the water displaced (make sure the moon is fully submerged). This mass will be equal to the mass of the moon.

Don’t forget to dry the moon thoroughly before returning it to orbit, as otherwise the excess water will increase the mass that you measured, rendering the experiment useless.

[Munch]

@Dead_Cat

That’s a silly way to do it - bathroom scales can be quite inaccurate (and think of how hard it would be to count the number of times the needle spins round!). As Archimedes could have told you, if you’re going to be in the bathroom anyway, just fill the bath with water and put the moon in the bath. Then measure the mass of the water displaced (make sure the moon is fully submerged). This mass will be equal to the mass of the moon.

Don’t forget to dry the moon thoroughly before returning it to orbit, as otherwise the excess water will increase the mass that you measured, rendering the experiment useless.

Both of you have left out a very important step - make sure you do it during a Full Moon, otherwise you’re going to just end up with a fraction of the weight.

[Ludovic]

1. Remove moon from orbit.
2. Place moon in an orbit around Uranus.
3. Measure how much Uranus is deformed by Mooning.

[Elendil_s_Heir]

I’m coming to realize just how little I knew about astronomy…

[Chronos]

That’s a silly way to do it - bathroom scales can be quite inaccurate (and think of how hard it would be to count the number of times the needle spins round!). As Archimedes could have told you, if you’re going to be in the bathroom anyway, just fill the bath with water and put the moon in the bath. Then measure the mass of the water displaced (make sure the moon is fully submerged). This mass will be equal to the mass of the moon.

That’d only work if the Moon floated. But since it’s more dense than water (unsurprisingly, since it’s basically made of rock), it’d sink, and you’d just get the Moon’s volume (which is easy to calculate anyway), not its mass.