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Postal76
2007-Aug-16, 08:18 PM
Other than putting a satellite in orbit around the moon and measuring its period, how can one calculate the moon's mass?

To use Newton's version of Kepler's Third Law you'd have to know the center of mass of the Earth-Moon system, right?

grant hutchison
2007-Aug-16, 09:14 PM
To use Newton's version of Kepler's Third Law you'd have to know the center of mass of the Earth-Moon system, right?Not necessarily. Knowing the orbital period, the separation, and Earth's mass would be one way of getting there.
But you might also consider examining how the moon's gravity affects things here at the surface of the Earth.

Sorry to be a little obscure, but this has the feel of a homework question ... :)

Grant Hutchison

Postal76
2007-Aug-16, 09:36 PM
Not necessarily. Knowing the orbital period, the separation, and Earth's mass would be one way of getting there.
But you might also consider examining how the moon's gravity affects things here at the surface of the Earth.

Sorry to be a little obscure, but this has the feel of a homework question ... :)

Grant Hutchison

I appreciate you trying to make me think more about it, but I can assure you this is not a homework question. It is in fact a matter of debunking some dubious theories regarding the moon.

Assuming that I can just use the mean separation between the Earth and the Moon, along with the moon's sidereal period and the Earth's mass, I get a lunar mass of about 7.28 x 10^22 kg (via Kepler's Third Law), which has an error of about %1 with respect to the accepted value of 7.36 x 10^22 kg.

Is this sound logic, and why would the calculation yield a lower than actual mass?

tony873004
2007-Aug-16, 09:38 PM
If you're given size and density, or size and surface gravity, you could do it that way too.

Jerry
2007-Aug-16, 09:43 PM
I appreciate you trying to make me think more about it, but I can assure you this is not a homework question. It is in fact a matter of debunking some dubious theories regarding the moon.

Assuming that I can just use the mean separation between the Earth and the Moon, along with the moon's sidereal period and the Earth's mass, I get a lunar mass of about 7.28 x 10^22 kg (via Kepler's Third Law), which has an error of about %1 with respect to the accepted value of 7.36 x 10^22 kg.

Is this sound logic, and why would the calculation yield a lower than actual mass?
Kepler's law is only an approximation. You have second, third and forth order effects: Variation of the barycenter about the sun, Venus, Jupiter, tidal disruption and so on.

Postal76
2007-Aug-16, 10:08 PM
Kepler's law is only an approximation. You have second, third and forth order effects: Variation of the barycenter about the sun, Venus, Jupiter, tidal disruption and so on.

Yes, but is Kepler's law a good approximation? Is it possible that the mass could be much greater than what Kepler's Third Law approximates?

grant hutchison
2007-Aug-16, 10:44 PM
I appreciate you trying to make me think more about it, but I can assure you this is not a homework question. Sorry, you showed some of the classic signs: first-time poster posing a brief physics problem and providing no context to the question. :)

Is this sound logic, and why would the calculation yield a lower than actual mass?Sound logic. If you've eliminated the usual problems with old data, rounding errors and significant figures, I think Jerry has provided the answer.
Jean Meeus describes the effects of perturbations from other bodies at some length in his Mathematical Astronomy Morsels. The major influence of relevance to your problem is the sun; its effect on the Kepler relationship you've used is to reduce the apparent force between Earth and moon by a factor of 1.002723 (Meeus provides a reference to a French paper from 1959). That apparent force reduction will trickle through your calculation and reduce the apparent mass of the moon.

Grant Hutchison

grant hutchison
2007-Aug-16, 11:05 PM
Oh.
And another point, also taken from Meeus, is that the "mean distance of the moon" is a pretty slippery concept. It varies according to which "theory of the moon" is used, the time over which the mean is taken, and whether we calculate the mean distance by treating the orbit as a curve in space or as a varying quantity over time.

Grant Hutchison

Hornblower
2007-Aug-16, 11:31 PM
I appreciate you trying to make me think more about it, but I can assure you this is not a homework question. It is in fact a matter of debunking some dubious theories regarding the moon.

What sort of dubious theories are you trying to debunk, and how does the Moon's mass (which now is very accurately known) figure in them?

Postal76
2007-Aug-17, 12:15 AM
Thanks for the replies.

Believe it or not, some people believe that the moon is much more massive than commonly thought and that we are either being lied to by agencies such as NASA or scientists are somehow mistaken in their calculations. The moon's supposed greater surface gravity lends to other zany ideas like the moon having a breathable atmosphere. Yes, there are people that believe this.

In Moongate: Suppressed Findings of the U.S. Space Program, William Brian claims that the moon's gravity is 64% of earth's gravity. I've seen this idea "debunked" on this very forum, but I don't really understand his explanation or the counter argument.

Anyways, I just wanted to see how easy it was for someone with a limited knowledge of physics and astronomy (like me) to disprove such an idea. It turns out it's a bit harder than I expected. I also find it strange that I couldn't find a method of calculating the moon's mass anywhere on the 'net.

grant hutchison
2007-Aug-17, 11:20 AM
Anyways, I just wanted to see how easy it was for someone with a limited knowledge of physics and astronomy (like me) to disprove such an idea. It turns out it's a bit harder than I expected. I also find it strange that I couldn't find a method of calculating the moon's mass anywhere on the 'net.I think the "hint" I gave in my first post probably addresses your problem: we can detect the moon's gravity here on Earth by looking at the tides. If the moon were 64% more massive, it would raise higher tides, which would be more dominant over solar tides: three times stronger rather than twice as strong. The only way to get the currently observed ratio of solar to lunar tide would be if the sun were also 64% more massive, which would affect the orbits of all the planets. The orbits of the planets are much more nearly Keplerian and the sun is by far the dominant mass in the equation, so there's less wiggle room when trying to determine the mass of the sun from the planets' orbits.

Grant Hutchison

Kiwi
2007-Aug-17, 03:22 PM
In Moongate: Suppressed Findings of the U.S. Space Program, William Brian claims that the moon's gravity is 64% of earth's gravity. I've seen this idea "debunked" on this very forum, but I don't really understand his explanation or the counter argument.

Perhaps I can help -- I don't know all the technical details so can only speak layperson's language.

William L. Brian's book was the first information I saw regarding a moonlanding "hoax". Being a photographer, I immediately looked at the photos before reading anything and, after seeing a few, wondered why Brian had picked so many with lens flare in them. After seeing more, I thought, "Don't say he thinks these are evidence of an atmosphere on the moon?" It turned out that he did.

Regarding his arguments about the neutral point and lunar gravity, I drew myself a rough scale drawing of a spacecraft's trajectory from the Earth to the Moon, plotted positions of the Moon at different stages of the journey, and immediately saw that the craft would never linger near an imaginary straight line between the two bodies. The further the craft was from that line, the further from the Moon the neutral point would be. Brian's major arguments didn't stand up to the simplest scrutiny.

This is the important difference: The conventional figures for the neutral point are based on this imaginary straight line between Earth and the Moon. When a spacecraft crosses its particular neutral point, it will be a long way from that line and travelling along part of an orbit, not in a straight line. Check it out for yourself.

mugaliens
2007-Aug-17, 06:53 PM
Simple. We can very accurately calculate the Earth's mass based on measurements at the "four corners," when the Moon is either 90 degrees ahead of us or behind us in our orbit around the sun.

These observations negate the effect of tides.

Combined with Keplerian motions, and the distance as determined by the laser distance reflectors set in place during Apollo, the mass of the Moon can be obtained to a rather detailed degree.