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kleindoofy
2007-Feb-12, 10:22 PM
Ok, this is a purely hypothetical question. The real question is at the end of the post.

My neighbor's little eight year old son asked me how much the moon weighs.

It was immediately clear to me that it would be useless to try and explain to him that the moon doesn't really weigh anything or that weight in itself is relative to mass and acceleration, yada, yada etc.

So, I tried to reduce the problem to something relative to our daily environment of experience here on Earth: how does one weigh the moon?

I thought, if one (theoretically) reduced the moon to rubble, in kernels roughly the size of beach sand, and spread it evenly over the surface of the earth in a layer 1mm thick (please, supposing the earth were evenly smooth surfaced - no oceans, like a billiard ball), one could weigh e.g. one square meter of the matter and multiply it by the total surface covered by the matter.

This brought me to my question:

How much of the earth's surface would the moon's matter cover? Or better: how much of the surface of a smooth ball with the same size as the earth would the moon's matter cover in the example explained above? Would a layer 1mm thick only cover part of it? Or would it cover the whole surface in even a thicker layer? If so, how thick?

And finally, using earthly standards and viewing the whole thing from a child's perspective: what does the moon weigh (using the method proposed in my example)?

Disclaimer: I used the search function to look for a similar question here on the board before asking, but failed.

aurora
2007-Feb-12, 10:28 PM
The way this usually gets answered, is to explain what the child would weigh on the moon, rather than what the moon would weigh on earth.

That said, the mass of the moon and earth can be found on this website:
http://nssdc.gsfc.nasa.gov/planetary/factsheet/index.html

Bjoern
2007-Feb-12, 10:35 PM
How much of the earth's surface would the moon's matter cover? Or better: how much of the surface of a smooth ball with the same size as the earth would the moon's matter cover in the example explained above? Would a layer 1mm thick only cover part of it? Or would it cover the whole surface in even a thicker layer? If so, how thick?


Nice question. ;)

The moon has a volume of V_m = 4 pi/3 r_m^3, where r_m is the radius of the moon.

A thick layer on the surface of the earth of thickness d would have a volume of V = 4 pi/3 ((r_e + d)^3 - r_e^3), where r_e is the radius of the earth.

Equate the two; that yields:
r_m^3 = (r_e + d)^3 - r_e^3 = 3 r_e^2 d + 3 r_e d^2 + d^3
Insert the known values for r_m and r_e and solve for d (this is a cubic equation for d, so that's not too easy - but I suspect that d is much smaller than r_e, and hence you can neglect d^3, leaving a quadratic equation).

I'm too lazy to do the calculation for myself right now... ;-) Have fun!

01101001
2007-Feb-12, 10:51 PM
The Moon's volume is about 1/50 that of Earth's. If we ignore compression and density change, and just slather the Moon's mass upon the Earth, it would add...

R radius of Earth in meters, 6378.1 kilometers
r unknown depth of Moon layer upon Earth in meters
V original Earth volume in cubic meters, 1083206246123080894852 m^3
v Moon volume, (1/50)V

Total volume of moon-covered Earth = Earth volume plus Moon volume

(4/3)pi(R + r)^3 = V + (1/50)V
(R + r)^3 = (3/4)(1 + 1/50)V/pi
R + r = cuberoot((3/4)(1 + 1/50)V/pi)
r = cuberoot((3/4)(1 + 1/50)V/pi) - R

I get, after asking Mr Google, (((3/4)(1 + 1/50)(1083206246123080894852 m^3)/pi)^(1/3) - (6378.1 km)

r = just over 35 kilometers, depth of uncompressed Moon matter covering uncompressed Earth

trinitree88
2007-Feb-12, 11:14 PM
101 nice. ....and for what it's worth, the moon itself is covered with a layer of compacted dust meters deep in most spots known as regolith. It's an accumulation of primarily cratering impact debris mixed with traces of the solar wind, comet dust, supernova dust, NASA rocKet fuel exhaust....conservation of mass is such a nice thing.:dance: pete

Bob
2007-Feb-12, 11:49 PM
I get a slightly different answer using a slightly different approach and numbers picked off Google:

t = thickness of a uniform layer of moon spread over the surface of the earth

t x (surface area of earth) = (volume of moon)

t = 2.2 E10 km**3) / 5.1 E8 km**2 = 43 km

01101001
2007-Feb-13, 12:26 AM
I get a slightly different answer using a slightly different approach and numbers picked off Google:
[...]
t = 2.2 E10 km**3) / 5.1 E8 km**2 = 43 km

Same ballpark. Good enough. Let's average them and force the kid to believe it!

kleindoofy
2007-Feb-14, 10:30 PM
A layer 35-43 km thick? Wow.

Well, I guess it sounds like a lot, but considering the diameter of the earth and when viewed from a few thousand km away, it would look more or less like a thick layer of wrapping paper around the globe.

Whatever, thank you very much for your answers. This is really going to impress the kid.

[note to self: when dealing with astronomical phenomena, think big!]