Moon's a Ballon

[paigetheoracle]

What if most, if not all moon's (I know some are not, I'm thinking of some of the larger, more obviously spherical ones) are actually the end product of planets that have burned themselves out? What is this is Mars eventual fate? (no atmosphere and maybe little or no core life). Looking at the moon I wondered, could it be a burned out shell - hence the Apollo experiments seeming to indicate it was hollow? Think of Plasticine. When you set it alight, the internal gases expand and it becomes a cavernous structure. Some scientists (sorry no references - not my field and it may be in the realm of common knowledge anyway) have proposed that the layer beside the core is spinning and that if it stopped, the magnetic shield above the Earth would disappear and so would the atmosphere. This would explain Mars's craters and if the moon had died earlier, perhaps it would explain why it is in such a terrible condition compared with the former? What if when the magnetic core dies, such a planetoid could not maintain its orbit and then ends up getting pulled into the field of another planet close by and that is what happened here? This does not dispute that the moon was part of the Earth at one time or spun off from the sun but that it may have been more alive in its youth as we are now and as other moons in the solar system may also have been: May be obvious. May have been posited before (old hat/standard theory) but I'm new here and this is just a thought in a field I'm no expert in, so I thought if it wasn't it might provoke some discussion and if not it could die the death and be swept under the carpet as passed its sell by date.

[Boerseun]

Highly unlikely, if I get you right.

 

The moon isn't 'hollow', seismic studies performed on the moon have proved that. The fact that it doesn't have a magnetic field of any importance is most likely due to the fact that its a much smaller body than the Earth, and smaller bodies lose heat a lot faster. Therefore, no molten core to produce a magnetic field.

 

The Earth, on the other hand, still retains heat from the formation of the planet, keeping the guts churning, producing a sufficient magnetic field to protect the atmosphere from the solar wind. In time, it too will cool down and solidify, and Earth will probably lose its atmosphere over thousands of years.

 

Mars, as well, seems to have solidified its interior due to being much smaller.

[paigetheoracle]

Highly unlikely, if I get you right.

 

The moon isn't 'hollow', seismic studies performed on the moon have proved that.

 

Well I meant cavernous - should choose my words more precisely!

 

As for the rest it was just amatuer conjecture (I stray into odd fields like a cow who can lift latches, just to see what is over the hill, if it isn't me).

 

Good hunting!:evil:

[Boerseun]

Fair enough.

 

I should think that if the moon was indeed hollow (I fail to see the distinction between hollow and cavernous...:evil:), the first thing we'd see would be a change in tides on Earth. Then, of course, we should be very thankful for the vacuum existing between us and the moon, because every time an asteroid or any other space debris whacks into the moon, the sucker'll ring like a bloody big bell! For years! Imagine that! And here are we, down on Earth, trying to sleep!

[Jay-qu]

sorry boerseun, but the sound waves wouldnt propagate to earth for us to hear..

 

I can assure you that they moon would not have any large caverns on the scale that I think you are proposing, for one it would mean the mass of the moon (which cant change) would have to be more dense in some parts - which I think we would have noticed. Also you would also notice that when an asteroid strikes it, it may break through into one of this carverns.. which is not observed.

[Boerseun]

sorry boerseun, but the sound waves wouldnt propagate to earth for us to hear..

'Course not! Because...

...we should be very thankful for the vacuum existing between us and the moon...

No moon-peals for you tonight.

[Jay-qu]

:hyper: I shouldnt make a habit of posting while I'm still reading..

[UncleAl]

hence the Apollo experiments seeming to indicate it was hollow?

Before you make a statement like this you had better have literature citations to back it up.

 

Lunar laser ranging has defined the moon's orbit to within a centimeter or so. Orbits of lunar satellites are well defined. Either one plus Green's function demands that the moon is a solid mass, neither hollow nor majorly Swiss-cheesed.

 

Google

moon density 7.6 million hits

 

Read them and get back to us.

[Qfwfq]

Green's function

Which one? Are you talking about the prei-mage of something? Are you sure deviations from inverse square law would be sufficient to rule out the radial distribution being mostly near the surface by orbital data alone?

[CraigD]

Are you sure deviations from inverse square law would be sufficient to rule out the radial distribution being mostly near the surface by orbital data alone?

It’s possible to demonstrate the principle of how one can determine possible internal distributions of mass in a central body from the observed orbital speeds of its satellites.

 

Consider the acceleration A of a test body due to n “point masses” (bodies with mass, but 0 diameter) arranged in a straight line:

[math]A = G(\frac{m_1}{r_1^2} + \frac{m_2}{r_2^2} + . . . \frac{m_n}{r_n^2})[/math]

For my demonstration, arbitrary units are assumed so that [math]G = 1[/math], [math]r_n = n[/math], and [math]m_n[/math] is given by an “mass distribution” array written as a delimited string. A line of demonstration output consists of this array followed by “=” and the calculated test body acceleration, followed by a comment (whitespaced so it aligns prettily).

                    1 1 1 1 1 1 1 1 1 1 2 2 3 4 5 6
 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
, , , , ,3.5,,, , , , , , , ,3.5                     =.1555555555555555556  ;thin, hollow shell
, , , , ,1,1,1,1,1,1,1,1,1,1,1                       =.1568291723338759635  ;uniform density
, , , , , , , , , ,15.6,,,, ,                        =.156                  ;point mass
, , , , , , , , , , , , , , ,3.5,,, , , , , , , ,3.5 =.02115555555555555555
, , , , , , , , , , , , , , ,1,1,1,1,1,1,1,1,1,1,1   =.02972756459045822703
, , , , , , , , , , , , , , , , , , , ,15.6,,,, ,    =.039                 

Note that, when the distance between the center of mass of the central body and the test body is increased from 10 to 20, only when the central body is a point mass does acceleration decrease in exact agreement with the inverse square (1/r^2), the thin shell deviating more than the uniform density mass.

 

Once acceleration is known, orbital speed can easily be calculated ([math]V = \sqrt{A*r}[/math])

 

The actual calculation of gravitational acceleration for spherical bodies with various mass distributions is more complicated than the above, because it must be done in 3, rather than 1 dimension, but this simple demonstration serves to illustrate how the mass distribution of a central body effect the speed of orbits at various distances from its center of mass.

 

PS: the M code for this demonstration:

F  R A," " Q:'$L(A)  S A=$P(A,"="),T=0 F D=2:1 S T=$P(A,",",D)/(D-1)/(D-1)+T I D'<$L(A,",") W "=",T,! Q

[Qfwfq]

You are using exact inverse square Craig, how does it work out for a spherically symmetric mass density distribution? That's what I was talking about.

[CraigD]

You are using exact inverse square Craig…

Correct. I’m modeling a body as a 1-dimensional, constant line length lattice of point masses, and applying the inverse square

… how does it work out for a spherically symmetric mass density distribution? That's what I was talking about.

You can roughly approximate a sphere by assigning each of the point masses a mass proportional to the area of equal spaced cross sections of it. The resulting acceleration, using distance 20 as above, is closer to a point source than the “1,1,1,1,1,1,1,1,1,1,1 ‘uniform density’” body above: 0.03279… vs. 0.02973… vs. the point source’s 0.039

                                     1                                       2
1   2   3   4   5   6   7   8   9   0   1   2   3   4   5   6   7   8   9   0   1   2   3   4   5
F  R A," " Q:'$L(A)  S A=$P(A,"="),T=0 F D=2:1 S T=$P(A,",",D)/(D-1)/(D-1)+T I D'<$L(A,",") W "=",T,! Q
,   ,   ,   ,   ,0.3,0.8,1.2,1.5,1.6,1.7,1.6,1.5,1.2,0.8,0.3                                         =.1550579694252479801
,   ,   ,   ,   ,   ,   ,   ,   ,   ,   ,   ,   ,   ,   ,0.3,0.8,1.2,1.5,1.6,1.7,1.6,1.5,1.2,0.8,0.3 =.0327869564202304354 

Note that this technique is very approximate – the array for the uniform density sphere is correct for particle many diameters distant, but should be slightly “flattened” when the test body is close to the central body’surface.

[Qfwfq]

My point was that, for a spherical distribution (uniform density for a given r) and inverse square law, all the mass at r can be considered at the centre, for points at greater distance, and has no effect for points internal to the sphere.

 

This theorem dates to Newton's Principia and makes things a lot simpler. It shows that if the force goes exactly by inverse square, you couldn't tell the radial distribution only by orbits, you could only distinguish angular non-uniformity, and especially that at the surface. The moon certainly has mounts and vales, and with some great differences in altitude.