Comments:Scientists say the moon is slowly shrinking
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|Comments from feedback form - "wat about the earth"||2||02:32, 5 February 2012|
|Is there any water involved?||3||15:46, 22 March 2011|
|Comments from feedback form - "So it may be shrinking. Its ma..."||12||15:42, 22 March 2011|
|Comments from feedback form - "i cant belive it!! :["||1||15:40, 22 March 2011|
|Just a pointer||2||10:45, 23 August 2010|
Does anybody know whether or not there is water involved in this shrinking process? If this SHOULD be the case, shrinking sure hasn't to do with the expanding phase change liquid-solid. However, it might be the phase change gaseous-liquid.
Not that I know of, the sources did not indicate that this may be the case. The sources indicated that the contraction could be caused by the cooling of the inner core of the Moon.
So it may be shrinking. Its mass still remains the same. What effect does this have on the planet earth?
Actually, I would think that the moon's mass is slowly increasing due to precipitation of interstellar dust and metoeroids, etc. With shrinkage and increased mass the net result should be a very gradual increase in density, hence a gradual increase in gravitation. One would expect that the earth and the moon are moving closer together, but the accepted view is that they are moving apart (or that the moon is moving gradually away from the earth).
Why would an increase in density lead to an increase in gravitation? Gravitation depends on mass, not density, though I agree with you that mass is probably increasing very slowly.
Indeed, an increase in density would not lead to an increase in gravitational force felt at any given distance from the moon, as long as the distance was measured from the centre of the moon, and not the surface. Gravity is dependant on mass, not on density. However, *surface* gravity on the moon would increase ever so slightly as a result of the moon's slow contraction over time. How much gravity an object possesses is dependant only on its mass, but the strength of the force of gravity tugging on you is dependant on both the mass of the object you're near and your distance from it.
Because the moon is more dense now than in the past, while standing on the surface you'd be closer to the centre-of-mass of the moon, and therefore experience more gravity. (Note that if you dig into the moon (or any other object) you don't experience increased gravity because the material on top of you is now pulling you in the opposite direction.)
The same effect is what makes black holes interesting. If, for instance, you crushed all the mass in the sun into a black hole, the orbit of the earth would be unaffected. The mass of the sun (now a black hole) would be the same, so the gravity output would be identical. But stars are huge and diffuse, so you can't get very close to them before you start burrowing into their "surface", at which point only the portion of the star's mass below you would be tugging you in the direction of the core. The stuff on top of you would be pulling you upward. But with a very dense object like a white dwarf, neutron star, or black hole, their high density allows you to get very close to the centre-of-mass without anything pulling you in the opposite direction. Because you can get so much closer to the center, you experience a greater force than would be possible if the object were less dense.
Referring to your quote "that if you dig into the moon (or any other object) you don't experience an increased gravity because the material on top of you is now pulling you in the opposite direction". Given that scenario, the mass at the center of the earth has all of the mass of the earth outside of it and should therefore be pulled "in the opposite direction". This implies that the earth should be hollow. Isn't this the kind of thinking that led to the "hollow earth" hypothesis. In fact, you can still become a member of the Hollow Earth Society. However, it is currently understood that the density of a massive object is highest at the center. Gravity is a function of both mass and density acting from the center of the mass "as a whole".
No, it doesn't imply that the Earth is hollow. What it implies is that if you dug a tunnel straight through the entire Earth (and used... I dunno... a forcefield to keep it open), at the very centre of the Earth you would be weightless, because all of the Earth's mass would be tugging on you equally in every direction.
It's a well established fact that gravity increases with density. In other words, if the earth was smaller in diameter, given the same mass (i.e., higher density) its gravitational field would be higher (assuming the field is measured at a fixed distance from its center). If you take this to the extreme, by compressing the earth to 1 mm in diameter, it's gravitational field would be phenomenally high. The earth's mass may even be a black hole at that size.
Mass always has density, even if it's just a bunch of hydrogen atoms loosely dispersed in space. When those hydrogen atoms condense into a star, that mass is now at a much higher density and consequently produces a much stronger gravitational field. If the star eventually collapses into a black hole, you wouldn't want to be anywhere in that vicinity of space. Time to dust off the physics and astronomy textbooks. The strength of a gravitation field is directly proportional to the increase in density of a given mass as measured from a fixed point to the center of that mass (and, outside that mass - not measured from somewhere inside the mass).
Please recommend a textbook reference for the proposition that field strength of a fixed mass at a fixed distance from the center increases with density.
As a counterexample, I have a copy of Stephen Eales' Planets and Planetary Systems to hand. Eales quotes the high school equation for Newton's law of universal gravitation without qualification, which I read as force per unit mass (outside the object) is proportional to mass divided by the square of the distance from the centre of mass. (ISBN 9780470016930, Wiley-Blackwell 2009, p.106)
InfantGorilla, thank you for your reference. I will seek out a counter reference and get back to you asap. We must keep in mind that Newtonian equations break down when it comes to explaining extreme phenomena, such as black holes. Within an Einsteinian curved, spacetime framework, gravity is no longer a force and can act instantaneously at a distance. This discovery has had severe consequences for modern cosmology. I'll get back to you, even if it is to prove myself wrong.
You're thinking of this backwards.
Earth's equatorial radius is currently 6,378 kilometres. If you crushed the Earth to the size of a pea, it would be a black hole. But someone orbiting from a distance of 6678 kilometres from the centre of mass (300 kilometres above the current surface, or Near Earth Orbit) would experience the same force of gravity in either circumstance. Regardless of whether you are 6678 kilometres from the centre of the Earth or 6678 kilometres from the centre of an Earth massed black hole, the gravity you experience will be identical. At a given distance from an object the force of gravity will be the same, regardless of the density of the object, provided that all the mass of the object is below you.
The reason that a black hole "has a stronger gravitational field" (even though it doesn't really) is because you can get much closer to the centre of mass without crashing into the surface. That's it.
Feet is not a volume unit. You should use either cubic feet, or better yet, cubic meters, as it is the standard in most places.
or if the figure is a diameter, in which case feet or metres is fine, then say so.