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Orbit Problem
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Lets say that we were to go to the moon, take some mass and bring it back to earth (i.e we transfered some of the mass from the moon to the Earth). The question is what would happend to the velocity of the moon and the radius between Earth and the moon? So far I have used the equation Fg = GMm/r^2 and rewrote it as Fg = G(M+x)(m-x)/r^2, x being the mass transfered. But I am unsure how the radius between the two and the velocity of the moon would be effected. Any help would be greatly appriciated |
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Re: Orbit Problem
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by
on 2008-10-23 17:45:31 |
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Interesting question. I think an additional piece of information is needed. You see, scooping up some moon rock and rocketing them back to Earth will push on the moon. The reason being, the rocket acts as a sort of propellant. Now, the angle of lift off could be a number of different orientation (headed strait for Earth, or pointed a little away, etc.) Let's ignore this effect, and simply imagine that the material was magically relocated to the surface of Earth. At first, the radius would be no different than it is now. However, since the masses have changed, the orbit will change, allowing the moon to spiral away from the Earth faster than it already is. (The moon is in fact falling away from the Earth, by the way, due to tidal forces.) http://en.wikipedia.org/wiki/Orbit_of_the_Moon 
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Re: Orbit Problem
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by nightsreach
on 2008-10-25 03:55:38
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Wouldn't the Earth's increased mass produce the opposite effect? Since gravitational force depends on the mass.  |
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Re: Orbit Problem
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by
on 2008-10-26 02:43:58 (edited 2008-10-26 02:45:30)
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The force on one body is the same as the force on the other body, in the opposite direction of course. This force is maximum when the two masses are identical, at a given radius. F=Gm1m2/r2 since G and r are constant: F~m1m2 Since mass is conserved, the maximum of the product of the two masses is maximized when they are equal. Since the Earth is already larger, adding to it's mass will decrease the force of attraction. You can see this is true with the example of 4 beans. Two beans in one pile, two in the other. Their product is 2*2=4. Move one bean from pile A to pile B, and the product is 1*3=3. Move the final bean and the product is 0*4=0. There is probably a great mathematical name for this effect. If not, I will call it the product of two bean piles.
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Re: Orbit Problem
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by SuicidopoliS
on 2008-10-27 04:37:38 (edited 2008-10-27 04:38:57)
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Gendou wrote, just 2 lines above this one: There is probably a great mathematical name for this effect. Isn't this what's referred to as "extremizing a product/function"? Although "extremize" doesn't appear to be a valid English verb, i'm pretty sure every mathematician and physicist on this planet understands what's being meant. So maybe it's time it gets added to the dictionary :) In this case, the function to be extremized is f(x,y)=x*y, with the bound condition that g(x,y)=x+y-C=0, with C a constant, C=4 in the case of the beans. Using the "Implicit Function Theorem", a.k.a. the "Big Bad Mean ...", we do indeed find that f is extremized for x=y=C/2. Wiiiiiiii...! Although "The Product Of Two Bean Piles" sounds pretty neat too... .oO° Life's THE CURE, the rest are details! °Oo. |
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Re: Orbit Problem
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by
on 2008-10-27 18:15:23 |
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Wow, thanks SuicidopoliS for the awesome maths. I like the sound of "big bad mean".
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