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iamnotaparakeet
Centurio Caesareae
Joined: Aug 01, 2007
Age: 22
Posts: 11615
Location: Domus Psittacorum
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 Posted: Wed Jun 04, 2008 8:54 am Post subject: BBCode and First Order Equations Involving Gravity |
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This is from my forum which allows advanced BBCode options. They wont show here, but if you want to see how it looks: http://conservativechristiansonly.freeforums.org/first-order-equations-involving-gravity-t49.html
| NonPsittacusSum wrote: | Let's work with algebra and circles to come up with approximate values, shall we?
Lets set some axioms:
1. Orbiting objects have mass and velocity.
2. Gravity holds these objects in their orbits.
With that said, the force of gravity is what keeps these objects in their, for this discussion, circles (i.e. gravity is the centripetal force.)
[align=center]F[sub]Gravity[/sub] = F[sub]Centripetal[/sub][/align]
The force of gravity is,
[align=center](G*m[sub]1[/sub]*m[sub]2[/sub])/r[sup]2[/sup][/align]
and centripetal force is,
[align=center](m[sub]2[/sub]*v[sup]2[/sup])/r[/align]
Together that is,
[align=center](G*m[sub]1[/sub]*m[sub]2[/sub])/r[sup]2[/sup] = (m[sub]2[/sub]*v[sup]2[/sup])/r[/align]
[align=center](G*m[sub]1[/sub]*[s]m[sub]2[/sub][/s])/r[sup][s]2[/s][/sup] = ([s]m[sub]2[/sub][/s]*v[sup]2[/sup])/[s]r[/s][/align]
[align=center](G*m[sub]1[/sub])/r = v[sup]2[/sup][/align]
Let's call m[sub]1[/sub] = m[sub]Sun[/sub] and the velocity that of Earth.
[align=center](G*m[sub]Sun[/sub])/r = v[sub]Earth[/sub][sup]2[/sup][/align]
For velocity of a planet with low eccentricity:
[align=center]v[sub]Earth[/sub][sup]2[/sup] = (G*m[sub]Sun[/sub])/r[/align]
For the orbit radius of such an object:
[align=center]r = (G*m[sub]Sun[/sub])/v[sub]Earth[/sub][sup]2[/sup][/align]
For the mass of the star that Earth orbits:
[align=center]m[sub]Sun[/sub] = (r*v[sub]Earth[/sub][sup]2[/sup])/G[/align]
For the Gravitational constant (which has been determined already by more feasible means):
[align=center]G = (r*v[sub]Earth[/sub][sup]2[/sup])/m[sub]Sun[/sub][/align]
Now, who would like to try these out?
G = 6.67*10[sup]-11[/sup] (N*m[sup]2[/sup])/kg[sup]2[/sup]
m[sub]Sun[/sub]= 1.99*10[sup]30[/sup] kg
r[sub]Earth[/sub] = 1.49*10[sup]11[/sup] m
r[sub]moon[/sub] = 1.74*10[sup]6[/sup] meters
m[sub]moon[/sub] = 7.35*10[sup]22[/sup] kg
Mass of Earth is derivable from the above, so I'll let that be figured out. |
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lau
Quinquaginta Novem! Male Gee-knee-us + silly bits.
Joined: Jun 18, 2006
Age: 59
Posts: 7042
Location: Somerset UK
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| Posted: Wed Jun 04, 2008 12:09 pm Post subject: |
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As you have over-specified the problem, why don't you give us the two values you "be figured out" for the mass of the Earth? Naturally, they will differ, and reflect (at least) the inaccuracies of treating a three body problem as if it were a two body one.
Quite a pretty free exclusive website.
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