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Grav field fun! 
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#1
Apr1305, 05:40 PM

P: 189

[tex] \vec{g}(x,y,z) = kG((x^3 y^2 z^2)\hat{e_x} + (x^2 y^3 z^2)\hat{e_y} + (x^2 y^2 z^3) \hat{e_z})[/tex] given this grav field (k is constant)
find the mass density of the source of this field, and what is the total mass in a cube of side 2a centered about the origin? hmmm well we all know...[tex]\int \int \vec{g} \bullet d\vec{a} = 4 \pi G m_{enclosed}[/tex] and density [tex] p = \frac{m}{V} [/tex] at least the overall density of it is (nondifferential) sooooo...[tex]\frac{1}{4 \pi G}\int\int\int (\nabla \bullet \vec{g}) dV = m_{source/enclosed}[/tex] now the limits i made a cube of side 2a, because the flux through a box is easier when g is given in cart coords....any way i get... [tex]m_{source} = \frac{2}{3 \pi}G k a^9[/tex] how do I get a neg mass (unless this is dark matter which it very well could be) and I'm thinking i missed something about density cause why would it ask that first and then the mass enclosed second?...a lil help? 


#2
Apr1305, 06:24 PM

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Are u sure u set the integratiing limits from a to +a for all three integrals?
Daniel. 


#3
Apr1305, 07:07 PM

P: 189

yes...i am sure



#4
Apr1305, 07:37 PM

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Grav field fun!
You forgot about the minus in the RHS of Gauss's theorem for the gravitostatic field...
[tex] \oint\oint_{\Sigma} \vec{g}\cdot d\vec{S}=\mbox{}4\pi G m_{\mbox{enclosed by}\ \Sigma} [/tex] Daniel. 


#5
Apr1305, 08:55 PM

P: 189

i see so that just answers the negative mass problem, now how is it that this grav field increases as you move farther away from the "source" this tells me that the "mass" enclosed is ever increasing i suppose so i guess having the mass enclosed as a function of the flux box side is alright what do you think?



#6
Apr1405, 12:25 AM

P: 189

anyone? anyone? buler buler?



#7
Apr1405, 12:28 AM

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Why did u conclude it increases?
Daniel. 


#8
Apr1405, 09:26 AM

P: 189

ok if [tex]\frac{1}{4 \pi G}\int\int\int (\nabla \bullet \vec{g}) dV = m_{source/enclosed}[/tex] then...ill just do it and show you
[tex]\nabla \bullet \vec{g} = 9kGx^2y^2z^2[/tex] [tex]\frac{9k}{4 \pi }\int^a_{a}\int^a_{a}\int^a_{a} (x^2y^2z^2) dxdydz = \frac{2}{3}k \pi a^9 = m_{source/enclosed}[/tex] so as my "flux cube" getts larger obviously it isn't limiting at some value as the function is ever increasing so, the mass should always be increasing yes? 


#9
Apr1405, 10:34 AM

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Of course the mass increases,but you said about the field.It's not the same thing...
Daniel. 


#10
Apr1405, 11:33 AM

P: 189

alright cool thanks man all that for yes u did it right except a negative sign lol



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