Calculating Mass Density & Enclosed Mass in Grav Field $\vec{g}(x,y,z)$

AI Thread Summary
The discussion revolves around calculating the mass density and total mass enclosed in a gravitational field defined by the equation \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}). The participants clarify that the mass density can be derived from the divergence of the gravitational field using Gauss's theorem, which involves integrating over a defined volume. A negative mass result arises from the gravitational field's properties, leading to a discussion about the implications of dark matter. The conversation also touches on how the mass enclosed within a growing volume increases without limit, despite the gravitational field's behavior. Ultimately, the calculations confirm the approach was correct, highlighting the importance of understanding the signs in gravitational equations.
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\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}) 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...\int \int \vec{g} \bullet d\vec{a} = 4 \pi G m_{enclosed}

and density p = \frac{m}{V} at least the overall density of it is (non-differential)

sooooo...\frac{1}{4 \pi G}\int\int\int (\nabla \bullet \vec{g}) dV = m_{source/enclosed}

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...

m_{source} = \frac{-2}{3 \pi}G k a^9 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?
 
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Are u sure u set the integratiing limits from -a to +a for all three integrals?

Daniel.
 
yes...i am sure
 
You forgot about the minus in the RHS of Gauss's theorem for the gravitostatic field...

\oint\oint_{\Sigma} \vec{g}\cdot d\vec{S}=\mbox{-}4\pi G m_{\mbox{enclosed by}\ \Sigma}

Daniel.
 
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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?
 
anyone? anyone? buler buler?
 
Why did u conclude it increases?

Daniel.
 
ok if \frac{1}{4 \pi G}\int\int\int (\nabla \bullet \vec{g}) dV = m_{source/enclosed} then...ill just do it and show you

\nabla \bullet \vec{g} = -9kGx^2y^2z^2
\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} 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?
 
Of course the mass increases,but you said about the field.It's not the same thing...:wink:


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