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

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Homework Help Overview

The discussion revolves around calculating mass density and the total mass enclosed in a gravitational field described by a vector function. The original poster presents a specific gravitational field and seeks to determine the mass density of its source and the total mass within a defined volume.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the application of Gauss's law in gravitation, questioning the integration limits and the implications of a negative mass result. There is discussion on how the gravitational field behaves with distance from the source and the relationship between mass and gravitational flux.

Discussion Status

Some participants have provided clarifications regarding the application of Gauss's theorem, particularly concerning the sign in the equation. The conversation reflects a mix of confirmations and further inquiries about the behavior of the gravitational field and the mass density calculations.

Contextual Notes

There is a focus on the assumptions made regarding the integration limits and the interpretation of the gravitational field's behavior as the volume increases. The original poster expresses confusion about the negative mass result and its implications.

Phymath
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[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 (non-differential)

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?
 
Last edited:
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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...

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

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 [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?
 
Of course the mass increases,but you said about the field.It's not the same thing...:wink:


Daniel.
 
  • #10
alright cool thanks man all that for yes u did it right except a negative sign lol
 

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