Integrating density function that is the inverse square law? [closed]

Question

If I want to calculate the mass of a sphere, and I know the density function is the inverse square law, then it is just a matter of using a volume integral to calculate the mass, that's no problem.

$M=\int_{V}^{}\rho(r)dV = \int_{}^{}\frac{k}{4\pi r^2}(4\pi r^2)dr = kR$

My question is, assuming my math is correct, since we are left with M = kR (k is a constant), what is the implication of that? Does a sphere of 1m radius have 1kg of mass for example? That doesn't seem right to be but maybe it is?

Answer 1

Given $R$ (in some units), the numerical value of $M$ (in some units) will depend on the value of $k$ (in consistent units).

For example, if $R=1\ {\rm m}$, and $k=1\ {\rm kg/m}$, then $M=kR = 1\ {\rm kg}$. However, if $k=2\ {\rm kg/m}$, then $M = kR = 2\ {\rm kg}$, even though $R$ is still $1\ {\rm m}$. Physically, $k$ is a measure of the density of the sphere; a denser sphere will have more mass for the same radius.

Possibly more interesting than the numerical value for fixed values of the parameters is the scaling of the mass with radius. Scaling tells us how the mass changes if you change the size of the sphere, keeping the same density profile. Because your relationship between $M$ and $R$ is linear (in other words, in the equation $M=kR^{\color{red} 1}$, the exponent of $R$ is ${\color{red} 1}$), then if you double $R$, $M$ will also double (assuming you add mass to the sphere following the same inverse square profile you had before). If you half $R$, then $M$ will be divided by two.