Confusion on product rule for mass of differential volume element

[TRB8985]

Homework Statement

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as earth, as having a density that decreases linearly with distance from the center. Let the density be 15.0E3 kg/m³ at the center and 2.0E3 kg/m³ at the surface. What is the acceleration due to gravity at the surface of this planet?

Relevant Equations

ρ = M/V ; g = GM/R²

Good evening,

I'm running into some trouble with this problem, and I have a hint as to why, but I'm not completely sure. Please see the steps below for context.
I've been able to set up the proper equation representing the density as a function of distance from the center which looks like this:
$$ \rho(r) = \rho_{center} - \frac {\rho_{center} - \rho_{surface}} {R_{earth}} \cdot r $$In order to calculate the acceleration due to gravity at the surface, I know that the mass of this planet is required. My plan was to start from the density equation (with the variables being functions of radial distance from the center): $$ \rho(r) = \frac {m(r)}{V(r)} $$Then multiply both sides by the volume as a function of r: $$ \rho(r) \cdot V(r) = m(r) $$Followed by taking a derivative of both sides with respect to r: $$ \frac {d}{dr} (\rho(r) \cdot V(r)) = \frac {d}{dr}(m(r)) $$This leads to:
$$ \frac {d\rho(r)}{dr} \cdot V(r) + \rho(r) \cdot \frac {dV(r)}{dr} = \frac {dm(r)}{dr}$$Normally in the case of constant density, we would trash the first term on the LHS and move forward with the rest. However, since I have the function for the density in terms of r, I figured keeping this term would be necessary.

My next line looked like this: $$ \frac {d}{dr}(\rho_{center} - \frac {\rho_{center} - \rho_{surface}} {R_{earth}} \cdot r) \cdot (\frac {4}{3}\pi r^3) + (\rho_{center} - \frac {\rho_{center} - \rho_{surface}} {R_{earth}} \cdot r) \cdot \frac {d}{dr} (\frac {4}{3} \pi r^3) = \frac {dm(r)}{dr} $$ Which becomes: $$ (- \frac {\rho_{center} - \rho_{surface}} {R_{earth}}) \cdot (\frac {4}{3}\pi r^3) + (\rho_{center} - \frac {\rho_{center} - \rho_{surface}} {R_{earth}} \cdot r) \cdot ( 4\pi r^2) = \frac {dm(r)}{dr} $$ At this point, I multiplied both sides by dr, then integrated r from 0 to R_earth and the mass from 0 to the total mass of the planet, M: $$ \int_{r=0}^{r=R_{earth}} [\frac {4}{3}\pi r^3(\frac {\rho_{surface} - \rho_{center}} {R_{earth}}) + 4\pi r^2(\rho_{center} - \frac {\rho_{center} - \rho_{surface}} {R_{earth}} \cdot r)]dr = \int_{m=0}^{m=M} dm(r) $$After integration and a long series of algebra, I ended up with this: $$ M = \frac {4}{3} \pi \rho_{surface} R_{earth}^3 \approx 2.18 \cdot 10^{24} kg $$ Unfortunately, this result is a factor of roughly 2.6 times smaller than the correct value: $$ M_{correct} \approx 5.71 \cdot 10^{24} kg $$ If I go back to the beginning and use $\rho(r)dV(r) = dm(r)$ instead, everything works out beautifully with no trouble. My question is... why?

In problems similar to this with constant density, it was abundantly clear that $d\rho V$ was zero because the derivative of a constant density is zero. But here, the derivative of $\rho$ is non-zero. So... aren't we obligated to include it?

 

[erobz]

Homework Statement: Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as earth, as having a density that decreases linearly with distance from the center. Let the density be 15.0E3 kg/m³ at the center and 2.0E3 kg/m³ at the surface. What is the acceleration due to gravity at the surface of this planet?
Relevant Equations: ρ = M/V ; g = GM/R²

Good evening,

I'm running into some trouble with this problem, and I have a hint as to why, but I'm not completely sure. Please see the steps below for context.
I've been able to set up the proper equation representing the density as a function of distance from the center which looks like this:
$$ \rho(r) = \rho_{center} - \frac {\rho_{center} - \rho_{surface}} {R_{earth}} \cdot r $$In order to calculate the acceleration due to gravity at the surface, I know that the mass of this planet is required. My plan was to start from the density equation (with the variables being functions of radial distance from the center): $$ \rho(r) = \frac {m(r)}{V(r)} $$Then multiply both sides by the volume as a function of r: $$ \rho(r) \cdot V(r) = m(r) $$Followed by taking a derivative of both sides with respect to r: $$ \frac {d}{dr} (\rho(r) \cdot V(r)) = \frac {d}{dr}(m(r)) $$This leads to:
$$ \frac {d\rho(r)}{dr} \cdot V(r) + \rho(r) \cdot \frac {dV(r)}{dr} = \frac {dm(r)}{dr}$$Normally in the case of constant density, we would trash the first term on the LHS and move forward with the rest. However, since I have the function for the density in terms of r, I figured keeping this term would be necessary.

My next line looked like this: $$ \frac {d}{dr}(\rho_{center} - \frac {\rho_{center} - \rho_{surface}} {R_{earth}} \cdot r) \cdot (\frac {4}{3}\pi r^3) + (\rho_{center} - \frac {\rho_{center} - \rho_{surface}} {R_{earth}} \cdot r) \cdot \frac {d}{dr} (\frac {4}{3} \pi r^3) = \frac {dm(r)}{dr} $$ Which becomes: $$ (- \frac {\rho_{center} - \rho_{surface}} {R_{earth}}) \cdot (\frac {4}{3}\pi r^3) + (\rho_{center} - \frac {\rho_{center} - \rho_{surface}} {R_{earth}} \cdot r) \cdot ( 4\pi r^2) = \frac {dm(r)}{dr} $$ At this point, I multiplied both sides by dr, then integrated r from 0 to R_earth and the mass from 0 to the total mass of the planet, M: $$ \int_{r=0}^{r=R_{earth}} [\frac {4}{3}\pi r^3(\frac {\rho_{surface} - \rho_{center}} {R_{earth}}) + 4\pi r^2(\rho_{center} - \frac {\rho_{center} - \rho_{surface}} {R_{earth}} \cdot r)]dr = \int_{m=0}^{m=M} dm(r) $$After integration and a long series of algebra, I ended up with this: $$ M = \frac {4}{3} \pi \rho_{surface} R_{earth}^3 \approx 2.18 \cdot 10^{24} kg $$ Unfortunately, this result is a factor of roughly 2.6 times smaller than the correct value: $$ M_{correct} \approx 5.71 \cdot 10^{24} kg $$ If I go back to the beginning and use $\rho(r)dV(r) = dm(r)$ instead, everything works out beautifully with no trouble. My question is... why?

In problems similar to this with constant density, it was abundantly clear that $d\rho V$ was zero because the derivative of a constant density is zero. But here, the derivative of $\rho$ is non-zero. So... aren't we obligated to include it?

The mass doesn't equal $\rho V$, it equals the $\int \rho ~dV$. When density is constant $\rho$ comes outside of the integral to yield the formula $\rho V$. So basically, you started with a false relationship for the mass, and it's still false after differentiation.

 

[Hill]

In $$\rho(r) = \frac {m(r)}{V(r)}$$ on LHS you have a density at radius $r$ while on the RHS you have an average density of a ball of radius $r$.

 

[TRB8985]

Ahh, now I see what's meant by starting with a false relationship! I'd totally agree, that's not right at all.

So really, it sounds like I should be doing this instead: $$ \rho(r) = \frac {dm(r)}{dV(r)} $$ .. and not even worry about a $d \rho \cdot V$ appearing at all.

 

[PeroK]

I suggest you look for the integral you need to evaluate: using the law of gravitation to give the surface gravity for a spherically symmetric object.

Hint: Newton's shell theorem.