- #1

TRB8985

- 74

- 15

- 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?

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?