Let

A = ρ_{c} + ρ_{m}, b = \frac{ρ _{m}}{R _{c}}, d = \frac{ρ _{m} R _{c}^{3}}{12} .

m (r) = ⎩ ⎨ ⎧ \frac{4 π}{3} ρ_{c} r^{3}, 4 π (\frac{A r ^{3}}{3} - \frac{b r ^{4}}{4} - d), 0 \leq r \leq R_{c}, R_{c} \leq r \leq R_{p} .

This is continuous at the core boundary. Hydrostatic equilibrium gives

d P / d r = - G m (r) ρ (r) / r^{2}

, with

P (R_{p}) = 0

F (r) = 4 π G (\frac{A ^{2} r ^{2}}{6} - \frac{7 A b r ^{3}}{36} + \frac{b ^{2} r ^{4}}{16} + b d lo g (r / R_{c}) + \frac{A d}{r}) .

Since

F^{'} (r) = G m (r) (A - b r) / r^{2}

, the mantle pressure is

P (r) = F (R_{p}) - F (r), R_{c} \leq r \leq R_{p} .

Inside the uniform core, the shell theorem gives

m (r) = 4 π ρ_{c} r^{3} /3

, so

P (r) = F (R_{p}) - F (R_{c}) + \frac{2 π G ρ _{c}^{2}}{3} (R_{c}^{2} - r^{2}), 0 \leq r \leq R_{c} .

ii

Solution

The mantle density decreases outward when

ρ_{m} \geq 0

. Requiring a nonnegative surface density gives the immediate physical bound

0 \leq ρ_{m} \leq \frac{ρ _{c} R _{c}}{R _{p} - R _{c}} .

The measured planetary mass also fixes

M_{p} = 4 π [\frac{( ρ _{c} + ρ _{m} ) R _{p}^{3}}{3} - \frac{ρ _{m} R _{p}^{4}}{4 R _{c}} - \frac{ρ _{m} R _{c}^{3}}{12}] .

Equivalently,

ρ_{m} = \frac{ρ _{c} R _{p}^{3} /3 - M _{p} / ( 4 π )}{R _{p}^{4} / ( 4 R _{c} ) - R _{p}^{3} /3 + R _{c}^{3} /12},

and a viable model requires this value to obey the boxed bound.

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