Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 315 / 3 / a / i

Table of contents
- Solution i

Solution

Let

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

The enclosed mass is

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} .

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version