Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 12 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

In spherical coordinates,

θ

is the polar angle from the positive

z

axis and

ϕ

the azimuthal angle. The Jacobian determinant of the coordinate transformation gives the volume element

d V = r^{2} sin θ d r d θ d ϕ, 0 \leq θ \leq π, 0 \leq ϕ < 2 π .

ii

Solution

Height above the surface is

h = r - R

. Integrate the density over all exterior spherical shells:

M = \int_{R}^{\infty} \int_{0}^{π} \int_{0}^{2 π} ρ_{0} e^{- (r - R) / h_{0}} r^{2} sin θ d ϕ d θ d r = 4 π ρ_{0} \int_{0}^{\infty} (R + h)^{2} e^{- h / h_{0}} d h .

Repeated integration by parts, or the substitution

h = h_{0} s

, gives

\int_{0}^{\infty} e^{- h / h_{0}} d h = h_{0}, \int_{0}^{\infty} h e^{- h / h_{0}} d h = h_{0}^{2}, \int_{0}^{\infty} h^{2} e^{- h / h_{0}} d h = 2 h_{0}^{3} .

Expand

(R + h)^{2}

and use these moments to obtain the mass of an exponential spherical atmosphere:

M = 4 π ρ_{0} h_{0} (R^{2} + 2 R h_{0} + 2 h_{0}^{2}) .

For

h_{0} ≪ R

, the leading term is surface area

4 π R^{2}

times column mass

ρ_{0} h_{0}

; the other terms retain the finite-curvature corrections.

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