IPhO 2013 · Theory problem 3

The Greenlandic Ice Sheet

Introduction

This problem deals with the physics of the Greenlandic ice sheet, the second largest glacier in the world, Fig. 3.1(a). As an idealization, Greenland is modeled as a rectangular island of width \(2L\) and length \(5L\) with the ground at sea level and completely covered by incompressible ice (constant density \(\rho_{\mathrm{ice}}\)), see Fig. 3.1(b). The height profile \(H(x)\) of the ice sheet does not depend on the \(y\)-coordinate and it increases from zero at the coasts \(x=\pm L\) to a maximum height \(H_{\mathrm m}\) along the middle north-south axis (the \(y\)-axis), known as the ice divide, see Fig. 3.1(c).

Figure 1

Figure 3.1 (a) A map of Greenland showing the extent of the ice sheet (white), the ice-free, coastal regions (green), and the surrounding ocean (blue). (b) The crude model of the Greenlandic ice sheet as covering a rectangular area in the \(xy\)-plane with side lengths \(2L\) and \(5L\). The ice divide, the line of maximum ice sheet height \(H_{\mathrm m}\) runs along the \(y\)-axis. (c) A vertical cut (\(xz\)-plane) through the ice sheet showing the height profile \(H(x)\) (blue line). \(H(x)\) is independent of the \(y\)-coordinate for \(0<y<5L\), while it drops abruptly to zero at \(y=0\) and \(y=5L\). The \(z\)-axis marks the position of the ice divide. For clarity, the vertical dimensions are expanded compared to the horizontal dimensions. The density \(\rho_{\mathrm{ice}}\) of ice is constant.

Two useful formulas

In this problem you can make use of the integral:

\[ \int_0^1\sqrt{1-x}\,\mathrm dx=\frac23 \]

and the approximation \((1+x)^\alpha\approx1+\alpha x\), valid for \(|\alpha x|\ll1\).

The height profile of the ice sheet

On short time scales the glacier is an incompressible hydrostatic system with fixed height profile \(H(x)\).

3.1 Write down an expression for the pressure \(p(x,z)\) inside the ice sheet as a function of vertical height \(z\) above the ground and distance \(x\) from the ice divide. Neglect the atmospheric pressure. 0.3

Consider a given vertical slab of the ice sheet in equilibrium, covering a small horizontal base area \(\Delta x\Delta y\) between \(x\) and \(x+\Delta x\), see the red dashed lines in Fig. 3.1(c). The size of \(\Delta y\) does not matter. The net horizontal force component \(\Delta F\) on the two vertical sides of the slab, arising from the difference in height on the center-side versus the coastal-side of the slab, is balanced by a friction force \(\Delta F=S_{\mathrm b}\Delta x\Delta y\) from the ground on the base area \(\Delta x\Delta y\), where \(S_{\mathrm b}=100\ \mathrm{kPa}\).

3.2a For a given value of \(x\), show that in the limit \(\Delta x\rightarrow0\), \(S_{\mathrm b}=kH\,\mathrm dH/\mathrm dx\), and determine \(k\). 0.9
3.2b Determine an expression for the height profile \(H(x)\) in terms of \(\rho_{\mathrm{ice}}\), \(g\), \(L\), \(S_{\mathrm b}\) and distance \(x\) from the divide. The result will show that the maximum glacier height \(H_{\mathrm m}\) scales with the half-width \(L\) as \(H_{\mathrm m}\propto L^{1/2}\). 0.8
3.2c Determine the exponent \(\gamma\) with which the total volume \(V_{\mathrm{ice}}\) of the ice sheet scales with the area \(A\) of the rectangular island, \(V_{\mathrm{ice}}\propto A^\gamma\). 0.5

A dynamical ice sheet

On longer time scale, the ice is a viscous incompressible fluid, which by gravity flows from the center part to the coast. In this model, the ice maintains its height profile \(H(x)\) in a steady state, where accumulation of ice due to snowfall in the central region is balanced by melting at the coast. In addition to the ice sheet geometry of Fig. 3.1(b) and (c) make the following model assumptions:

1) Ice flows in the \(xz\)-plane away from the ice divide (the \(y\)-axis). 2) The accumulation rate \(c\) (m/year) in the central region is a constant. 3) Ice can only leave the glacier by melting near the coasts at \(x=\pm L\). 4) The horizontal (\(x\)-)component \(v_x(x)=\mathrm dx/\mathrm dt\) of the ice-flow velocity is independent of \(z\). 5) The vertical (\(z\)-)component \(v_z(z)=\mathrm dz/\mathrm dt\) of the ice-flow velocity is independent of \(x\).

Consider only the central region \(|x|\ll L\) close to the middle of the ice sheet, where height variations of the ice sheet are very small and can be neglected altogether, i.e. \(H(x)\approx H_{\mathrm m}\).

3.3 Use mass conservation to find an expression for the horizontal ice-flow velocity \(v_x(x)\) in terms of \(c\), \(x\), and \(H_{\mathrm m}\). 0.6

From the assumption of incompressibility, i.e. the constant density \(\rho_{\mathrm{ice}}\) of the ice, it follows that mass conservation implies the following restriction on the ice flow velocity components

\[ \frac{\mathrm dv_x}{\mathrm dx}+\frac{\mathrm dv_z}{\mathrm dz}=0. \]

3.4 Write down an expression for the \(z\) dependence of the vertical component \(v_z(z)\) of the ice-flow velocity. 0.6

A small ice particle with the initial surface position \((x_i,H_{\mathrm m})\) will, as time passes, flow as part of the ice sheet along a flow trajectory \(z(x)\) in the vertical \(xz\)-plane.

3.5 Derive an expression for such a flow trajectory \(z(x)\). 0.9

Age and climate indicators in the dynamical ice sheet

Based on the ice-flow velocity components \(v_x(x)\) and \(v_z(z)\), one can estimate the age \(\tau(z)\) of the ice in a specific depth \(H_{\mathrm m}-z\) from the surface of the ice sheet.

3.6 Find an expression for the age \(\tau(z)\) of the ice as a function of height \(z\) above ground, right at the ice divide \(x=0\). 1.0

An ice core drilled in the interior of the Greenlandic ice sheet will penetrate through layers of snow from the past, and the ice core can be analyzed to reveal past climate changes. One of the best indicators is the so-called \(\delta^{18}\mathrm O\), defined as

\[ \delta^{18}\mathrm O =\frac{R_{\mathrm{ice}}-R_{\mathrm{ref}}}{R_{\mathrm{ref}}}\,1000\ \text{‰}, \]

where \(R=[{}^{18}\mathrm O]/[{}^{16}\mathrm O]\) denotes the relative abundance of the two stable isotopes \({}^{18}\mathrm O\) and \({}^{16}\mathrm O\) of oxygen. The reference \(R_{\mathrm{ref}}\) is based on the isotopic composition of the oceans around Equator.

Figure 2

Figure 3.2 (a) Observed relationship between \(\delta^{18}\mathrm O\) in snows versus the mean annual surface temperature \(T\). (b) Measurements of \(\delta^{18}\mathrm O\) versus depth \(H_{\mathrm m}-z\) from the surface, taken from an ice core drilled from surface to bedrock at a specific place along the Greenlandic ice divide where \(H_{\mathrm m}=3060\ \mathrm m\).

Observations from the Greenland ice sheet show that \(\delta^{18}\mathrm O\) in the snow varies approximately linearly with temperature, Fig. 3.2(a). Assuming that this has always been the case, \(\delta^{18}\mathrm O\) retrieved from an ice core at depth \(H_{\mathrm m}-z\) leads to an estimate of the temperature \(T\) near Greenland at the age \(\tau(z)\).

Measurements of \(\delta^{18}\mathrm O\) in a 3060 m long Greenlandic ice core show an abrupt change in \(\delta^{18}\mathrm O\) at a depth of 1492 m, Fig. 3.2(b), marking the end of the last ice age. The ice age began 120,000 years ago, corresponding to a depth of 3040 m, and the current interglacial age began 11,700 years ago, corresponding to a depth of 1492 m. Assume that these two periods can be described by two different accumulation rates, \(c_{\mathrm{ia}}\) (ice age) and \(c_{\mathrm{ig}}\) (interglacial age), respectively. You can assume \(H_{\mathrm m}\) to be constant throughout these 120,000 years.

3.7a Determine the accumulation rates \(c_{\mathrm{ia}}\) and \(c_{\mathrm{ig}}\). 0.8
3.7b Use the data in Fig. 3.2 to find the temperature change at the transition from the ice age to the interglacial age. 0.2

Sea level rise from melting of the Greenland ice sheet

A complete melting of the Greenlandic ice sheet will cause a sea level rise in the global ocean. As a crude estimate of this sea level rise, one may simply consider a uniform rise throughout a global ocean with constant area \(A_{\mathrm O}=3.61\times10^{14}\ \mathrm{m^2}\).

3.8 Calculate the average global sea level rise, which would result from a complete melting of the Greenlandic ice sheet, given its present area of \(A_{\mathrm G}=1.71\times10^{12}\ \mathrm{m^2}\) and \(S_{\mathrm b}=100\ \mathrm{kPa}\). 0.6

The massive Greenland ice sheet exerts a gravitational pull on the surrounding ocean. If the ice sheet melts, this local high tide is lost and the sea level will drop close to Greenland, an effect which partially counteracts the sea level rise calculated above.

To estimate the magnitude of this gravitational pull on the water, the Greenlandic ice sheet is now modeled as a point mass located at the ground level and having the total mass of the Greenlandic ice sheet. Copenhagen lies at a distance of 3500 km along the Earth surface from the center of the point mass. One may consider the Earth, without the point mass, to be spherically symmetric and having a global ocean spread out over the entire surface of the Earth of area \(A_{\mathrm E}=5.10\times10^{14}\ \mathrm{m^2}\). All effects of rotation of the Earth may be neglected.

3.9 Within this model, determine the difference \(h_{\mathrm{CPH}}-h_{\mathrm{OPP}}\) between sea levels in Copenhagen (\(h_{\mathrm{CPH}}\)) and diametrically opposite to Greenland (\(h_{\mathrm{OPP}}\)). 1.8

Data sheet: Table of physical parameters

Speed of light in vacuum \(c=2.998\times10^8\ \mathrm{m\,s^{-1}}\)
Planck's constant over \(2\pi\) \(\hbar=1.055\times10^{-34}\ \mathrm{J\,s}\)
Gravitational constant \(G=6.67\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}\)
Gravitational acceleration \(g=9.82\ \mathrm{m\,s^{-2}}\)
Elementary charge \(e=1.602\times10^{-19}\ \mathrm C\)
Dielectric permittivity of vacuum \(\epsilon_0=8.854\times10^{-12}\ \mathrm{C^2\,J^{-1}\,m^{-1}}\)
Electron mass \(m_e=9.109\times10^{-31}\ \mathrm{kg}\)
Avogadro constant \(N_{\mathrm A}=6.022\times10^{23}\ \mathrm{mol^{-1}}\)
Boltzmann constant \(k_{\mathrm B}=1.381\times10^{-23}\ \mathrm{J\,K^{-1}}\)
Stony meteorite, specific heat \(c_{\mathrm{sm}}=1.2\times10^3\ \mathrm{J\,kg^{-1}\,K^{-1}}\)
Stony meteorite, thermal conductivity \(k_{\mathrm{sm}}=2.0\ \mathrm{W\,m^{-1}\,K^{-1}}\)
Stony meteorite, density \(\rho_{\mathrm{sm}}=3.3\times10^3\ \mathrm{kg\,m^{-3}}\)
Stony meteorite, melting point \(T_{\mathrm{sm}}=1.7\times10^3\ \mathrm K\)
Stony meteorite, specific melting heat \(L_{\mathrm{sm}}=2.6\times10^5\ \mathrm{J\,kg^{-1}}\)
Silver, molar mass \(M_{\mathrm{Ag}}=1.079\times10^{-1}\ \mathrm{kg\,mol^{-1}}\)
Silver, density \(\rho_{\mathrm{Ag}}=1.049\times10^4\ \mathrm{kg\,m^{-3}}\)
Silver, specific heat capacity \(c_{\mathrm{Ag}}=2.40\times10^2\ \mathrm{J\,kg^{-1}\,K^{-1}}\)
Water, molar mass \(M_{\mathrm{wa}}=1.801\times10^{-2}\ \mathrm{kg\,mol^{-1}}\)
Water, density \(\rho_{\mathrm{wa}}=0.998\times10^3\ \mathrm{kg\,m^{-3}}\)
Water, specific heat capacity \(c_{\mathrm{wa}}=4.181\times10^3\ \mathrm{J\,kg^{-1}\,K^{-1}}\)
Water, heat of vaporization \(L_{\mathrm{wa}}=2.260\times10^6\ \mathrm{J\,kg^{-1}}\)
Water, boiling temperature \(T_{100}=100\,^\circ\mathrm C=373.15\ \mathrm K\)
Ice, density of glacier \(\rho_{\mathrm{ice}}=0.917\times10^3\ \mathrm{kg\,m^{-3}}\)
Steam, specific heat capacity \(c_{\mathrm{st}}=2.080\times10^3\ \mathrm{J\,kg^{-1}\,K^{-1}}\)
Earth, mass of the \(m_{\mathrm E}=5.97\times10^{24}\ \mathrm{kg}\)
Earth, radius of the \(R_{\mathrm E}=6.38\times10^6\ \mathrm m\)
Sun, mass of the \(m_{\mathrm S}=1.99\times10^{30}\ \mathrm{kg}\)
Sun, radius of the \(R_{\mathrm S}=6.96\times10^8\ \mathrm m\)
Average Sun-Earth distance \(a_{\mathrm E}=1.50\times10^{11}\ \mathrm m\)

Official English text of the problem.

Answers

  • 3.1: \(\displaystyle p(x,z)=\rho_{\mathrm{ice}}g[H(x)-z]\), for \(0\le z\le H(x)\).
  • 3.2a: \(\displaystyle S_{\mathrm b}=-\rho_{\mathrm{ice}}gH\frac{\mathrm dH}{\mathrm dx}\), so \(\boxed{k=-\rho_{\mathrm{ice}}g}\) on the \(x>0\) half.
  • 3.2b: \(\displaystyle \boxed{H(x)=\sqrt{\frac{2S_{\mathrm b}}{\rho_{\mathrm{ice}}g}(L-|x|)}}\), with \(\displaystyle H_{\mathrm m}=\sqrt{\frac{2S_{\mathrm b}L}{\rho_{\mathrm{ice}}g}}\).
  • 3.2c: \(\boxed{\gamma=\tfrac54}\); specifically, \(\displaystyle V_{\mathrm{ice}}=\frac23 AH_{\mathrm m}\).
  • 3.3: \(\displaystyle \boxed{v_x(x)=\frac{c}{H_{\mathrm m}}x}\).
  • 3.4: \(\displaystyle \boxed{v_z(z)=-\frac{c}{H_{\mathrm m}}z}\).
  • 3.5: \(\displaystyle \boxed{z(x)=H_{\mathrm m}\frac{x_i}{x}}\) for \(x_i\ne0\). For \(x_i=0\), the trajectory is vertical along \(x=0\).
  • 3.6: \(\displaystyle \boxed{\tau(z)=\frac{H_{\mathrm m}}c\ln\!\frac{H_{\mathrm m}}z}\).
  • 3.7a: \(\displaystyle \boxed{c_{\mathrm{ia}}=0.123\ \mathrm{m/year},\qquad c_{\mathrm{ig}}=0.175\ \mathrm{m/year}}\).
  • 3.7b: \(\boxed{\Delta T\approx+10\ \mathrm K}\), approximately \(10\,^\circ\mathrm C\) of warming.
  • 3.8: \(\boxed{\Delta h_{\mathrm{mean}}\approx8.8\ \mathrm m}\).
  • 3.9: \(\displaystyle \boxed{h_{\mathrm{CPH}}-h_{\mathrm{OPP}}\approx4.5\ \mathrm m}\), with Copenhagen higher.

Solution

3.1

Hydrostatic balance in the vertical direction gives

\[ \frac{\partial p}{\partial z}=-\rho_{\mathrm{ice}}g. \]

The pressure at the surface is zero when atmospheric pressure is neglected:

\[ p(x,H(x))=0. \]

Integrating downward from the surface therefore gives

\[ \boxed{p(x,z)=\rho_{\mathrm{ice}}g[H(x)-z]}. \]

3.2a

Consider the right-hand half of the ice sheet, \(x>0\). The pressure force on a vertical face of height \(H(x)\) and width \(\Delta y\) is

\[ F(x)=\Delta y\int_0^{H(x)} \rho_{\mathrm{ice}}g[H(x)-z]\,\mathrm dz =\frac12\rho_{\mathrm{ice}}gH(x)^2\Delta y. \]

The force on the center-side face acts toward increasing \(x\), while that on the coastal-side face acts toward decreasing \(x\). Their resultant is

\[ \Delta F =\frac12\rho_{\mathrm{ice}}g \left[H(x)^2-H(x+\Delta x)^2\right]\Delta y. \]

To first order in \(\Delta x\),

\[ \Delta F =-\rho_{\mathrm{ice}}gH(x)\frac{\mathrm dH}{\mathrm dx} \,\Delta x\Delta y. \]

Balancing this outward force against the basal friction \(S_{\mathrm b}\Delta x\Delta y\) yields

\[ \boxed{S_{\mathrm b} =-\rho_{\mathrm{ice}}gH\frac{\mathrm dH}{\mathrm dx}}, \qquad \boxed{k=-\rho_{\mathrm{ice}}g}. \]

The derivative is negative on this half, so \(S_{\mathrm b}>0\), as required. On either half, the corresponding magnitude relation is \(S_{\mathrm b}=\rho_{\mathrm{ice}}gH|\mathrm dH/\mathrm dx|\).

3.2b

On \(0\le x\le L\), the result above can be written

\[ \frac{\mathrm d(H^2)}{\mathrm dx} =-\frac{2S_{\mathrm b}}{\rho_{\mathrm{ice}}g}. \]

Using the coastal boundary condition \(H(L)=0\),

\[ H(x)^2=\frac{2S_{\mathrm b}}{\rho_{\mathrm{ice}}g}(L-x). \]

Reflection about the ice divide gives the complete profile:

\[ \boxed{ H(x)=\sqrt{\frac{2S_{\mathrm b}}{\rho_{\mathrm{ice}}g}(L-|x|)} },\qquad -L\le x\le L. \]

At the divide,

\[ \boxed{ H_{\mathrm m}=\sqrt{\frac{2S_{\mathrm b}L}{\rho_{\mathrm{ice}}g}} }, \]

so \(H_{\mathrm m}\propto L^{1/2}\) when the basal friction, density, and gravitational acceleration are fixed.

3.2c

The island area is

\[ A=(2L)(5L)=10L^2. \]

Since the profile is independent of \(y\), the ice volume is

\[ V_{\mathrm{ice}} =5L\int_{-L}^{L}H(x)\,\mathrm dx =10L H_{\mathrm m}\int_0^L \sqrt{1-\frac{x}{L}}\,\mathrm dx. \]

With \(u=x/L\), the supplied integral gives

\[ V_{\mathrm{ice}} =10L^2H_{\mathrm m}\int_0^1\sqrt{1-u}\,\mathrm du =\frac{20}{3}L^2H_{\mathrm m} =\boxed{\frac23 AH_{\mathrm m}}. \]

Thus

\[ V_{\mathrm{ice}}\propto L^{2+1/2} =L^{5/2} \propto A^{5/4}, \]

and

\[ \boxed{\gamma=\frac54}. \]

3.3

Take a central control volume between the divide and a position \(x>0\), with width \(\Delta y\). In steady state, the accumulation over its surface must equal its horizontal outflow.

The accumulated mass per unit time is

\[ \dot m_{\mathrm{in}} =\rho_{\mathrm{ice}}cx\Delta y. \]

There is no horizontal flow through the divide by symmetry. Since \(v_x\) is independent of height, the outflow through the face at \(x\) is

\[ \dot m_{\mathrm{out}} =\rho_{\mathrm{ice}}v_x(x)H_{\mathrm m}\Delta y. \]

Equating these gives

\[ \boxed{v_x(x)=\frac{c}{H_{\mathrm m}}x}. \]

The same signed expression applies for \(x<0\), where the flow is toward negative \(x\). This result uses the central-region approximation \(H(x)\approx H_{\mathrm m}\).

3.4

Incompressibility requires

\[ \frac{\mathrm dv_z}{\mathrm dz} =-\frac{\mathrm dv_x}{\mathrm dx} =-\frac{c}{H_{\mathrm m}}. \]

Integration gives

\[ v_z(z)=-\frac{c}{H_{\mathrm m}}z+C. \]

There is no flow through the ground, so \(v_z(0)=0\), implying \(C=0\). Therefore,

\[ \boxed{v_z(z)=-\frac{c}{H_{\mathrm m}}z}. \]

In particular, \(v_z(H_{\mathrm m})=-c\): downward motion at the surface balances the addition of new ice.

3.5

For a particle with \(x_i\ne0\), its trajectory satisfies

\[ \frac{\mathrm dz}{\mathrm dx} =\frac{\mathrm dz/\mathrm dt}{\mathrm dx/\mathrm dt} =\frac{v_z}{v_x} =-\frac zx. \]

Separating variables and applying the initial condition \(z(x_i)=H_{\mathrm m}\),

\[ \int_{H_{\mathrm m}}^z\frac{\mathrm dz'}{z'} =-\int_{x_i}^x\frac{\mathrm dx'}{x'}, \]

so

\[ \ln\frac{z}{H_{\mathrm m}} =-\ln\left|\frac{x}{x_i}\right|. \]

The particle remains on its original side of the divide; hence \(x/x_i>0\), and

\[ \boxed{xz=x_iH_{\mathrm m}}, \qquad \boxed{z(x)=H_{\mathrm m}\frac{x_i}{x}}. \]

The trajectory is a branch of a rectangular hyperbola, along which the particle moves outward and downward. Equivalently,

\[ x(t)=x_i e^{ct/H_{\mathrm m}}, \qquad z(t)=H_{\mathrm m}e^{-ct/H_{\mathrm m}}. \]

For \(x_i=0\), the particle remains at \(x=0\) and descends vertically according to the same expression for \(z(t)\). These trajectories apply while the particle remains in the central region.

3.6

At the ice divide, the horizontal velocity vanishes. The age is the elapsed time since the particle was deposited at \(z=H_{\mathrm m}\):

\[ \tau(z)=\int_{H_{\mathrm m}}^z\frac{\mathrm dz'}{v_z(z')} =-\frac{H_{\mathrm m}}c \int_{H_{\mathrm m}}^z\frac{\mathrm dz'}{z'}. \]

Therefore,

\[ \boxed{\tau(z)=\frac{H_{\mathrm m}}c \ln\frac{H_{\mathrm m}}z}. \]

This is zero at the surface and increases with depth. In this idealized model, it diverges as \(z\to0\), because the vertical velocity approaches zero at the bed.

3.7a

For this ice core, \(H_{\mathrm m}=3060\ \mathrm m\). The present heights above the bed of the two dated layers are

\[ z_1=3060-1492=1568\ \mathrm m, \]

\[ z_2=3060-3040=20\ \mathrm m. \]

The transition layer has experienced only the \(11\,700\) years of interglacial flow. Thus

\[ \ln\frac{3060}{1568} =\frac{c_{\mathrm{ig}}(11\,700\ \mathrm{year})}{3060\ \mathrm m}, \]

giving

\[ \boxed{ c_{\mathrm{ig}} =\frac{3060}{11\,700} \ln\frac{3060}{1568} =0.175\ \mathrm{m/year} }. \]

For the older layer, the accumulation rate changed during its descent. With constant \(H_{\mathrm m}\),

\[ \frac{\mathrm d\ln z}{\mathrm dt} =-\frac{c(t)}{H_{\mathrm m}}. \]

It experienced ice-age flow for

\[ 120\,000-11\,700=108\,300\ \mathrm{years}, \]

followed by interglacial flow for \(11\,700\) years. Consequently,

\[ \ln\frac{3060}{20} =\frac{ c_{\mathrm{ia}}(108\,300\ \mathrm{year}) +c_{\mathrm{ig}}(11\,700\ \mathrm{year}) }{3060\ \mathrm m}. \]

Subtracting the equation for the transition layer eliminates the interglacial contribution:

\[ \ln\frac{1568}{20} =\frac{c_{\mathrm{ia}}(108\,300\ \mathrm{year})}{3060\ \mathrm m}. \]

Hence

\[ \boxed{ c_{\mathrm{ia}} =\frac{3060}{108\,300} \ln\frac{1568}{20} =0.123\ \mathrm{m/year} }. \]

Accounting for the interglacial compression of the older ice is essential.

3.7b

Figure 3.2(b) gives representative isotope values of approximately

\[ \delta_{\mathrm{ia}}^{18}\mathrm O\approx-42\ \text{‰}, \qquad \delta_{\mathrm{ig}}^{18}\mathrm O\approx-35\ \text{‰}. \]

Thus the transition corresponds to an increase of roughly \(7\ \text{‰}\).

The approximately linear trend in Fig. 3.2(a), in the relevant temperature range, has slope about

\[ \frac{\Delta(\delta^{18}\mathrm O)}{\Delta T} \approx0.7\ \text{‰}\,\mathrm{K^{-1}}. \]

Therefore,

\[ \Delta T \approx\frac{7\ \text{‰}} {0.7\ \text{‰}\,\mathrm{K^{-1}}} \approx10\ \mathrm K. \]

The change is a warming:

\[ \boxed{T_{\mathrm{ig}}-T_{\mathrm{ia}}\approx+10\ \mathrm K}. \]

The precision is limited by the graphical readings.

3.8

Use the island geometry and height profile to determine the present model volume. Since \(A_{\mathrm G}=10L^2\),

\[ L=\sqrt{\frac{A_{\mathrm G}}{10}} =\sqrt{\frac{1.71\times10^{12}}{10}} =4.14\times10^5\ \mathrm m. \]

The maximum height predicted by the model is

\[ H_{\mathrm m} =\sqrt{\frac{2S_{\mathrm b}L}{\rho_{\mathrm{ice}}g}} =\sqrt{ \frac{2(1.00\times10^5)(4.1352\times10^5)} {(917)(9.82)} } =3.03\times10^3\ \mathrm m. \]

Using part 3.2c,

\[ V_{\mathrm{ice}} =\frac23 A_{\mathrm G}H_{\mathrm m} =3.45\times10^{15}\ \mathrm{m^3}. \]

Melting conserves mass, so the resulting water volume is

\[ V_{\mathrm{water}} =\frac{\rho_{\mathrm{ice}}}{\rho_{\mathrm{wa}}}V_{\mathrm{ice}}. \]

Spreading this water uniformly over the constant ocean area gives

\[ \Delta h_{\mathrm{mean}} =\frac{V_{\mathrm{water}}}{A_{\mathrm O}} =\frac{\rho_{\mathrm{ice}}}{\rho_{\mathrm{wa}}} \frac{2A_{\mathrm G}H_{\mathrm m}}{3A_{\mathrm O}}. \]

Numerically,

\[ \Delta h_{\mathrm{mean}} =\frac{917}{998} \frac{3.4548\times10^{15}}{3.61\times10^{14}} =8.79\ \mathrm m. \]

Thus

\[ \boxed{\Delta h_{\mathrm{mean}}\approx8.8\ \mathrm m}. \]

3.9

An ocean at rest has an equipotential free surface. We therefore compare the gravitational potentials at Copenhagen and at the point opposite Greenland.

The point mass representing the ice sheet is

\[ M=\rho_{\mathrm{ice}}V_{\mathrm{ice}} =(917)(3.4548\times10^{15}) =3.17\times10^{18}\ \mathrm{kg}. \]

Let \(R\) denote Earth’s radius and \(h\) a small sea-surface displacement from a reference sphere. Earth’s contribution to the gravitational potential per unit mass is

\[ \Phi_{\mathrm E}(R+h) =-\frac{Gm_{\mathrm E}}{R+h} \approx-\frac{Gm_{\mathrm E}}R+gh, \]

where \(h\ll R\). The ice mass contributes

\[ \Phi_{\mathrm{ice}}=-\frac{GM}{d}, \]

with \(d\) the straight-line distance to the ice mass.

To first order, \(d\) can be evaluated on the reference sphere: changes of a few metres are negligible compared with distances of millions of metres. Equal surface potentials therefore require

\[ gh_{\mathrm{CPH}}-\frac{GM}{d_{\mathrm{CPH}}} = gh_{\mathrm{OPP}}-\frac{GM}{d_{\mathrm{OPP}}}. \]

Thus

\[ h_{\mathrm{CPH}}-h_{\mathrm{OPP}} =\frac{GM}{g} \left(\frac1{d_{\mathrm{CPH}}} -\frac1{d_{\mathrm{OPP}}}\right). \]

The given \(3500\ \mathrm{km}\) is an arc length \(s\), so the central angle and chord distance are

\[ \theta=\frac{s}{R}, \qquad d_{\mathrm{CPH}}=2R\sin\frac{\theta}{2} =2R\sin\frac{s}{2R}. \]

At the opposite point,

\[ d_{\mathrm{OPP}}=2R. \]

Consequently,

\[ \boxed{ h_{\mathrm{CPH}}-h_{\mathrm{OPP}} =\frac{GM}{2gR} \left[ \frac1{\sin(s/(2R))}-1 \right] }. \]

The stated Earth area gives \(R=\sqrt{A_{\mathrm E}/(4\pi)}\approx6.37\times10^6\ \mathrm m\), consistent with the supplied radius. Using \(R=6.38\times10^6\ \mathrm m\),

\[ \theta=0.5486, \qquad d_{\mathrm{CPH}}=3.456\times10^6\ \mathrm m, \qquad d_{\mathrm{OPP}}=1.276\times10^7\ \mathrm m. \]

Hence

\[ h_{\mathrm{CPH}}-h_{\mathrm{OPP}} = \frac{(6.67\times10^{-11})(3.168\times10^{18})}{9.82} \left( \frac1{3.456\times10^6} -\frac1{1.276\times10^7} \right) =4.54\ \mathrm m. \]

Therefore,

\[ \boxed{h_{\mathrm{CPH}}-h_{\mathrm{OPP}}\approx4.5\ \mathrm m}. \]

The positive sign means that Greenland’s gravitational attraction raises the equilibrium ocean surface at Copenhagen relative to the opposite side of Earth. Any common offset needed to conserve ocean volume cancels from this difference.

The solution was checked against the official solution.