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7.7.2 WISP Thermal Considerations

In order for the WISP to melt through the ice fast enough to reach the water/ice interface in a year, the RTG must produce enough heat. The heat required to melt a differential mass of ice starting from temperature T is:

\begin{displaymath}dq = \left( \int_T^{273.15\,\mathrm{K}}c_v(\tau)\,d\tau\,+\Delta H
\right) dm,
\end{displaymath} (12)

where cv is the specific heat of ice and is a function of temperature, and $\Delta H$ is the heat of fusion of water.

Because $dm=\rho\,dV$, where $\rho$ is density and V is volume, this equation can be integrated over the crust thickness. If the cross-section of the probe is A, then

 \begin{displaymath}Q = \rho A \int_0^h\left( \int_T^{273.15\,\mathrm{K}}
c_v(\tau)\,d\tau\,+\Delta H \right) dz
\end{displaymath} (13)

In Equation 13, h is the thickness of the ice crust, and z is the coordinate in the vertical direction. The water/ice interface is defined to be at z=0, and the surface is defined to be z=h.

In order to evaluate Equation 13, we need to know cv, T, $\rho$ and $\Delta H$. cv is a function of T, and T is a function of z. Reference [8] gives some specific heats of ice at various temperatures. Linear regression yielded the relationship cv=172.21+6.9697T, with cv in joules per kilogram, and T in kelvin. For the lack of a better model, the temperature distribution through Europa's crust is assumed to vary linearly from the surface to the water/ice interface. At the surface, T=100 K. At the interface, T=273.15 K. Therefore, T=273.15-173.15(z/h), where T is in kelvin. $\rho$ is estimated to be 1000 kg/m3. Finally, Reference [8] lists $\Delta H$ for water as 6010 J/mol = 334,000 J/kg.

With these values, the Equation 13 reduces to

 \begin{displaymath}Q=(4.79\times 10^8 \textrm{J}/\textrm{m}^3)Ah
\end{displaymath} (14)

Equation 14 gives the heat required as a function of the volume of ice melted. The cross-sectional area is about 100 cm \ensuremath{^2}, or 0.01 m \ensuremath{^2}. We guessed the ice thickness to be around 10 km. The heat required to melt through this is $4.79\times 10^{10}$ J.

The lander is to transmit data for at least one year; therefore, the RTG should produce the heat necessary to melt 10 km in a year. This rate is 1520 watts. Asterius' baseline uses a 2000 W RTG because some heat will certainly leak through the side walls, and the assumptions used to derive 1520 W were very conjectural. The 2000 W RTG increases the chance of mission success by allowing the probe to descend faster should the crust be thicker than 10 km.