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\section{Creation of a Magnetosphere}  In order to retain this atmosphere, we will need to create a magnetosphere to shield Mars from solar wind and harmful cosmic radiation. Assuming Mars has constant density and a solid inner core and liquid outer core proportionally identical to the Earth's, we can create a dynamo by detonating nuclear weapons in Mars' inner core, thus inducing convection in Mars outer core.  \\  To give Mars the same amount of protection from solar wind as the Earth, we want Mars' magnetopause to be approximately 20 Mars radii away from its center: about the same distance from Earth's center to its magnetopause.  \\  $R_{Magnetopause}=(\frac{2B_0^2}{\mu_0\rho_{sw}v_{sw}^2})^{1/6}$  \\ magnetopause.\\  $R_{Magnetopause}=(\frac{2B_0^2}{\mu_0\rho_{sw}v_{sw}^2})^{1/6}$\\  where $B_0$ is the magnetic dipole moment of the planet and $\rho_{sw}$ and $v_{sw}$ are the density and velocity of solar wind at Mars' orbit respectively.Because the convecting currents which generate the magnetosphere are complex and occur within the spherical outer core, we will simplify our calculations by approximating Mars' outer core as a single gigantic rotating cylinder with radius ${R_{outer core} - R_{inner core}}/2$ and height $2R_{outer core}$. With this approximation, we find Mars' magnetic dipole moment to be  \\[h!] be\\  $B_0 = \frac{\rho_c\omega\mu_0}{16\pi}(\frac{R_{outer core}-R_{inner core}}{2})^4*2R_{outer core}$  \\[h] core}$\\  where $\rho_c$ is the charge density of the outer core and $\omega = 2\pi/2t_{con}$ where $t_{con}$ is the convective timescale.  \\ timescale.\\  $t_{con}$ is mathematically related directly to the adiabatic gradient, and using this we can find the necessary inner core temperature $T_{inner}$ as a function of the the outer core temperature $T_{outer}$.  \\  Consolidating the necessary equations we find that