Assuming that 80% of the Mars surface is covered with water at an average
depth of 3.5 miles and the radius of the earth is given by 5.97x106 meter,
estimate the mass of the water on the Mars in kilogram
m=0.8ρV=0.8ρ43π(R3−(R−r)3)=60⋅1018 kg.m=0.8\rho V=0.8\rho\frac43\pi(R^3-(R-r)^3)=60\cdot 10^{18}~kg.m=0.8ρV=0.8ρ34π(R3−(R−r)3)=60⋅1018 kg.
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