Help Notes: (a) Calculate the total mass of a uniform layer of water covering the entire Martian surface to a depth of 3 m (see Section 6.8). (b) Compare it...

Sunday, 1 February 2015

(a) Calculate the total mass of a uniform layer of water covering the entire Martian surface to a depth of 3 m (see Section 6.8). (b) Compare it...

For this question we need Martian radius, which is about 3390 km (see the link attached).

Actually, enough accuracy is reached if we assume the layer is flat with the given thickness and with the surface area as of Mars.

The formula for the surface of a sphere is $\displaystyle{4}\pi{{R}}^{{3}}.$ Numerically we obtain $\displaystyle{0.003}\cdot{4}\cdot\pi\cdot{{3390}}^{{2}}\approx{433242}{k}{{m}}^{{3}}.$

The mass of one cubical kilometer of water is $\displaystyle{{10}}^{{9}}$ kg, so the total mass is about...

For this question we need Martian radius, which is about 3390 km (see the link attached).

Actually, enough accuracy is reached if we assume the layer is flat with the given thickness and with the surface area as of Mars.

The formula for the surface of a sphere is $\displaystyle{4}\pi{{R}}^{{3}}.$ Numerically we obtain $\displaystyle{0.003}\cdot{4}\cdot\pi\cdot{{3390}}^{{2}}\approx{433242}{k}{{m}}^{{3}}.$

The mass of one cubical kilometer of water is $\displaystyle{{10}}^{{9}}$ kg, so the total mass is about $\displaystyle{4.3}\cdot{{10}}^{{14}}{k}{g{.}}$ This is about $\displaystyle{{10}}^{{6}}$ times less than the given mass of Venus's atmosphere.

Help Notes

Sunday, 1 February 2015

(a) Calculate the total mass of a uniform layer of water covering the entire Martian surface to a depth of 3 m (see Section 6.8). (b) Compare it...

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