91Ó°ÊÓ

• The distance in ocean where Mars was found is, h = 0.500 km(100 m) = 500 m
• Acceleration due to gravity on Mars,$\mathrm{g}=3.71\mathrm{m}/{\mathrm{s}}^2$
• Density of freshwater,$\mathrm{ÒÏ}=1.00×{10}^3\mathrm{kg}/{\mathrm{m}}^3$
• Density of seawater,${\mathrm{ÒÏ}}_0=1.03×{10}^3\mathrm{kg}/{\mathrm{m}}^3$

Gauge pressure is a measure of how close a fluid's pressure is to that of the atmosphere. The atmospheric pressure increases the pressure of any unenclosed fluid.

$p-p_0=\mathrm{ÒÏ}gh....\left(i\right)$

Evaluate gauge pressure at bottom of ocean. Substitute the value in equation (i),

$\begin{array}{r}p−p_0=\left(1.00×{10}^3\mathrm{kg}/{\mathrm{m}}^3\right)\left(3.71\mathrm{m}/{\mathrm{s}}^2\right)\left(500\mathrm{m}\right)\left(\frac{\mathrm{Pa}}{\mathrm{kg}/{\mathrm{ms}}^2}\right)\\ p−p_0=1.86×{10}^6\mathrm{Pa}\end{array}$

Evaluate depth in earth’s ocean that will experience gauge pressure,

Rearranging equation (i) and substituting value in equation,

$\begin{array}{r}h=\frac{p−p_0}{\mathrm{ÒÏ}g}\\ h=\frac{1.86×{10}^6\mathrm{Pa}}{\left(1.03×{10}^3\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}\left(\frac{\mathrm{m}}{\mathrm{Pa}/{\mathrm{kgs}}^2}\right)\\ h=184\mathrm{m}\end{array}$

Thus, cylinder of water of a particular height weighs more on Earth than it does on Mars, pressure at given depth is higher on Earth.