Question

An astronaut landed on a far away planet that has a sea of water. To determine the gravitational acceleration on the planet's surface, the astronaut lowered a pressure gauge into the sea to a depth of 28.6 m. If the gauge pressure is measured to be 2.4 atm, what is the gravitational acceleration on the planet's surface?

Answer 1

The concept required to solve this problem is hydrostatic pressure. From the theory and assuming that the density of water on that planet is equal to that of the earth $(1000kg / m ^ 3)$we can mathematically define the pressure as

$P = \rho g h$

Where,

$\rho$ = Density

h = Height

g = Gravitational acceleration

Rearranging the equation based on gravity

$g = \frac{P_h}{\rho h}$

The mathematical problem gives us values such as:

$P = 2.4 atm (\frac{101325Pa}{1atm}) = 243180Pa$

$\rho = 1000kg/m^3$

$h = 28.6m$

Replacing we have,

$g = \frac{243180}{(1000)(28.6)}$

$g = 8.5m/s^2$

Therefore the gravitational acceleration on the planet's surface is $8.5m/s^2$