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 12.3 m. If the gauge pressure is measured to be 1.1 atm, what is the gravitational acceleration on the planet's surface? Use 1000 kg/m3 as the mass density of water.

Answer 1

To solve this problem it is necessary to apply the concepts related to hydrostatic pressure or pressure due to a fluid.

Mathematically this pressure is given under the formula

`P_h = \rho g h`

Where,

`\rho` = Density

h = Height

g = Gravitational acceleration

Rearranging in terms of g

`g = (P_h)/(\rho h)`

our values are given as

`P_h = 1.1 atm ((101325Pa)/(1atm)) = 111457.5Pa`

`\rho = 1000kg/m^3`

`h = 12.3m`

Replacing we have

`g = (111457.5)/((1000)(12.3))`

`g = 9.061m/s^2`

Therefore the gravitational acceleration on the planet's surface is
`9.061m/s^2` (Almost the gravity of the Earth)