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 = (P_h)/(\rho h)`

The mathematical problem gives us values such as:

`P = 2.4 atm ((101325Pa)/(1atm)) = 243180Pa`

`\rho = 1000kg/m^3`

`h = 28.6m`

Replacing we have,

`g = (243180)/((1000)(28.6))`

`g = 8.5m/s^2`

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