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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.

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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

[tex]P_h = \rho g h[/tex]

Where,

[tex]\rho[/tex] = Density

h = Height

g = Gravitational acceleration

Rearranging in terms of g

[tex]g = \frac{P_h}{\rho h}[/tex]

our values are given as

[tex]P_h = 1.1 atm (\frac{101325Pa}{1atm}) = 111457.5Pa[/tex]

[tex]\rho = 1000kg/m^3[/tex]

[tex]h = 12.3m[/tex]

Replacing we have

[tex]g = \frac{111457.5}{(1000)(12.3)}[/tex]

[tex]g = 9.061m/s^2[/tex]

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

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