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 24.3 m. If the gauge pressure is measured to be 3.8 atm, what is the gravitational acceleration on the planet's surface?

Answer 1

Answer:

Acceleration due to gravity will be $g=17.3m/sec^2$

Explanation:

We have given gauge pressure P = 3.8 atm = 3.8×101325 = 385035 Pa

Depth h = 24.3 m

Density $\rho =1000kg/m^3$

We have find the acceleration due to gravity at the surface of planet

We know that pressure is given by

$P=\rho gh$

So $385035=1000\times g\times 24.3$

$g=17.3m/sec^2$

Acceleration due to gravity will be $g=17.3m/sec^2$