Gauge Pressure at the Bottom of Ocean on Mars

[forestmine]

Homework Statement

Scientists have found evidence that Mars may once have had an ocean .5km deep. The acceleration due to gravity on Mars is 3.71m/s^2.

(a) What would be the gauge pressure at the bottom of such an ocean, assuming it was freshwater?
(b) To what depth would you need to go in the Earth's ocean to experience the same gauge pressure?

Homework Equations

p=p_{0}+ρgh

gauge pressure = absolute pressure - atmospheric pressure

The Attempt at a Solution

What I'm thinking is that I can calculate the absolute pressure by using p=ρgh, so p=(1000)(3.71)(500)=1855000 Pa.

But I think I'm getting a little confused about my absolute and atmospheric pressures. Absolute should be the pressure at the bottom of 500m, I believe. What should I be using for my atmospheric pressure?

 

[rock.freak667]

Homework Equations

p=p_{0}+ρgh

gauge pressure = absolute pressure - atmospheric pressure

if P = absolute pressure, then P-P₀ would be the gauge pressure wouldn't it? (or just 'ρgh'.

The Attempt at a Solution

What I'm thinking is that I can calculate the absolute pressure by using p=ρgh, so p=(1000)(3.71)(500)=1855000 Pa.

But I think I'm getting a little confused about my absolute and atmospheric pressures. Absolute should be the pressure at the bottom of 500m, I believe. What should I be using for my atmospheric pressure?

 

[forestmine]

Ok, so you're saying I've already calculated the gauge pressure by neglecting p_{0} in my earlier equation?

If that's the case, how would I go about solving part b?

 

[BetoG93]

I am pretty sure that gauge pressure doesn't involve atmospheric pressure. Since Mars' atmosphere is quite different to Earth's, atmospheric pressure there has a value different than 101 kPa.
Anyway, the only equation you need is P = pgh (p is density; my cell phone doesn't have rho).

 

[rock.freak667]

Ok, so you're saying I've already calculated the gauge pressure by neglecting p_{0} in my earlier equation?

If that's the case, how would I go about solving part b?

Right, well you'd use the gauge pressure you got earlier and using the same formula just with Earth values i.e. the value of 'g' on Earth.