Question

Scientists have found evidence that Mars may once have had an ocean 0.500 km deep. The acceleration due to gravity on Mars is 3.71 m/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?

Short answer

(a) 1855000 Pa; (b) 189.19 m

Step-by-step solution

Gauge Pressure Formula

The formula to calculate gauge pressure (\(P_g\)) at a depth is given by \(P_g = \rho g h\), where \(\rho\) is the density of the fluid, \(g\) is the acceleration due to gravity, and \(h\) is the depth. We know that for freshwater, \(\rho = 1000 \, \text{kg/m}^3\).

Calculate Gauge Pressure on Mars

Substitute the given values for Mars into the gauge pressure formula: \(\rho = 1000 \, \text{kg/m}^3\), \(g = 3.71 \, \text{m/s}^2\), \(h = 0.500 \, \text{km} = 500 \, \text{m}\). Therefore, \(P_g = 1000 \, \text{kg/m}^3 \times 3.71 \, \text{m/s}^2 \times 500 \, \text{m}\). Calculate to find \(P_g = 1855000 \, \text{Pa}\).

Finding Equivalent Depth on Earth

To find the depth on Earth (let's denote it as \(h_{earth}\)) where the gauge pressure equals that found on Mars, use the same gauge pressure formula: \(P_g = \rho g h\). For Earth, \(\rho = 1000 \, \text{kg/m}^3\), \(g = 9.81 \, \text{m/s}^2\), and \(P_g = 1855000 \, \text{Pa}\). Thus, \(1855000 \, \text{Pa} = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times h_{earth}\). Solve for \(h_{earth}\): \(h_{earth} = \frac{1855000}{1000 \times 9.81} \, \text{m}\).

Calculate Equivalent Depth on Earth

Calculate \(h_{earth} = \frac{1855000}{9810} \, \text{m} \approx 189.19 \, \text{m}\). So, the equivalent depth on Earth where the gauge pressure is the same as at the bottom of a 0.5 km Martian ocean is approximately 189.19 meters.

Key concepts

Gravity on Mars

Mars, our neighboring planet, has a distinct gravitational pull compared to Earth. The acceleration due to gravity on Mars is about 3.71 meters per second squared (m/s\(^2\)). This means gravity on Mars is weaker than on Earth, where the acceleration due to gravity is 9.81 m/s\(^2\). Gravity plays a crucial role in various physical phenomena, including the pressure exerted by fluids. If there had been an ocean on Mars, as some evidence suggests, this weaker gravitational force would affect how the water exerts pressure in that ocean.
Understanding Martian gravity helps scientists predict how objects behave on Mars and is vital for mission planning and scientific calculations, such as pressure in potential ancient Martian oceans.
In summary:

• Mars' gravitational acceleration is 3.71 m/s\(^2\), weaker than Earth's 9.81 m/s\(^2\).
• This weaker gravity influences the pressure exerted by fluids on the Martian surface.

Freshwater Density

Freshwater density is an important factor when calculating pressure. The standard density of freshwater is 1000 kilograms per cubic meter (kg/m\(^3\)). This means that in calculations involving water, such as finding gauge pressure, this density is often used as a default value.
When calculating the gauge pressure at the bottom of a body of water, knowing the density is crucial as it affects the result significantly. Denser fluids will exert more pressure, so always using the correct density value is important for accurate results.
For students working with pressure equations:

• Standard freshwater density is usually taken as 1000 kg/m\(^3\).
• Density impacts how fluids exert pressure at a given depth.

Pressure Calculations

Gauge pressure is a measure of pressure at a certain depth in a fluid. To find it, the formula used is \(P_g = \rho g h\), where \(P_g\) is the gauge pressure, \(\rho\) is the fluid density, \(g\) is the acceleration due to gravity, and \(h\) is the depth of the fluid.
This formula helps calculate the pressure within a fluid at any given depth, and is widely used in fluid mechanics. By substituting known values into the formula, students can solve for the gauge pressure in different scenarios, including those on planets with different gravitational forces, like Mars.
Some key points to remember:

• The formula \(P_g = \rho g h\) is used to calculate gauge pressure.
• Substitute known values to find the pressure exerted by a fluid.
• It's applicable even on planets with different gravitational accelerations.

Acceleration Due to Gravity

The acceleration due to gravity, often just called gravity, is the force that attracts a body towards the center of the Earth, or any other celestial body. On Earth, this force is approximately 9.81 meters per second squared (m/s\(^2\)).
This value is critical in physics as it affects everything from how objects fall to how pressure is calculated in fluids. On other planets, the acceleration due to gravity could be different. For example, on Mars, gravity is weaker at 3.71 m/s\(^2\).
This discrepancy must be considered in pressure calculations or any situation involving gravitational forces, as it alters how forces, including fluid pressure, are calculated:

• Earth's gravity is 9.81 m/s\(^2\).
• Mars has weaker gravity at 3.71 m/s\(^2\).
• Gravity affects movement, fall rates, and pressure calculations.