A Windy Day on Mars

Given that the maximum wind speed measured on Mars is approximately 60 miles per hour (mph), find an approximation to the corresponding wind speed on Earth, at standard sea level pressure, which would give rise to the same pressure on a surface perpendicular to the wind direction.

A typical density of air on Mars is 0.0163 times the density of air at standard temperature and pressure at sea level on Earth.

7.7 mph 23.2 mph 51.6 mph 137 mph

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2 solutions

Rohit Gupta
Jan 10, 2017

Imagine a tube of fluid (liquid or air) of density ρ \rho is moving at a speed of v v towards a wall. The cross section of the tube is A A as shown below.

The volume of fluid that hits the wall in time dt will be, d V = A v d t dV=Av \, dt
The momentum of the fluid that hits the wall in time dt is ρ ( A v d t ) v \rho(Av\, dt)v
If the fluid stops and then falls freely just after hitting the wall, then the change in momentum in time dt will be d p = ρ ( A v d t ) v dp=\rho(Av\, dt)v
(Note: For gasses, the collision is nearly perfectly elastic, resulting in twice the impulse due to Newton's third law. d p = 2 ρ ( A v d t ) v dp=2\rho(Av\, dt)v )
Force on the wall F = d p d t = ρ A v 2 F=\dfrac{dp}{dt} = \rho Av^2
Pressure generated on the wall will be
P = F A = ρ v 2 P=\frac{F}{A} = \rho v^2

Density of air is different on the Mars and the Earth, therefore, for the same pressure on the wall
ρ Earth v Earth 2 = ρ Mars v Mars 2 \rho_\text{Earth}v^2_\text{Earth}=\rho_\text{Mars}v^2_\text{Mars}
Plugging in the values, we get,
v E a r t h = 7.7 mph . v_{Earth}=\boxed{7.7 \text{ mph}}.

Your expression for pressure differs from mine because you assumed a totally inelastic collision of the fluid and the wall. For gasses, the collision is nearly perfectly elastic, resulting in twice the impulse due to Newton's third law. The missing factor of 2 doesn't affect your answer because it cancels out after equating the pressures on Earth and Mars. Whichever model is used, the important feature is that the pressure is proportional to the square of the wind speed.

Dale Gray - 4 years, 5 months ago

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Agreed, but yes I mentioned it and it will not really change the answer to the question. Still let me mention it as a note for clarity.

Rohit Gupta - 4 years, 5 months ago

Perhaps you would like to change the v on the left side of your first equation to capital v considering it represents volume and you are using lower case v as speed.

Dale Gray - 4 years, 5 months ago

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Point noted. Thanks for pointing out.

Rohit Gupta - 4 years, 5 months ago
Dale Gray
Jan 8, 2017

Great question, however you may want to give density values so that people don't have to search for them.

Shaun Leong - 4 years, 5 months ago

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