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This question already has an answer here:

Spacecrafts and rovers tend to have dust settled on their panel which leads to reduced power generation.

If wind turbines could power the instruments, then the seasonality of solar power generation can be avoided.

Given the reduced atmospheric pressure and density on Mars, how feasible is wind power generation? Has it been attempted earlier or are there any future missions to test it?

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marked as duplicate by karthikeyan, peterh, Fred, Jan Doggen, Organic Marble Dec 3 '18 at 15:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Partial answer.

Windy day on Mars, remember that Mars' gravity is less than Earth's so the wind isn't as strong as it might appear.

enter image description here

Archived from the Telltale Procjet at Mars Simulation Laboratory: https://web.archive.org/web/20120220080017/http://www.marslab.dk/TelltaleProject.html

Borrowed from (currently unanswered) Was the telltale on the Mars Phoenix Lander used for meteorology? Why not a hot wire anemometer instead?

Development of the Telltale

The air pressure on Mars is less that 1% of the Earths, and this means that the instrument had to be extremely sensitive. The forces due to the wind are of the order of few millionths part of a Newton (The instrument weighs 20 grams, meaning that the gravitational pull is 0.2 Newton). This was accomplished by making the active part of the instrument as lightweight as possible (about 10 thousands of a gram). After intensive testing to document that the experiment would survive the launch and landing vibrations, it was calibrated in the Wind tunnel at the Mars Simulation Laboratory


The power per unit area (Watts/m^2) in wind is given by the kinetic energy density times the velocity:

$$P = \frac{1}{2} \rho v^2 v = \frac{1}{2} \rho v^3;$$

it's linearly proportional to density and cubic with velocity.

There is going to be an extraction factor of order 0.59 or less. See Betz Law limit of 16/27 in:

So at about 0.6% pressure of Earth's, or about 1% of the density, you'd need wind 5x faster to get the same power, assuming the same extraction efficiency.

The problem is that losses due to friction (mechanical & aerodynamic) and possibly more strength to deal with longer blades and/or higher tension when rotating, and the likely very low probability that the wind will be 5x faster than on Earth means this is not going to be easy.

However, there may be other systems to extract energy that are better for Mars than those optimized for Earth.

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    $\begingroup$ Great answer. InSight has wind sensors, so someday we will be able to put hard numbers into the formulas you provide. $\endgroup$ – DrSheldon Dec 3 '18 at 16:48
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    $\begingroup$ @DrSheldon Thanks! This is just fyi: I posted this and then noticed the question would soon be voted to be closed as duplicate, so I also posted this answer on the earlier question as well. If this question is closed, I'll delete this instance (no need for two copies) or if it's edited and changed I'll modify this copy. $\endgroup$ – uhoh Dec 4 '18 at 1:11

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