What if you tried to fly a kite on Mars?

Question

I wonder what kite flying might be like on Mars, in one per cent the atmospheric pressure of Earth, about two per cent the Earth's atmospheric density and 38% the Earth's surface gravity. Are there local and/or temporal conditions making kite flying possible on Mars? What would it take for an astronaut to fly a kite there? Due to Mars' thin atmosphere, I suppose astronauts would wait for a sand storm to come which would lift the kite.

Answer 1

This is an interesting question that got me thinking — and calculating, during a much-needed break from designing a rotating space station!

In short, you probably could fly a kite at Mars! Probably not the old stick-and-paper kites of 50-100 years ago (low L/D and high $\beta$; see below), but maybe a parafoil-style one.

If you're math-challenged, skip down below Eq. 12 for the summary.

It boils down to two parameters important in aerodynamics: Lift-to-drag ratio (L/D), and ballistic coefficient ($\beta$). L/D is fairly simple: for a body within a fluid (gas) flow, L/D is the ratio of the lift force the body generates divided by the drag force it generates. $\beta$ is the body's mass divided by the effective area the body presents to the flow, given by $\beta=\frac{M}{C_{D}A}$, where M is the body's mass, $C_D$ is the coefficient of drag, a measure of how "draggy" a body's shape is relative to its size, and A is the area of the body's shape projected onto a plane perpendicular to the fluid flow. $\beta$ is essentially a measure of how mass-efficient the body's structure is: how much mass does it take to make the lifting body with that effective area?

The general formula for drag force $F_D$ is

$$F_{D}=\frac{C_D}{2}A\rho V^2\tag{1}$$

and for the lift force it is just L/D times the drag force:

$$F_{L}=\frac{L}{D}\frac{C_D}{2}A\rho V^2\tag{2}$$

A kite is moored by a tether (the kite string) that overcomes the drag force, preventing the kite from flying away when, in the absence of the retarding force of the string, it comes to the local fluid velocity — as happens when the kite-flyer accidentally drops the string spool! The tension in the tether has the anti-flow-ward component to cancel the drag force, but also has a downward component because it is not parallel to the ground: it rises at some angle α from the horizontal. For this simplified derivation I'll ignore sag of the tether due to gravity and drag force on the tether, and just deal with α as the tether angle where it connects to the kite. The anti-flow-ward component, $F_{AF}$, is given by

$$F_{AF}= F_{T}\cos\alpha\tag{3}$$

where $F_T$ is the tension force in the tether. $F_{AF}$ is equal and opposite to the drag force, so

$$F_{T}\cos\alpha+\frac{C_D}{2}A\rho V^2 = 0\tag{4}$$

or

$$F_{T}\cos\alpha=-\frac{C_D}{2}A\rho V^2\tag{5}$$

so the magnitude of the tether tension is

$$F_{T} =-\frac{C_D}{2\cos\alpha}A\rho V^{2}\tag{6}$$

The magnitude of the downward component $F_P$ of the tether tension is

$$F_{P}= F_{T}\sin\alpha\tag{7}$$

acting in the same direction as gravity. This could be made zero by having α = 0, but then the kite is on the ground, a distinctly suboptimal situation for a kite! The downward force of gravity is given by

$$F_{G}= Mg\tag{8}$$

where M is the mass of the body and g is the local gravitational acceleration. The sum of those two downward forces must be counteracted by the upward force of lift, so, in terms of magnitude,

$$F_{G}+ F_{P}= F_{L}\tag{9}$$

Substituting from Eqs. 2, 6, 7, & 8,

$$Mg+\frac{C_D}{2\cos\alpha}A\rho V^{2}\sin\alpha=\frac{L}{D}\frac{C_D}{2}A\rho V^2\tag{10}$$

Combining trig terms,

$$Mg+\frac{C_D}{2}A\rho V^{2}\tan\alpha=\frac{L}{D}\frac{C_D}{2}A\rho V^2\tag{11}$$

This can be solved for V, which tells you how fast the wind must be blowing to maintain the kite's altitude:

$$V=\sqrt{2\frac{g}{\rho}\frac{M/(C_{D}A)}{(L/D)-\tan\alpha}}\tag{12}$$

If you assume for now that α is small enough that $F_P$ is much smaller than gravity and tan $\alpha$ is insignificant, this boils down to the ratio of two parameters associated with Mars, the gravitational acceleration and the air density, and the ratio of the two parameters I mentioned earlier, L/D and $\beta$. Since L/D is in the denominator, the higher the L/D (i.e., the more efficient is the airfoil), the slower is the wind speed necessary. $\beta$ is in the numerator, so the more mass it takes to implement the airfoil, the higher the necessary wind speed.

This breaks down as V approaches the speed of sound, where for most shapes $C_D$ exhibits very non-linear behavior, turbulence from local supersonic flow causes buffeting, and you can lose a significant part of the lift that's so critical to staying aloft. But specifying subsonic flow, small $\alpha$, plugging in parameters typical of Mars: $g = 3.711 \frac{m}{s^2}$, $\rho = ~0.015\frac{kg}{m^3}$; and plugging in 2.7 for L/D (appropriate for a small parafoil-type kite) and 0.25 $\frac{kg}{m^2}$ for $\beta$ (maybe a bit sporty!), you get a wind speed of ~6.8 m/s (~15 MPH) to just barely break even, with no tether weight considered, and $\alpha$ close to zero. To get it to fly at a higher $\alpha$ and at an altitude where the tether weight is no longer insignificant, you need either a higher L/D, a lower $\beta$, or both.

Wind speeds of 7 m/s are not uncommon at Mars, and sometimes are significantly higher than that, so at times (catch as can — try predicting the winds at Mars!) this might actually work.