Terminal velocity on the moon? Is it possible and how long would it take?

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Solution 1

The aerodynamic drag at high speeds is an inertial effect. Air has a mass, so pushing it out of your way takes energy and the faster you push it the more energy it takes. That's why drag increases with speed.

I won't go through the derivation, but the equation that relates the aerodynamic drag to velocity is:

$$ F = \tfrac{1}{2}C_d\,A\,\rho\,v^2 \tag{1} $$

where $\rho$ is the density of the atmosphere, $A$ is the cross sectional area and $C_d$ is a fudge factor that factors in the effect of turbulence.

The parameter $A$ is the same on Earth and the Moon because it's just down to the shape of a human. The parameter $C_d$ won't be the same because the Moon's atmosphere is too thin to develop turbulence on a short length scale, but $C_d$ is generally of order one so let's just ignore it and bear in mind that our answer is going to be approximate.

If we start on Earth then terminal velocity is when the aerodynamic force given by equation (1) equals the gravitational force $mg$, so we get:

$$ mg = \tfrac{1}{2}A\,\rho_e\,v_e^2 \tag{2} $$

where the subscript $e$ refers to the earth and remember that we are ignoring the parameter $C_d$. On the Moon the surface acceleration is about $0.165g$ so for the Moon we get:

$$ 0.165mg = \tfrac{1}{2}A\,\rho_m\,v_m^2 \tag{3} $$

We use equation (2) to substitute for $mg$ in equation (3) to get:

$$ 0.165 \left(\tfrac{1}{2}A\,\rho_e\,v_e^2\right) = \tfrac{1}{2}A\,\rho_m\,v_m^2 $$

And we can rearrange this to give an equation for the terminal velocity on the Moon, $v_m$, which is what we're after:

$$ v_m^2 = 0.165 \left(\frac{\rho_e}{\rho_m}\,v_e^2\right) \tag{4} $$

Now the great thing about this equation is we known the terminal velocity for a human on Earth is around $v_e = 55$ m/sec in the belly flop position, and the density of air on Earth is around $\rho_e = 1.25$ kg/m$^3$. So if we knew the density of the atmosphere on the Moon, $\rho_m$, we could just plug the numbers into equation (4) to give $v_m$.

But this is the problem. I can't find definitive values for the density of the Lunar atmosphere. From the data on the NASA Moon fact sheet I estimate the ratio $\rho_m/\rho_e$ is in the range $10^{-14}$ to $10^{-15}$. If we take the high end of this range we get:

$$ v_m = \sqrt{0.165 \times 10^{14}\times (55)^2} \approx 2 \times 10^8 \text{m/sec} $$

which is a goodly fraction of the speed of light! You're obviously never going to reach that speed by freefalling from rest on the Moon.

Solution 2

Terminal velocity on Earth is achieved when the acceleration due to Earth's gravitational field is balanced out by air drag (which increases as a function of velocity). Due to the fact that the Moon essentially has no atmosphere (you have a few gas molecules but it is nearly a vacuum), there wouldn't be a terminal velocity on the Moon because there's essentially no force counteracting the pull of its gravity. You'd just keep accelerating until you smacked into it.

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Updated on March 24, 2020

Comments

  • World Walker
    World Walker over 3 years

    Is there a way to find this out? If so please tell me the formula or just tell me how much time would it take for a 100 lb human to reach terminal velocity

    • Admin
      Admin over 6 years
      what causes terminal velocity to occur on Earth?
  • Admin
    Admin over 6 years
    +1 for converting the aural apparatus of a porcine related beast into a small container composed from the liquid output of an oriental worm. Lovely answer.