# Terminal velocity on the moon? Is it possible and how long would it take? [closed]

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

## closed as unclear what you're asking by Yashas, ZeroTheHero, Jon Custer, user259412, AccidentalFourierTransformMar 29 '17 at 11:13

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• what causes terminal velocity to occur on Earth? – user146020 Mar 22 '17 at 17:33

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.

• +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. – user146020 Mar 22 '17 at 22:43

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.

John Rennie has computed the terminal velocity due to the Moon's very thin atmosphere, but realistically there's another constraint on how fast the Moon's gravity can make an initially stationary object. In the moon's gravitational field, the gravitational potential energy per unit mass is reduced during a fall from $-\frac{GM}{r}$ to $-\frac{GM}{R}$, where the moon has mass $M$ and radius $R$, and the falling mass starts a distance $r$ from the moon's centre. The speed at which the mass hits the surface is then at most $\sqrt{\frac{2GM}{R}}=2376\text{m/s}$ (computed here), and even that can only be reached if (a) the object starts at infinity and (b) other masses don't spoil the calculation. However, we can get almost as high a final speed upon hitting the ground provided $r>>R$, which is realisable even if the test mass starts far closer to the Moon's centre than to the Earth's.

Think about it this way - what causes an object to have terminal velocity? What reason is there for gravity to suddenly "stop" accelerating an object? Terminal velocity on Earth is roughly 120 mph, yet we know objects in space, such as planets, move far faster than that due to gravitational attraction towards each other. What's the difference between Earth and space?

An atmosphere that creates air resistance, or drag. When an object falls in Earth, it is affected by air resistance. Eventually, this air resistance "catches up" with the velocity, and they match each other. But at his point, gravitational acceleration doesn't suddenly just stop. Instead, the velocity and air resistance both increase at the same rate, which is why it looks like the object does not continue to gain speed once reaching terminal velocity. There is no net acceleration, since gravity and air resistance both exert equal amounts of force that cancel out each other.

The moon does not have an atmosphere, meaning there is no air resistance. Thus, there is no terminal velocity - an object falling on the moon will continue to gain speed until hitting the surface. This is the same case with objects in space or in a vacuum... no air resistance or drag force = no terminal velocity.

If you want to read more on air resistance and terminal velocity, here is a link to the a Physics Classroom lesson on it:

http://www.physicsclassroom.com/class/newtlaws/Lesson-3/Free-Fall-and-Air-Resistance