# If I toss a coin, vertically, on the surface of Mars, will it land back in my hand?

When I toss a coin in Mars, is the planets atmosphere rare enough that I'd rotate with the planet (at its angular velocity), but not the coin?

• @Sathyam How small is small? I would assume that for the realm of this problem, the differences will be negligible. – HDE 226868 Dec 31 '15 at 18:58
• @Sathyam All I'm saying is that you're not going to see any major effect. We can extrapolate from here that the effect will be extraordinarily small. A coin will travel much higher than a person can jump. But it won't go high enough to make the journey long enough for the forces at work to make it go any significant distance parallel to the ground. – HDE 226868 Dec 31 '15 at 19:07
• – HDE 226868 Dec 31 '15 at 19:11
• @Sathyam The coin absolutely is in an elliptical orbit. It's approximately parabolic over the small distances involved, assuming the toss is not too high. But the height of the toss plays an essential role in the analysis. I'll wait for the OP to clarify his intent before commenting further. – garyp Dec 31 '15 at 19:18
• @garyp Agreed that parabola is an approximation. But the orbit is elliptical only in the absence of air drag. – Sathyam Dec 31 '15 at 19:25

It depends on where on Mars you toss the coin, and how high you toss it.

In a rotating frame of reference, an object in motion appears to be affected by a pair of fictitious forces - the centrifugal force, and the Coriolis force. Their magnitude is given by

$$\mathbf{\vec{F_{centrifugal}}}=m\mathbf{\vec\omega\times(\vec\omega\times\vec{r})}\\ \mathbf{\vec{F_{Coriolis}}}=-2m\mathbf{\vec\Omega}\times\mathbf{\vec{v}}$$

The question is - when are these forces sufficient to move the coin "away from your hand" - in other words, for what initial velocity $v$ is the total displacement of the coin greater than 10 cm (as a rough estimate of what "back in your hand" might look like; obviously you can change the numbers).

The centrifugal force is only observed when the particle is rotating at the velocity of the frame of reference - once the particle is in free fall, it no longer moves along with the rotating frame of reference and the centrifugal force "disappears". For an object moving perpendicular to the surface of the earth, the Coriolis force is strongest at the equator, becoming zero at the pole; it is a function of the velocity of the coin. We will calculate the expression as a function of latitude - recognizing that it will be a maximum at the equator.

As a simplifying assumption, we assume the change in height is sufficiently small that we ignore changes in the force of gravity; we also ignore all atmospheric drag (in particular, the wind; if the opening scene of "The Martian" were to be believed, it can get pretty windy on the Red Planet.) Finally we will assume that any horizontal velocity will be small - we ignore it for calculating the Coriolis force, but integrate it to obtain the displacement.

The vertical velocity is given by

$$v = v_0 - g\cdot t$$

and the total time taken is $t_t=\frac{2v_0}{g}$. At any moment, the Coriolis acceleration is

$$a_C=2\mathbf{\Omega}~v\cos\theta$$

Integrating once, we get

$$v_h = \int a\cdot dt \\ = 2\mathbf{\Omega}\cos\theta\int_0^t(v_0-gt)dt\\ = 2\mathbf{\Omega}\cos\theta\left(v_0 t-\frac12 gt^2\right)$$

And for the displacement

$$x_h = \int v_h dt \\ = 2\mathbf{\Omega}\cos\theta\int_0^t \left(v_0 t-\frac12 gt^2\right)dt\\ = 2\mathbf{\Omega}\cos\theta \left(\frac12 v_0 t^2-\frac16 gt^3\right)$$

Substituting for $t = \frac{2v_0}{g}$ we get

$$x_h = 2\mathbf{\Omega}\cos\theta v_0^3\left(\frac{4}{g^2} - \frac{4}{3 g^2}\right)\\ = \frac{16\mathbf{\Omega}\cos\theta v_0^3}{3g^2}$$

The sidereal day of Mars is 24 hours, 37 minutes and 22 seconds - so $\Omega = 7.088\cdot 10^{-5}/s$ and the acceleration of gravity $g = 3.71 m/s^2$. Plugging these values into the above equation, we find $x_h = 2.75\cdot 10^{-5}v_0^3 m$, where velocity is in m/s. From this it follows that you would have to toss the coin with an initial velocity of about 15 m/s for the Coriolis effect to be sufficient to deflect the coin by 10 cm before it comes back down.

On Earth, such a toss would result in a coin that flies for about 3 seconds, reaching a height of about 11 m. It is conceivable that someone could toss a coin that high - but I've never seen it.

AFTERTHOUGHT

Your definition of "vertical" needs to be carefully thought through. There is a North-South component of the centrifugal "force" that is strongest at 45° latitude, and that will cause a mass on a string to hang in a direction that is not-quite-vertical. If you launch your coin in that direction, you will not observe a significant North-South deflection during flight, but if you were to toss the coin "vertically" (in a straight line away from the center of Mars), there will in fact be a small deviation. The relative magnitude of the centrifugal force and gravity can be computed from

\begin{align}a_c &= \mathbf{\Omega^2}R\sin\theta\cos\theta \\ &= \frac12 \mathbf{\Omega^2}R\\ &= 8.5~\rm{mm/s^2}\end{align}

If you toss the coin at 15 m/s, it will be in the air for approximately 8 seconds. In that time, the above acceleration will give rise to a displacement of about 27 cm. This shows that your definition of "vertical" really does matter (depending on the latitude - it doesn't matter at the poles or the equator, but it is significant at the intermediate latitudes, reaching a maximum at 45° latitude).

• Beautiful. Thoroughly and fully answered, making no assumptions of the asker's intent. – Dewi Morgan Dec 31 '15 at 22:38
• @DewiMorgan thank you. Normally when a question has three answers I let it go - but I felt there was "something missing". – Floris Dec 31 '15 at 22:49
• @casey it is the vector product of velocity and rotation vector that matters for Coriolis effect. For motion perpendicular to the surface (a tossed coin) that makes the equator the strongest and the poles the weakest. The same is not true for wind that flows along the surface. Your point is valid for wind but not coins tossed. – Floris Jan 1 '16 at 5:01
• Nicely answered indeed. I was wondering where the coin would hit the ground if we let it from a certain high ,in order to make a free fall. – user98038 Jan 1 '16 at 8:43
• @aK1974 thanks. The problem you ask about is described in the "further reading" link I gave. It travels half as far as when tossed as it is in the air only half as long (assuming it is dropped from the max height reached in the toss) – Floris Jan 1 '16 at 14:48

The coin will come back to your hand just like it would on the earth. The effect of atmosphere is negligible comparing to the coin's inertia, so the horizontal position of the coin relative to your hand will hardly be affected. The rareness of the atmosphere will only affect the vertical motion of the coin, like how quickly the coin will fall into your hand.

• For run-of-the-mill coin tosses, you are correct, but only approximately so. The approximation, however, is very good. For very high coin tosses you have to show that atmosphere is negligible, and you have to show that it is valid to neglect the difference in relative horizontal motion of the coin and the earth's surface. – garyp Dec 31 '15 at 19:28

Yes, for the simple reason that you're not tossing the coin very high (presumably, anyway). You seem to think that on Earth, atmospheric drag is what keeps the coin "glued" to the tossing frame of reference, but that isn't really a factor at all.

Say that you're on Earth, at sea level, on the equator, and you toss the coin 3 meters straight up. Neglecting drag, the coin will be in the air for 1.56 seconds. The earth is rotating under your feet at 463 m/s, and has a radius of 6.37 * 10^6 m. The coin is gaining an altitude of 3 m, which is 4.71 * 10^-7 earth radii, so the rotational speed at that height will be different by an equal proportion, which works out to 0.00022 m/s. Getting an upper bound by assuming the coin spends the whole time at the maximum height because I'm lazy, and we end up with a deflection of 0.34 mm, which is less than the thickness of the coin, let alone its diameter. Anywhere away from the equator or above sea level, and the number would come out lower still.

Doing the same experiment on Mars, we'll suppose that you can give the coin the same initial velocity, and that you're on the equator at mean elevation. Mars's surface gravity is lower (3.71 m/s^2), so the coin will reach an impressive 7.92 m in height and stay in the air for 4.14 seconds. Mars is rotating under your feet at 241 m/s (less than Earth because it has a smaller circumference but a similar day length) and has a radius of 3.39 * 10^6 m. The coin then gains 2.34 * 10^-6 mars radii, and the rotational speed at that height is different by 0.00056 m/s. Making the same (over)estimate as before, we get 4.14 s * 0.00056 m/s = 2.33 mm. About one coin thickness, but not much more. Certainly not enough to miss your hand on the way down.

Basically, the heights you're dealing with when tossing a coin are just too small, compared to the size of a planet, to make much difference, atmosphere or not. Try lobbing a cannonball 1km up instead, and you'd be more likely to notice an effect. I haven't worked out the math, but I still don't think that the horizontal component of atmospheric drag would contribute much at all; the atmosphere would be more likely to make a difference by reducing the maximum height reached.

The coin comes down for sure unless you tossed it with an escape velocity and escape velocity depends on mass of planet, mass of coin etc. Even if you are well inside the escape velocity limit, it may not reach your hand. Your observation is correct that the drag force plays a role in its tangential velocity, but for all small distance tosses the coin will reach your hand regardless of the atmospheric rarity compared to earth.

I think yes,like as if you were on earth. The reason is that the coin has the same velocity as you and the surface of planet Mars, it doesn't have to do with atmosphere.

• The coin have the same velocity and angular acceleration at the instant of detachment from your hand. After deployment there is no longer a centripetal acceleration . There is a radial acceleration towards the center of the planet combined with a tangential velocity and the atmosphere does play a role in the form of a drag force. – Sathyam Dec 31 '15 at 19:16
• You missed the point of the question. It's not "will it come back down", it's "will it come back down in my hand" — and not, e.g. ten feet to the left. – hobbs Dec 31 '15 at 20:31

Yes you are moving with the surface but you are not accelerating. You are moving at a constant speed. If you toss in what you think is vertical (call it Y direction) the coin will have the same rotational velocity (call it X direction) as you. Even the atmosphere is moving with you so for a short distance there is no wind resistance in the X direction. The X velocity of the coin will remain constant and will be exactly your X velocity.

The combination of your two questions leads me to believe that what you are really asking is, does the density (or lack of it) of the atmosphere have an effect on the horizontal position of an object that initially is given only vertical momentum? If this interpretation of your question is correct, then I want to start by telling you that you have a reverse conception of the effect density has. The drag force of an atmosphere is proportional to its density, therefore as its density goes to zero (no atmosphere), the drag force goes to zero. So, the "rarer" the atmosphere, the less effect it has on the motion of the coin. Since we are only interested in the coin returning to the same spot (little or no horizontal displacement), differences in the vertical direction caused by the drag and different Martian gravity, are of no consequence. Therefore, since there are little or no horizontal forces acting on the coin, it will return to the same spot it was tossed from (your hand).

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