# Throwing an object around the Earth. Is it possible?

Some of you might be familiar with the cartoon called Astérix. In one of the episodes, Obelix (the really strong fat guy in the stylish striped pants) enters a spear throwing competition. He beats the other contestant by literally throwing his spear around the world. Is this possible?

• If an object is thrown at an angle (not vertically upwards) with a velocity slightly lesser than the escape velocity, it will orbit around earth. Escape_velocity#Orbit – Sreekumar R Jan 26 '15 at 21:47
• Yes, it's possible. And these objects are called satellites. – user71669 Jan 26 '15 at 21:49
• @SreekumarR: that totally ignores air resistance, which is wrong. – smci Jan 27 '15 at 3:54
• Also note that this is a modification of a classic Newtonian argument, which is what you're questioned abaout, the "Newton orbital cannon" astronautix.com/lvs/newannon.htm – vicenteherrera Jan 27 '15 at 11:59
• This feels like the sort of question that you'd find on what-if.xkcd.com. – Ajedi32 Jan 27 '15 at 14:12

TL,DR: it cannot be done because there is no spear strong enough to withstand the acceleration.

There are two orbits that would give you this result. The first is one where you throw the spear "slightly up" - it would rise out of the atmosphere as drag slows it down, and slowly descend at ever-steeper angle upon re-entry. With the right launch parameters (which are a strong function of the mass/drag ratio of the spear) such an orbit can be found.

The second orbit is "almost straight up and down". Taking advantage of the Coriolis effect, if you throw a spear (any object) vertically up (at a slight angle, removing the horizontal component of velocity due to rotation which is about 460 m/s at the equator), then the earth will rotate underneath. The object that takes 24 hours to return will fall back to the same place on earth after a complete lap. It's actually relatively easy to estimate the parameters of this "orbit" (mostly because the drag phase will be quite short compared to the total orbit). The force of gravity is

$$F_g = mg\left(\frac{R}{h+R}\right)^2$$

where $h$ is the height, and $R$ is the radius of the earth. The equation of motion is then

$$\frac{d^2h}{dt^2} = g \left(\frac{R}{R+h}\right)^2$$

That's a really tricky one to integrate, but numerical integration tells me that if you use an initial velocity of around 14,600 m/s, the object will rise to a height of $1.9\cdot 10^8 m$ and return in almost exactly 24 hours. Clearly you can't do this without taking account of the moon (since it would reach almost halfway to the moon, you can't ignore its gravity) but we are talking about cartoon physics. And the answer is quite different than if you ignored the way that gravity drops off with height (assuming constant pull, you would need an initial velocity of 86400 * 9.8 / 2 ~ 423 km/s vs 14.6 km/s).

Let's estimate the force of air friction next. I will assume a tungsten spear, 3 cm diameter. The whole spear is 2 m long - worthy of a great warrior like Obélix. The mass of such a mighty weapon is about 25 kg, and the drag very roughly given by

$$F_d = \frac12 \rho v^2 A C_D = \mathrm{0.5 * 1.2 * 14600^2 * \pi * 0.03^2 * 0.2 \approx 2\cdot 10^4 N}$$

Note that I am using a very approximate supersonic drag coefficient of 0.2 - reference. With that much drag, the spear will lose 5% of its momentum in the first second of flight - but that's about all, as it will quickly be in outer space ("thin air"). We might want to increase the initial velocity by about 10% to be on the safe side, taking it to 16 km/s.

So can Obélix throw a 25 kg spear at that velocity? Assuming he has a 1 m "arm", and constant acceleration, getting to 16 km/s in 1 m requires an acceleration

$$a = \frac{v^2}{2d} = \mathrm{128\cdot10^6 m/s^2}$$

This requires a force $F=m\cdot a \approx 3 GN$. A guy that casually can toss a menhir into a bunch of Roman soldiers is obviously very strong - but giga Newtons is a lot of force. Tossing a 1000 kg stone over 100 m requires a launch velocity of 31 m/s, and so a force that is orders of magnitude smaller (500 kN). But let's focus on the forces on the spear. And that's where it gets interesting... because we're at the limit of the mechanical properties of the spear.

The buckling strength of a rod that is free to move at one end is given (near enough) by

$$F_{max} = \frac{\pi^2EI}{(0.7\cdot \ell)^2}$$

For the magnificant spear, E=400 GPa, $I = \frac{\pi}{4}r^2 = 4\cdot 10^{-8} m^4$, and we take $\ell = 1m$ since we assume it is held in the middle.

$$F_{max}\approx 240 kN$$

That seems pretty strong - you could balance 4 elephants on top of the spear, and it will be fine. But here we are talking about a LOT more force - giganewtons, not kilonewtons.

And so - regardless of how strong Obélix is, there is no material in the world strong and dense enough to withstand being accelerated to the kinds of speed needed for this feat.

Note - I did this calculation for the "vertical toss". We could repeat for the "orbital toss", but in that case you have to account for a lot of atmospheric friction. Assuming that you could toss at an angle of 10 degrees to the horizontal to escape the atmosphere, the above calculation suggests that you might need to double the initial velocity to "get out". If you started at 8 km/s (the approximate "surface orbit" velocity calculated by Rob Jeffries) then doubling that to account for drag gets you to the same value. And even if you tried to get the spear up to 8 km/s "only", the above calculation still tells us "no can do". Of course the force needed drops 4x if the velocity is down 2x (simple energy argument) but that gets us to a force on the order of hundreds of MN - still about 4 orders of magnitude too large.

Oh - and there's one more thing: the tensile force at the back of the spear will be half the total force (1.5 GN), and with a surface area of $\mathrm{7 cm^2}$ that's a stress of 200 GPa. The ultimate strength of tungsten is about 1.5 GPa source, so it will rip.

I conclude that this feat is physically impossible - there is no spear that could withstand being thrown like this.

• strangehorizons.com/2002/20021021/manhole.shtml – Joshua Jan 27 '15 at 22:20
• Using some numerical code I wrote a while back (an implementation of en.wikipedia.org/wiki/Dormand%E2%80%93Prince_method), the simplified drag model you used, and a table of atmospheric density with altitude, I got a ~400 km/s launch speed needed if you were to try to do this horizontally from on top of a 1 km tall mountain at the equator. – cartographer Jan 28 '15 at 0:20
• @cartographer that's interesting. What launch angle did you assume for that? And what atmospheric model? Density change is close to exponential - there is almost nothing left at 25km. – Floris Jan 28 '15 at 0:31
• @Floris I just used a linear interpolation of this table: engineeringtoolbox.com/standard-atmosphere-d_604.html launch angle was completely horizontal (but effectively launched from on top of a 1km tall pole). – cartographer Jan 28 '15 at 0:38
• @cartographer - what altitude did your projectile reach? I imagine a slightly tilted launch angle would reduce the launch velocity by a lot. – Floris Jan 28 '15 at 1:36

Yes it is, if you can throw it hard enough. Not bothering with things like air resistance etc. (I think this is the least of the plausibility problems) you need to put the spear into a low-Earth orbit, such that the centripetal force is provided by gravitational acceleration.

$$\frac{v^2}{R} = \frac{GM}{R^2}$$

Using $R=6400\ km$ and $M= 6\times10^{24}\ kg$, gives $v= 7.9\ km/s$. Fast enough to get around the Earth in 85 minutes. You would have to launch the spear almost horizontally. NB: This assumes no assistance from the rotation of the Earth, which could subtract about 0.46 km/s from the initial velocity requirement in the most favourable launch direction at the equator.

Of course there is no way you can throw something ballistically at that speed without using some sort of launch system (and you can't really neglect air resistance).

EDIT: In response to a request(!) - the current world record for men's javelin is about 100m. Now there is a lot of aerodynamics in throwing a javelin, but let's neglect that, in which case the maximum range of a projectile is $v^2/g$. Thus the javelin would be launched at about 30 m/s. To launch at 7.9 km/s requires 69,000 times as much kinetic energy. Assuming fixed length arms then that factor is also the increased force factor over a world record javelin thrower.

EDIT2: Joshua's answer is correct. All the above can only be true for zero air resistance.

• I think you can improve your answer by adding how much force you need to do it, and then compare it to other tasks for reference (A professional boxer's hook is x amount of force, a spaceship's force on the earth on take off is y amount of force, throwing a spear around the earth is z amount of force) – Zaenille Jan 27 '15 at 0:26
• Oops. sorry. I'm not a physics expert/major. Haha. Thanks @RobJeffries. I just thought it would be interesting to know by using references. – Zaenille Jan 27 '15 at 0:42
• @MarkGabriel See edit. – ProfRob Jan 27 '15 at 0:43
• Yep, I saw. +1'd for the answer. :D – Zaenille Jan 27 '15 at 0:43
• I know this is not the point here, but I thought I'd make a small remark. Even ignoring air resistance, you'd have to launch it straight east or west while standing at the top of the highest mountain on the equator; otherwise, it's going to run into the ground. – Javier Jan 27 '15 at 3:17

No. No matter how hard you throw. Since orbits are ellipses, all trajectories meeting the criteria must pass through the ground at one point except for the surface-grazers.

Air resistance will not be negligible so there's no point in assuming it will be. The effect of air resistance on any shape other than a lifting body is a drag force straight backwards so the trajectory is now a descending spiral. Lifting shapes cannot be used here because of the necessity of keeping the drag profile small enough to avoid burning up. End of story.

• Since when are all orbits ellipses? – Cole Johnson Jan 27 '15 at 2:52
• So long as Newton's formulation of physics is sufficient (A circle is a special case of ellipse). – Joshua Jan 27 '15 at 2:53
• Not all ellipses pass through the ground or graze it. If the spear is thrown hard enough, the point at which it leaves will be the perigee, which will be above the ground (at the height of the hand). The orbit will decay while it is in significant atmosphere, so you must simply add enough extra velocity to account for this. So yes, it is possible. – Alex Jan 27 '15 at 5:07
• I like that this answer straightforwardly states that air resistance makes it impossible, but as Alex said, it is theoretically possible (without air resistance) to propel an object fast enough that it would make it around the world. The thing about ellipses always intersecting the Earth's surface isn't correct. – David Z Jan 27 '15 at 6:01
• @DavidZ I disagree that its impossible to throw an object and have it make it around the world even when taking into account air resistance. A stable orbit within the atmosphere would be impossible yeah (without some sort of energy input), but just one revolution is a different story. – cartographer Jan 27 '15 at 6:57

No, you cannot throw a spear hard enough to circle the earth.

The impact depth $D$ of a wood spear of length $L=2.5m$ (density $d_1$ below $1000\frac{\mathrm{kg}}{\mathrm{m}^3}$) in air (density $d_2$ about $1.2\frac{\mathrm{kg}}{\mathrm{m}^3}$) is about 2km (using Newton's approximation $D = L \frac{d_1}{d_2}$). Note that the velocity does not enter here, so throwing harder does not help.

After this length, the spear has transferred its impulse onto the air, and will drop. 2km obviously does not suffice to circle the earth.

I believe at the speeds Rob Jeffries calculated, we can safely ignore the aerodynamics of a spear (which would make Newton's approximation invalid).

• I guess you mean "wood spear of length 2.4 m", not "wood spear of length L". It would be interesting to explore lengths (or densities) that make impact depth greater than 8.5 km, so the spear can actually travel outside the atmosphere. – user27542 Jan 27 '15 at 16:11
• So you need a denser spear. With tungsten (18 g/cc) you can escape. Or a longer spear. I do like the approach though. I had to think about it for a minute - but I like it. – Floris Jan 27 '15 at 23:47

According to this paper (note: there's probably a paywall), I can draw three important reasons why you can not actually throw a spear around Earth. The paper, for those who can't access it, is titled "Effect of Vibrations on Javelin Lift and Drag". In it, they show that the amplitude of induced vibrations in a javelin is greater with greater throwing velocity. They also show that the drag force on a javelin due to these vibrations is an integral function of the square of its transverse velocity (basically, the RMS velocity of the vibrational modes). Furthermore, they show that drag and crossflow forces on the javelin vary significantly in magnitude and direction along the length of it. These differences included greater lift at the head and greater drag in the middle. This means that the javelin is initially prone to increase its angle of attack, which presents a greater cross section towards the direction of motion.

This analysis was done for velocities up to around $30m/s$; not the $7900m/s$ necessary to circle Earth. They simulated for vibrational amplitudes up to $0.1m$. This brings us to the first reason this is impossible; for such a great throwing velocity, the induced vibrations would have an amplitude far greater than $0.1m$. This would have the effect of increasing the differences between the magnitude of the forces at each point during flight as well as increasing the magnitudes overall (the magnitude and variability was linked to amplitude of vibrations. Think shear stresses). Under these conditions, the wooden spear would simply shatter (or rather, spectacularly explode).

If it remained intact, the increased amplitude vibrations would mean an increase in the transverse velocity, which drastically increases the drag on the spear. $7900m/s$ would cease to be nearly sufficient to go around the world.

The third reason this isn't possible has to do with the angle of attack increasing. At such high speeds above the speed of sound the compression of air causes a massive increase in temperature. This is mitigated if the is a smaller cross section exposed to the direction of motion, but an increased angle of attack added to the large amplitude vibrational modes makes the spear present a larger surface. I'm relatively certain this would be true even without vibrations (although they do help it be more true), but the compression and heating of air in this case would completely burn up the spear before it made a single trip around the world. Try to recall how much atmosphere it goes through. A meteor often doesn't survive to hit the ground, and that travels through far less atmosphere than this spear would. A wooden spear with a bronze(? iron? I'm not a historian) head would not endure.

Thank you, physics! Once again you bring the cold splash of reality to ruin another cartoon shenanigan. What would we do without you?

If it's going too fast, it'll wind up escaping Earth altogether. Obelix might be able to throw it hard enough it spiraled around at least once (for sufficiently large values of force imparted by magic potion). Sadly, once the spear has made its first lap, we know its speed has fallen so much it is now back at head height. It won't make another lap at the same height, and it would have to be implausibly smooth to make it back at all. But Obelix can still get it around the Earth if Toutanis (who, on clear days, works a second job as the god of meteors and space junk) is in a good mood. The nice thing about almost leaving the Earth's gravitational field is that space gets very empty if you go up high enough. We're making very extreme assumptions not only about magic potions but about Obelix's control or luck here, but if he throws the spear hard enough it will go into orbit, if it doesn't burn up or hit something first. Then it will go around the Earth many times before it gets back down to head height from essentially-negligible air resistance. If we assume changes in air resistance happen at scales much larger than the length of the spear, then since there is a trajectory where the spear goes not far enough before reaching head height, and there's also a trajectory where the spear goes too far before getting back to head height, there must also be a trajectory where the spear gets back at pretty much exactly head height.

• Some paragraphs divisions would be nice. – Brandon Enright Jan 27 '15 at 7:35
• "Then it will go around the Earth many times before it gets back down to head height from essentially-negligible air resistance." This is wrong. If no impulse is applied then the orbit will not change, a consequence of this is that the orbit will pass through the last point impulse was applied. So the Orbit will always pass through Obelix(ignoring air resistance). – Taemyr Jan 27 '15 at 8:41
• @Taemyr I hope Obelix will have the good sense to take a step to the side before the javelin passes through him. – Reinstate Monica -- notmaynard Jan 27 '15 at 15:50
• @iamnotmaynard But it's coming from behind him, how is he supposed to see it? Other than, you know, the fireball visible for thousands of miles around. – corsiKa Jan 27 '15 at 17:45