# Physics geeks: Galileo and cannonballs **URL:** \ **Category:** Factual Questions\ **Created:** [January 23, 2003, 9:11pm UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539 "2003-01-23T21:11:30Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![bruntilda](https://avatars.discourse-cdn.com/v4/letter/b/8dc957/32.png) [@bruntilda](https://boards.straightdope.com/u/bruntilda)\ **Post date:** [January 23, 2003, 11:47pm UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/21 "2003-01-23T23:47:13Z")
A cannonball fired straight up will have a vertical velocity from the firing of the cannon, and a horizontal velocity equal to the tangential velocity of the earth. The cannonball will follow a roughly parabolic path, with a constant horizontal velocity. However, the rotation of the earth is defined by its angular velocity, and to stay above the same point on the earth, the cannonball would have to travel faster the farther it went into the air. Think of it this way. Tie a weight to the end of a string and spin it over your head. The weight at the end has to travel much faster than the part of the string close to your hand. Therefore, the cannonball would not land in the same place that it was fired. It would land slightly westward of the firing point. ---
**Author:** ![Skeptico](https://avatars.discourse-cdn.com/v4/letter/s/e19b73/32.png) [@Skeptico](https://boards.straightdope.com/u/Skeptico)\ **Post date:** [January 24, 2003, 12:18am UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/22 "2003-01-24T00:18:46Z")
There is no deflection. > [@](#): > > Are you seriously suggesting that if you punted a cannonball straight up at, say, 11 kilometers a second, exactly vertically, at the equator, just below escape velocity, that while you waited for it to come back to Earth that it would stay exactly above the point of the Earth from which it left? Let me elaborate a bit more. Yes, the cannonball will return to the exact same point once it touches down on earth again (due to the conservation of angular momentum), but it will not stay at that same point during flight. Chrono’s formula does apply because it neglects the initial angular momentum of the ball. Having the ball drop from a static point, and having the ball go down (after going up) from a rotating earth are two different things. The latter has energy it picked up from the rotation of the earth, the former doesn’t. I suggest this [interactive experiment](http://liftoff.msfc.nasa.gov/Academy/ROCKET_SCI/ORBMECH/angular_momentum.html). When the cannonball goes up, it will “deflect” in flight, reaching a maximum “deflection” at apex. Once it starts falling down again, it “deflects backwards” (or if you prefer, “in reverse”), effectively correcting the original “deflection”. In other words, as it goes up, its angular velocity goes down, but when it drops again, its angular velocity goes up. In the ends, it lands at the same point it launched from. ---
**Author:** ![Skeptico](https://avatars.discourse-cdn.com/v4/letter/s/e19b73/32.png) [@Skeptico](https://boards.straightdope.com/u/Skeptico)\ **Post date:** [January 24, 2003, 12:25am UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/23 "2003-01-24T00:25:04Z")
I meant to say that Chrono’s formula **DOES NOT** apply. (I don’t know why I can’t edit my own posts…) ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [January 24, 2003, 12:25am UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/24 "2003-01-24T00:25:12Z")
> [@](#): > > \*Originally posted by Skeptico \* > Let me elaborate a bit more. Yes, the cannonball will return to the exact same point once it touches down on earth again (due to the conservation of angular momentum), but it will not stay at that same point during flight. Angular momentum will conserve the quantity omega × r[sup]2[/sup], where r is the distance to the center of the earth. As r goes up, omega can _only_ go down, and so it can only be less than the omega for the Earth. Thus it will fall behind. The farther it gets from the ground, the faster it will fall behind. And there’s no way it can catch up, if angular momentum is conserved. ---
**Author:** ![Skeptico](https://avatars.discourse-cdn.com/v4/letter/s/e19b73/32.png) [@Skeptico](https://boards.straightdope.com/u/Skeptico)\ **Post date:** [January 24, 2003, 12:28am UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/25 "2003-01-24T00:28:37Z")
> [@](#): > > As r goes up, omega can only go down, and so it can only be less than the omega for the Earth. Thus it will fall behind. The farther it gets from the ground, the faster it will fall behind. And there’s no way it can catch up, if angular momentum is conserved. You are absolutely correct, as it goes up, it will “fall behind”. There is no way it can catch up. However, once it starts falling, it will “speed up”. The scenario is akin to playing that ascent tape backwards. Thus it will fall right into the spot it launched from. ---
**Author:** ![William\_Ashbless](https://avatars.discourse-cdn.com/v4/letter/w/a87d85/32.png) [@William\_Ashbless](https://boards.straightdope.com/u/William_Ashbless)\ **Post date:** [January 24, 2003, 12:33am UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/26 "2003-01-24T00:33:28Z")
No, it will not. It cannot catch up to the point it left for all possible muzzle velocities, as I already proved. Period. To do so as per my example involving the cannonball at Pluto would involve it exceeding the speed of light long before it tried to ‘play catch-up’. A more rigorous mathematical analysis will show what has already been said – that it will fall behind, but by a very very little for short-distance firings. ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [January 24, 2003, 12:33am UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/27 "2003-01-24T00:33:42Z")
> [@](#): > > \*Originally posted by Scarlett67 \* > Wouldn’t the cannonball’s motion also have a component equal to the rotation of the earth, because the cannon from which it was fired was also moving with the earth at the time? And thus wouldn’t the cannonball indeed land at the spot from where it was fired? For all practical purposes this is correct. You would need a cannon of gigantic proportions and a projectile achieving a height comparable to the radius of the earth (6366Km) for other factors to be significant. The quote about Galileo is flat out wrong. ---
**Author:** ![William\_Ashbless](https://avatars.discourse-cdn.com/v4/letter/w/a87d85/32.png) [@William\_Ashbless](https://boards.straightdope.com/u/William_Ashbless)\ **Post date:** [January 24, 2003, 12:35am UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/28 "2003-01-24T00:35:00Z")
_Excellent._ 😃 ---
**Author:** ![Skeptico](https://avatars.discourse-cdn.com/v4/letter/s/e19b73/32.png) [@Skeptico](https://boards.straightdope.com/u/Skeptico)\ **Post date:** [January 24, 2003, 12:40am UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/29 "2003-01-24T00:40:15Z")
> [@](#): > > No, it will not. It cannot catch up to the point it left for all possible muzzle velocities, as I already proved. Period. To do so as per my example involving the cannonball at Pluto would involve it exceeding the speed of light long before it tried to ‘play catch-up’. Your proof has the **hidden assumption that the ball will have the same angular velocity as the Earth throughout the flight**. That is impossible, and nobody is arguing about that. Conservation of angular momentum does NOT mean that the angular velocity will stay constant, rather (as Achernar notes) that the quantity (angular velocity)x(radius^2) stays constant. ---
**Author:** ![William\_Ashbless](https://avatars.discourse-cdn.com/v4/letter/w/a87d85/32.png) [@William\_Ashbless](https://boards.straightdope.com/u/William_Ashbless)\ **Post date:** [January 24, 2003, 12:47am UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/30 "2003-01-24T00:47:39Z")
That’s a nice problem with my proof, Skeptico, but I don’t even require that it maintains its angular rate. If it doesn’t maintain its angular rate, then Earth is rotating ahead of it. This means that the cannonball must play catch-up, even to the point of going around the planet many many times to strike the Earth at the same spot it left (assuming Earth got so far ahead it rotated many many times). How much you want to bet that if you worked out how much extra distance it had to travel just around the planet to play catch up, after being all the way out at Pluto, that it would _still_ exceed the speed of light? ---
**Author:** ![bonzer](https://avatars.discourse-cdn.com/v4/letter/b/45deac/32.png) [@bonzer](https://boards.straightdope.com/u/bonzer)\ **Post date:** [January 24, 2003, 12:52am UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/31 "2003-01-24T00:52:06Z")
As to whether Galileo could measure the size of the effect involved, no, he couldn’t. Indeed pretty much exactly this question - in the form of whether a cannon ball dropped from a tower would fall to the east or the west - was the subject of a somewhat inconclusive argument between Newton and Hooke. ---
**Author:** ![SCSimmons](https://avatars.discourse-cdn.com/v4/letter/s/e495f1/32.png) [@SCSimmons](https://boards.straightdope.com/u/SCSimmons)\ **Post date:** [January 24, 2003, 3:36pm UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/32 "2003-01-24T15:36:58Z")
Aside from the physics questions already answered, there’s a fundamental historical question that hasn’t been addressed … > [@](#): > > \*Originally posted by Scarlett67 \* > **Did Galileo really show this?** I’ve done a fair bit of reading in history of science and history of physics, and I don’t recall any mention of Galileo having addressed this question, much less having answered it in this way. Does your author give a citation we could try to check on, Scarlett? ---
**Author:** ![SCSimmons](https://avatars.discourse-cdn.com/v4/letter/s/e495f1/32.png) [@SCSimmons](https://boards.straightdope.com/u/SCSimmons)\ **Post date:** [January 24, 2003, 3:42pm UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/33 "2003-01-24T15:42:01Z")
Oh, and BTW-there’s no question about Galileo detecting the effect by experiment. He did not have instruments of anywhere near the sensitivity required to detect this effect, and in any case was not as much of an experimentalist as he is often depicted. Many of the ‘experiments’ he describes are actually what we would now call ‘thought-experiments’ … ---
**Author:** ![Scarlett67](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/scarlett67/32/7387_2.png) [@Scarlett67](https://boards.straightdope.com/u/Scarlett67)\ **Post date:** [January 24, 2003, 5:05pm UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/34 "2003-01-24T17:05:44Z")
> [@](#): > > \*Originally posted by SCSimmons \* > \*\*Does your author give a citation we could try to check on, Scarlett? \*\* Nope, he just made the bald statement. No cite. I’ll query, and then it’s out of my hands. On this ms., I have no direct contact with the author – I just write up a list of queries for each chapter and send it in, and then I’ll never see it again. So I won’t find out the outcome. ☹ ---
**Author:** ![bobk2](https://avatars.discourse-cdn.com/v4/letter/b/6de8d8/32.png) [@bobk2](https://boards.straightdope.com/u/bobk2)\ **Post date:** [January 24, 2003, 5:27pm UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/35 "2003-01-24T17:27:14Z")
I hope it is permissible to share… My son’s High School Physics teacher claimed that one could get from New York to London in (about) five hours by rising up from New York in a helicopter, waiting for the earth to spin under you, and then setting down in London. Setting aside lattitude, and the idea that “the Earth spins while you don’t”, this instructor also managed to get the direction of the Earth’s rotation wrong. The instructor was quite hard on a student who tried explaining this to her. ---
**Author:** ![Pythagras](https://avatars.discourse-cdn.com/v4/letter/p/45deac/32.png) [@Pythagras](https://boards.straightdope.com/u/Pythagras)\ **Post date:** [January 24, 2003, 5:29pm UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/36 "2003-01-24T17:29:02Z")
Ah, I stand corrected by the logic of my peers. If the ball did stay above the same point, than the rotational speed of the ball as it went further out would increased constantly until it exceeded the speed of light. Somehow magically gaining momentum from nowhere. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [January 24, 2003, 6:16pm UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/37 "2003-01-24T18:16:21Z")
> [@](#): > > \*Originally posted by William\_Ashbless \* > \*\*That’s a nice problem with my proof, Skeptico, but I don’t even require that it maintains its angular rate. If it doesn’t maintain its angular rate, then Earth is rotating ahead of it. This means that the cannonball must play catch-up, even to the point of going around the planet many many times to strike the Earth at the same spot it left (assuming Earth got so far ahead it rotated many many times). > \*\* The mathematics isn’t all that hard under the simplifying assumptions that Galileo used. In the example the path is a parabola. In actual fact, in a vacuum, it is an ellipse. The assumption in the example is that the force of gravity always acts parallel to itself. Actually, as the shell moves along the force of gravity points toward the center of mass of the earth and so doesn’t stay parallel, making the ellipse. The launch site is moving eastward at a certain velocity. When the shell is fired straight up it also has that eastward velocity. When it comes to earth it has traveled eastward the same distance as the launch point which is the horizontal velocity times the time. Meanwhile the vertical distance is the inital velocity, times the time minus the distance the shell falls under the influence of gravity in the same time. These parametric equations are the same ones that Galileo used in geometric form in his original paper. The math assumes no air resistance and **a constant acceleration due to gravity**. This latter assumption isn’t true for shells that go away by an appreciable fraction of the earth diameter, as **sailor** pointed out. In that case the acceleration from gravity decreases with distance by the inverse square law, the math is a lot more complicated and the path is altered. Here is a [drawing of the trajectory](http://photos.yahoo.com/bc/dwsimm2000/vwp?.dir=/galileo&.src=ph&.dnm=shell.jpg&.view=t&.done=http%3a//photos.yahoo.com/bc/dwsimm2000/lst%3f%26.dir=/galileo%26.src=ph%26.view=t) as seen by someone who stays in the orientation existing at the time the shell was fired. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [January 24, 2003, 11:06pm UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/38 "2003-01-24T23:06:13Z")
Now that I think about it, Galileo wouldn’t have known enough about gravity to predict that the cannonball would have gone off course. He certainly didn’t know about stuff like conservation of angular momentum. ---
**Author:** ![Shade](https://avatars.discourse-cdn.com/v4/letter/s/2bfe46/32.png) [@Shade](https://boards.straightdope.com/u/Shade)\ **Post date:** [January 25, 2003, 1:56am UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/39 "2003-01-25T01:56:30Z")
_Could_ Galileo have measured the deflection? Based on Chronos’s calculations (which I’ll try and reconstruct when I’m less tired), a fall of 50m at lat=45[sup]0[/sup] would deflect a cannon ball about 5mm, assuming air resistance is negligible. OK, that’s miniscule. But, say, have a cannon ball in a clamp, and hang wto long pieces of string with a bell on each barely touching it on either side. Open clamp. Repeat many times, turning equipment round, etc, replacing it, etc. Hope west bell rings more often? ---
**Author:** ![Nanu](https://avatars.discourse-cdn.com/v4/letter/n/dec6dc/32.png) [@Nanu](https://boards.straightdope.com/u/Nanu)\ **Post date:** [January 25, 2003, 2:44am UTC](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539/40 "2003-01-25T02:44:31Z")
there is one problem with Skeptico’s logic, by conservation of angular momentum, the fastest angular speed the canonball can reach is the original angular speed of earth. What I am trying to say is that the cannonball would decrese its angular speed as it went up and thus fall behind, but it would always keep on falling behind because it will never reach an angular speed FASTER than that of earth, and thus never make up for the lost distance. [Previous page](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539.md?page=1) [Next page](https://boards.straightdope.com/t/physics-geeks-galileo-and-cannonballs/150539.md?page=3)