Newton Cannonball Brain Teaser

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Newton Cannonball Brain Teaser

Postby BurtJordaan on March 30th, 2017, 1:05 am 

You surely know about Newton's through experiment called Newton's cannonball. Ignoring air resistance, a gunner could shoot himself in the back, provided that the cannon is on a high enough mountain or tower to clear other mountains.

If we also ignore the rotation of Earth, the gunner could even shoot the cannonball over some higher mountains (that are not too close to him) and still achieve this "Darwin awards" feat. This is done by still shooting horizontally, but at a muzzle velocity higher than circular orbit requirement, but lower than escape velocity, e.g. (Wiki article):

Image

But what if we do take the rotation of Earth into account for this scenario? Can he still do it?

No need to calculate - just give a qualitative scheme for how the gunner should aim and fire the gun. We can work out the orbital details later, if required. We obviously assume that the cannonball has no thrusters available for course correction - just purely inertial flight. Also ignore the gravity effects of the Sun and the Moon.

In order to give all interested parties a chance to voice their views, I will not reply to each answer.
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Re: Newton Cannonball Brain Teaser

Postby Dave_Oblad on March 30th, 2017, 12:06 pm 

Hi Jorrie,

Quick drop by..

Assuming the Earth doesn't wobble and has perfect Gravitational Geometry and we have perfect control over Cannonball Velocity and Trajectory at firing:

A. From Earths poles it's a straight Trajectory-Velocity problem.
B. From the Equator, Trajectory & Velocity must be adjusted for East or West bound Cannonballs.
C. Anywhere else becomes a real problem:

The Rotation of the Earth can be plotted as a Sine wave.
The Orbit of the Cannonball can be plotted as a Sine wave.
These will not be the same frequency:
It could take many many many Earth rotations and Cannonball orbits for both to realign or fall back into phase such that the Trajectory, Velocity and Phase Matching accomplish the desired feat.

Gotta run..

Regards,
Dave :^)
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Re: Newton Cannonball Brain Teaser

Postby BurtJordaan on April 1st, 2017, 5:39 am 

Hi Dave, you gave a very nice summary and your "quick drop by" is very close to right. With no other takers, I owe you a reply.

A (poles): You are right - he can shoot it at just like for the non-rotating earth, at circular or parabolic orbital velocity.

B (equator): Yes, works for east/west circular orbits, and as you implied, he must add (west) or subtract (east) Earth's rotation speed for the top of the tower, in order to get the precise circular orbit.

C (anywhere else): Yes, it's more complicated, but he doesn't need multiple orbits, just one elliptical orbit for the ball, with orbital period precisely one sidereal day (23 hours, 56 minutes, 4.0916 seconds). This is the time it takes for any non-polar position on Earth being rotated to the same position relative to the distant stars. Remember that after one orbital period, objects in orbit also return to the same position relative to the same stars, so the gunner can still 'shoot himself in the back'.

This orbit will require a very high apogee, so it will need a beefy cannon, firing horizontally and due east, in order to take full advantage of Earth's rotational speed at the site. To find the exact muzzle velocity is not trivial, but rather straightforward when using Kepler's third orbital law and Newton's angular momentum conservation. More on that later.
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Re: Newton Cannonball Brain Teaser

Postby Dave_Oblad on April 2nd, 2017, 1:54 am 

Hi Jorrie,

Like Newton's boat, just shoot straight up and it will come straight back down.. or will it?
Yes.. at the poles and yes.. at the equator (mostly).. so we can take those off the table.

And, in theory, it may be possible to do so from any other place for a full rotation of the Earth and multiple orbits of the Ball.

Here is a possible issue: No matter what semi-orbital angle one shoots the Ball, the center of the Earth represents a point on a plane that contains all points of the orbit of the Ball (We can't tilt that Plane). The Earth takes 24 hours to rotate but the Ball must always be below escape velocity of about 11.2 km/s or about 25,000 miles per hour.

The Earth is about 25,000 miles in circumference. So at just below escape velocity it will do 1 orbit per hour (or less). So after the first orbit, the Launch/Target will be many miles away from the plane of the orbit. Thus, anywhere the Ball is fired, (except equator and poles) it will always take multiple orbits of the Ball to realign the orbital plane and our Target. We can't get around that fact. However.. I'm visualizing a near Earth Surface Parallel shot.

Ok.. Jorrie is correct, I hadn't considered a single very elongated orbital shot.

Or we must use Newton's Boat concept and merely lob the Ball high enough that it falls where we want it to, or where we expect to be, when it falls back to Earth. But technically, that's not shooting one direction and getting hit in the back. (unless you lean forward.. lol)

But just for fun: (Shooting somewhat straight up variations)

This would seem to have a limited range because when the Ball falls back, it may miss the horizon, where we expect to be (or sooner). If we are over the horizon, the Ball will miss the Earth and be in some eccentric semi-orbit that will decay eventually. That Math I don't really want to tackle. (as is true for all math really.. lol)

The other issue is.. suppose I want to launch the Ball over the North Pole, such that I shoot it at Noon and hope to receive it at Midnight. How high do I have to shoot it.. such that it takes twelve hours to complete the journey? Remember, even a partial lob still must describe a plane that includes the center of the Earth and all points the Ball will travel through on its journey. Also, I must not shoot directly over the Pole, because I will be imparting an Eastward Angular momentum to the Ball and the Ball must land on the opposite side that's moving in the opposite direction from my launch location.

So, the shorter the Lob, the less height we will need. But keep in mind, this is not shooting California from New York. This is shooting New York from New York. So we have limits such that we don't go so high as to exceed escape velocity and we also don't over shoot the Horizon. Thus.. it would seem that Newton's Boat has some limits.

Note: Not sure if I have the reference correct.. is it Newton's Boat or Newton's Ball? (Momentum) There is an old thread around here somewhere but can't find it. (tossing a ball up and down while in a moving boat.)

Question: What is the maximum Height we can shoot a Ball without it exceeding escape velocity and what would the maximum Round Trip time be for that maximum Height limit?

Question: Other than at the Poles.. will we still be under the impact Site for a Ball shot Straight Up to near Maximum Height? (I don't think so.. even if it's technically not an orbit, a long lob can have issues.)

Also.. for an equatorial shot, could we be over the horizon by the Ball's return time on a long shot? Given, Earth surface velocity is about 1000 miles per hour and if we extend that same angular momentum out to several thousand Miles, we will effectively be moving faster than the Momentum speed imparted to the Ball, because we are comparing the surface velocity of two different sized spheres (even if one is not actually a sphere).

I mean for a Geostationary Orbit, it has to be out about 22,000 miles and yet still circumnavigate the Earth in 24 hours. This means it must be moving about 6800 miles per hour to stay over one spot on the Earths Equator. That spot on the Earth is only moving about 1000 miles per hour. Thus the angular velocity imparted to a straight up shot can't keep up with the launch site below if shot too high. Did that make any sense?

Just some quick thoughts...

Weird fun problem.. thanks Jorrie. (never given such much thought before)

Best Regards,
Dave :^)
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Re: Newton Cannonball Brain Teaser

Postby BurtJordaan on April 2nd, 2017, 5:33 am 

Hi Dave, you have a very good intuition for this problem ;)

I think it is possible to do a 'lob' along the equator, but not for off-equator sites, as you have said. The only possibility that I have investigated and solved for off-equator is the 'sidereal day' orbit, but it may be possible that there is a 'half sidereal day' solution as well. A specific point on earth must surely go through the plane of the orbit twice per day.

Will have a (mathematical ;-) look at that one...
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Re: Newton Cannonball Brain Teaser

Postby BurtJordaan on April 2nd, 2017, 5:19 pm 

Hi Dave, I had a look at the 'lob' (or 'toss') issue again and am now of the opinion that even with the rotating earth, there must be an orbital solution for any location to have the cannonball come down on the cannon in less than one sidereal day. The math will be much tougher than for the sidereal day solution, because you cannot start at the perigee of the orbit and it is not a full orbit either.

I will continue to tinker with the math, but in the meantime, I will post the sidereal solution tomorrow (my time, UT+2).
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Re: Newton Cannonball Brain Teaser

Postby BurtJordaan on April 3rd, 2017, 12:39 am 

OK, let's attempt to get the muzzle velocity required for the 'one sidereal day orbit' required for the ball to return to the launching cannon on a rotating earth. Make the tower adjustable so that we can always find a place where the cannon is fired at exactly 6400 km from the center of earth, some 30 km ASL.

The first thing to do is to find, from Newton's gravity, the escape speed at 6400 km from Earth's center. Copy and paste the following well known formula into Google search:

sqrt(2*mass of Earth * gravitational constant)/ 6400 km) =

It should give you 11 160.5995 m / s. As the name implies, this speed will obviously never let the ball return to Earth - we need somewhat less than that. For that value, we need to find at least one other parameter of the sidereal day orbital ellipse. This is where Kepler's 3rd law comes in: "The square of the orbital period of a planet is proportional to the cube (3rd power) of the semi-major axis (a) of the orbital ellipse".

The formula for (a) (from most textbooks) can be pasted directly into Google as follows:

(mass of Earth * gravitational constant / (2 * pi)^2 * sidereal day^2)^(1 / 3) =

to get a = 42 163.757 kilometers. This is the semi-major axis of the required orbital ellipse. From this one can get the circular orbital velocity at distance (a) from earth by Googling:

sqrt(mass of Earth * gravitational constant / 42163.757 km) =

giving 3 074.63001 m / s. **

Then do a vector subtraction of the two velocities, i.e. Google:

sqrt (11160.5995^2 - 3074.63001^2) =

yielding the required eastwards perigee velocity of 10728.7292584 m / s.

Now all that remains is to subtract earth's rotational speed at your location, load the cannon with the precise charge required for that muzzle velocity and fire it horizontally, due east. If done correctly, duck tomorrow, in 23 hours, 56 minutes, 4.0916 seconds, or risk being self-shot in the back.

;^)

** Note, no spaces allowed in the numerical inputs, although Google is inclined to insert them into his answer. So remove spaces if you copy and past any value for a follow-on calculation. Google also tends to insert superfluous brackets for its own sanity and mankind's confusion. If it does not understand your formula, it will do a normal search, with your input as a search term. You may not find the answer among the search results, so rather try fixing the formula.

PS: It is one massively large orbit, with an apogee much higher than the geostationary satellites, which sit around 42,000 km radius. See attachment. The brown circle represents Earth.
Attachments
CannonballOrbit.png
Toy Cannonball Orbit, scale in km.
Last edited by BurtJordaan on April 3rd, 2017, 4:21 am, edited 2 times in total.
Reason: PS, attachment
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Re: Newton Cannonball Brain Teaser

Postby Dave_Oblad on April 3rd, 2017, 1:14 pm 

Hi Jorrie,

Fascinating.

As can probably be seen from my Lob insertion: Is it possible for a shot to require a 24 hour Period without exceeding Escape Velocity? Seems you have that covered.. Thanks.

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Dave :^)
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Re: Newton Cannonball Brain Teaser

Postby BurtJordaan on April 5th, 2017, 8:51 am 

BurtJordaan » 02 Apr 2017, 23:19 wrote:Hi Dave, I had a look at the 'lob' (or 'toss') issue again and am now of the opinion that even with the rotating earth, there must be an orbital solution for any location to have the cannonball come down on the cannon in less than one sidereal day. The math will be much tougher than for the sidereal day solution, because you cannot start at the perigee of the orbit and it is not a full orbit either.

Indeed, there are solutions, but the math is horribly tough. I have laboriously done one for a 45 degree elevation lob at the equator (for no good reason other than curiosity).

CannonballOrbit_45-deg.png
Cannonball Orbit 45-deg, distances in km


The surface to surface orbit time is about 10 hrs 50 min. This time was the tricky part and then even more so finding a solution where the earth has also rotated by exactly the same time, so that the ball falls on the cannon again (from behind at 45 deg).

In the end I had to run a simulation on a spreadsheet and let Excel 'Goal Seek' for equal times. For a 45 deg shot, it turned out to require around 10000 m/s muzzle velocity, so not a great saving compared to the simpler horizontal, 24 hour perigee shot (10728.72 m/s).

It should require less speed for an non-equator 45 deg elevation shot, but that's even more tricky, because the complication of 3D with intersecting planes then creeps into the works. Anyway, I've learned a few new orbital tricks in the process, e.g. http://bado-shanai.net/Platonic%20Dream/pdAreaofEllipticSector.htm.
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Re: Newton Cannonball Brain Teaser

Postby Dave_Oblad on April 5th, 2017, 1:30 pm 

Hi Jorrie,

Looks good for an Equatorial shot, but a single Lop shot, say from North America, would define a Travel Plane that has the Ball return to South America.. in a period less than a sidereal day. This is why I said we can't Tilt the Plane.

Highest Regards,
Dave :^)
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Re: Newton Cannonball Brain Teaser

Postby BurtJordaan on April 5th, 2017, 3:37 pm 

No, not if you shoot it north-easterly. Orbits don't care about the equator or even about the fact that earth rotates. It only cares about earth's center of gravity. The rotation does influence where you hit in 10.9 hours (for a 45 deg shot), but it does not have to fall in the southern hemisphere.

It is true that you cannot shoot due east (along a the same lat), but you can take a more northern shortcut. So the Cape can shoot the Cape...
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Re: Newton Cannonball Brain Teaser

Postby Dave_Oblad on April 5th, 2017, 5:25 pm 

Hi Jorrie,

Yep, you are right of course. If the Travel Plane intersects the Center of Gravity and it can be lobbed far enough to take the exact right amount of travel time, you should be able to hit New York from New York or LA from LA in less than a full orbit.

I think I need more sleep.. lol. Bravo.

Regards,
Dave :^)
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