### Re: Newton Cannonball Brain Teaser

by **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 :^)