What Would It Take to Shoot a Cannonball Into Orbit?

Gravity is pretty complicated if you think about it. The motion of a ball falling on the surface of the Earth is caused by the same interaction as the moon orbiting the Earth. That's crazy. It's even crazier to realize that humans figured out that these two motions (falling ball and moon) are from the same gravitational force. It sure doesn't look the same.

Now imagine that you are around during the time of Isaac Newton (let's say early 1700s). How do you make this model of universal gravity? I don't know how he did it, but Newton finally made the connection between the motion of planets (and moons) and the motion of objects on the surface of the Earth. He explains this connection with his famous thought experiment of a cannon firing a ball from a tall mountain. Here is his diagram from A Treatise of the System of the World.

The diagram shows that an object moving on the surface of the Earth could eventually become an object orbiting the Earth. He does this by imagining a cannonball—a super fast cannonball fired from a super high mountain. The actual range of this ball would be farther than a normal cannon ball because you would have take into the curvature of the Earth. Oh, right—you have to ignore air resistance. In fact, if you shoot the ball fast enough it will "miss" the Earth entirely and enter a low Earth orbit.

There you have it. Gravity on the surface of the Earth is the same as the gravity between the moon and Earth. Like I said, this is a big deal.

Analysis of the Image

I don't know who drew this diagram. Maybe it's been modified over time. However, I'm going to start with a version of it and then check it for accuracy. In particular, I want to know the height of the mountain and I want to check the trajectory of the cannon ball paths. It's just what I do.

Although it's just an image, it is still useful to video analysis software to analyze this image. Of course I'm going to use the free (and awesome) Tracker Video Analysis. From the image, I want to find the height of the "mountain". That's fairly straightforward. I can just set the radius of the Earth diagram to the radius of the Earth and then measure the height. This puts the top of the mountain at 1.198 times the radius of the Earth from the center of the Earth (it's easier to deal with this in terms of the Earth's radius). Oh, that makes the height of the mountain 19.8 percent the radius of the Earth—see how nice that is?

Just for comparison, Mount Everest has a height of 8848 meters. In terms of the Earth, this is 0.139 percent of the Earth's radius. Or to put this another way, Newton's mountain is seriously ginormous. I guess there are a bunch of other cool things we could consider regarding a mountain as high as Mt. Newton (that's what I'm calling it now)—but I will just leave those questions alone for now.

What about the trajectory of these cannonballs launched from Mt. Newton? Let's consider the trajectories of the first three cannonballs (the three lowest speeds). I don't know anything about the time for these motions, but I can get the x and y positions. Since this doesn't really fit with your normal projectile motion plots (y vs. x), I'm going to instead plot r vs. θ where r is the distance of the ball from the center of the Earth and θ is the angular position from the center of the Earth. Yes, this is polar coordinates.

Since the angle is measured from the horizontal x-axis, these cannonball trajectories start from the left and move right. I hope that's not too confusing. But the real question: are these trajectories real? I could show that an object interacting gravitationally with the Earth should have an elliptical trajectory, but I'm not going to do that. Nope. Instead I am going to make a numerical model and adjust the starting velocity until I get something close to one of these. It's going to be fun.

Here's the plan (it's the same plan as most numerical calculations). But here are the basic steps.

• Break the problem into very small steps of time (about 1 second in this case).
• Calculate the gravitational force vector on the cannon ball based on the position of the ball with respect to the Earth.
• Use this force to update the momentum of the ball.
• Use the momentum (and thus velocity) to update the position of the ball.
• Repeat until you want to stop.

That's it. Here's what you get. Yes, this is an actual and real numerical calculation. You can see the code by clicking the "pencil" icon. In the code view, you can change the starting velocity. If you don't change that velocity, you are only cheating yourself. Seriously. Try changing the velocity.

OK, that's starting to look like the Newton picture. But can I get a trajectory that is JUST like the picture? Let me adjust the starting velocity to see if I get data that is very similar to the above three shots (from the picture). Here's what I get (as a plot of radial distance vs. θ).

This is for the first three cannon shots and you can see the trajectories don't quite match up. With my best estimates, these are cannonball speeds of 2800 m/s, 4200 m/s, and 6200 m/s. Oh, just for fun the orbital velocity at the height of Mount Newton would be 7252 m/s. Go ahead and use that velocity in the code example above. It should make a nice orbit.

So, let me summarize what we have. The diagram of Newton's cannonball seems nice, but it's just a sketch. If only Newton had python.