# Throwing Something Into Orbit

There's a lot of stuff in the news lately about asteroids, what with the Dawn mission orbiting Vesta, and the talk of a manned asteroid mission as a possible future step for NASA. Prompted by this, I'm going to dip into the territory usually occupied by Matt and Rhett, and ask a somewhat silly question:

What size asteroid would you be able to throw a baseball into orbit, a la Bugs Bunny?

(Sadly, probably for copyright reasons, I couldn't quickly find a YouTube video of the cartoon where Bugs throws a baseball all the way around the world. But you can probably picture it, even if you're too young to remember the actual cartoon.)

Now, I'm not interested in the feasibility of this on Earth, or any specific astronomical object. I'm asking what size an imaginary object would need to be in order for this to be possible. We can get an idea from introductory physics.

In order to be in a stable orbit, a thrown object needs to be constantly changing the direction of its velocity, so that it eventually ends up back where it started. The force causing this change is the gravitational attraction between the thing in orbit and the asteroid it's orbiting. If we take the magnitude of the centripetal acceleration needed (which is v2/R, where v is the speed and R the orbit) and multiply it by the mass of the orbiting object (m), we can set these equal:

This is a nice, simple equation, where G is Newton's gravitational constant and M is the asteroid mass. The immediate thing to notice is that both sides have an m and an R, so a little basic algebra shows that the final answer doesn't depend on the mass. You could even solve this for R, but then you run into a problem, which is that you need a value for M as well. And we don't want to just guess values randomly.

So, what do we do? Well, we make two simple assumptions about the asteroid: first, that it's spherical, and second, that it has the same density all the way through, which we'll call ρ, because for some reason physicists like using the Greek letter rho to mean density. I'm not sure why.

If we make those assumptions, we can replace M in our equation with the density multiplied by the volume of our spherical asteroid, giving:

This is an equation we can really like, because the only thing in it that we don't know is R. We solve it for R and get:

So, our radius depends only on the speed of the throw, the density of the asteroid, and the gravitational constant. We can just plug in some reasonable-ish numbers for these, and get the size we're after. If we imagine a good-but-not-great baseball player throwing the ball at about 30 m/s (~67 mph), and call the density of the asteroid 3 g/cm3 (on the high side for the density of rock, but a nice round number that isn't hugely unreasonable), the answer is about 33 km, or a bit less than 10% the radius of Vesta. So, if you could land a person on a rock with a radius of 33km, they could then throw a baseball into a circular orbit around that rock.

Could you really stand there and do that, though? Well, if we plug our density and volume in to get the mass, we find that it's around 4.5x1017kg. The acceleration due to gravity for an object dropped near the surface of such a body would be:

Plug those numbers in, and you find an acceleration of about 0.03 m/s/s, or a bit less than 1/300th of Earth's gravity. So, you probably wouldn't notice much gravity at all, making it a little tough to stand there and throw a ball.

(If you're wondering, the escape velocity for such a rock would be around 43 m/s. so a really good major league pitcher throwing straight up would be able to throw a ball clean off the asteroid, never to return.)

So, if you've ever wondered about what it would take to be able to throw a ball into orbit, now you know. And knowing is half the battle. Though, to be honest, I have no idea what you'd be fighting against that could be defeated by solving this problem twice...

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Minor nit: it doesn't matter what direction the major league pitcher throws the ball, as long as its path doesn't intersect the surface of the asteroid.

well, if you're stuck on an asteroid and have no other means of propulsion... :-P

By Nomen Nescio (not verified) on 27 Jul 2011 #permalink

I was really hoping for an article about an orbital pinwheel.

But this was pretty cool. I love doing stuff like this.

I love/hate to pick nits, but an asteroid much much smaller than Vesta probably won't be remotely spherical. While this doesn't change the forces/velocities much, it does create a potential problem: a "mountain" that obstructs the orbit - the periapsis will be the height of the thrower's arm (this is why Real Rockets(tm) usually launch into an high-eccentricity ellipse, then restart engines at apogee to circularise it).

Though, to be honest, I have no idea what you'd be fighting against that could be defeated by solving this problem twice...

Knowing that is clearly the other half. :-)

By Lurker #753 (not verified) on 28 Jul 2011 #permalink

I love to pick nits, so I'll notice that no significant error is introduced by your failure to use R+h in the denominator of your second equation to account for a small orbital altitude. Of course, this is easily justified post hoc by the size of your answer for R, which shows the value of estimating first and asking questions later.

The other would be to show the effect of the square root on how R scales with density.

BTW, did you see the news that the Earth has a trojan asteroid in a cool orbit around a Lagrange point (L4)?

I love to pick nits, so I'll notice that no significant error is introduced by your failure to use R+h in the denominator of your second equation to account for a small orbital altitude. Of course, this is easily justified post hoc by the size of your answer for R, which shows the value of estimating first and asking questions later.

True. I should've mentioned that in the original post. See, this is why I normally leave this kind of post to Matt and Rhett...