How about a massive catapult to replace the space shuttle

I recently saw a comment on a blog somewhere about putting satellites into space (I think it was from a post about a rocket that blew up). The poster suggested using a giant catapult to put things in space instead of rockets. Maybe he or she was kidding, or maybe not. But I have heard this idea before. Would it work?


First, how do things get in orbit? Orbit is a motion in which the gravitational force on an object makes it move in a circle around another object. In order to be in a circular orbit, it takes a certain amount of energy.

Let me just start with some calculations. How fast would a satellite need to be going to be in orbit at the same level as the space station (I just picked that because I know it is about 300 km above the surface of the Earth) Here is a simplified diagram of an object in orbit (not drawn to scale):
![fixed orbit](http://blog.dotphys.net/wp-content/uploads/2008/08/fixed-orbit.jpg)
Here the only force on the object (satellite) is gravity. The gravitational force can be modeled as:
![gravity1](http://blog.dotphys.net/wp-content/uploads/2008/08/gravity1.jpg)
So, what does this force do? One way to think of this is that it makes the satellite accelerate. Don’t think of acceleration as increase in speed, acceleration is a change in velocity. Since the satellite is moving in a circle, it is accelerating (because its direction of velocity is changing). When an object is accelerating only due to moving in a circle, the acceleration is:
![circular acceleration](http://blog.dotphys.net/wp-content/uploads/2008/08/circular-acceleration.jpg)

Note that r-hat is in the direction of the center of the circle. Newton’s Second law says:

![Newton 2](http://blog.dotphys.net/wp-content/uploads/2008/08/newton-2.jpg)

Notice that there is only one force (gravity) and that both the force and the acceleration are in the same direction. For simplicity, I will drop the vector notation and just deal with the magnitudes (this is ok since there is no vector addition). This gives:
![gravity and circular](http://blog.dotphys.net/wp-content/uploads/2008/08/gravity-and-circular.jpg)
Now, I want the speed (*v*):
![speed](http://blog.dotphys.net/wp-content/uploads/2008/08/speed.jpg)
Here G is the gravitational constant, ME is the mass of the Earth, and r is the distance between the satellite and the center of the Earth (why the center – maybe that will be a story for another day). So, the energy needed to obtain this speed would be the kinetic energy:
![kinetic energy 1](http://blog.dotphys.net/wp-content/uploads/2008/08/kinetic-energy-1.jpg)
Note that really this is the CHANGE in kinetic energy that I am interested in, but since it starts at rest, it doesn’t matter. UNLESS you start close to the equator, then you already have some energy. I will (for now) neglect this contribution. The reader (you) can add that in as a homework exercise. Or maybe I will put that on the test.

Great, now that I have the velocity I can design my launching catapult – right? No. Not only does the satellite have to have that speed, but it must also but at that height. How much energy does that take? It is the change in gravitational potential energy (that is one way to look at it). Gravitational potential energy is:
![gravity potential](http://blog.dotphys.net/wp-content/uploads/2008/08/gravity-potential.jpg)
Why is it negative? How about I just say that it is negative because it’s an attractive force? That should be good enough for now. The change in potential energy will be:
![fixed delta u](http://blog.dotphys.net/wp-content/uploads/2008/08/fixed-delta-u.jpg)
The two “r’s” are the distance from the center of the Earth to the orbit, and the radius of the Earth.

Putting all this together, the energy needed to achieve orbit:
![e total](http://blog.dotphys.net/wp-content/uploads/2008/08/e-total.jpg)
Simplifying:
![energy total 2](http://blog.dotphys.net/wp-content/uploads/2008/08/energy-total-2.jpg)

Ok, so where is this energy going to come from? It is going to come from the catapult. Maybe it is a lever type catapult, maybe a big sling shot. Either way, I can look at this in terms of the work-energy relationship:
![work energy](http://blog.dotphys.net/wp-content/uploads/2008/08/work-energy1.jpg)
If I assume that the force is in the same direction as the displacement for this catapult launching device, then I can write:
![work](http://blog.dotphys.net/wp-content/uploads/2008/08/work1.jpg)
Where the delta E is from before. Now what do I do? Well, let me make some estimates of how far this catapult will exert a force on the satellite. After that, I can estimate the acceleration of the satellite.
I am thinking about a HUGE catapult. A catapult to end all wars. One so big that it will be legendary. How about a catapult the size of the Sears Tower in Chicago?
![256px Sears Tower ss](http://blog.dotphys.net/wp-content/uploads/2008/08/256px-sears-tower-ss.jpg)
The Sears Tower is 527 meters to the top of the antenna. What the heck, I will make it twice as tall as that (because I really want this thing to be GRAND). I will go with an even 1000 meters. Now I can determine the average force exerted on the satellite (if it were a sling shot type catapult, the force is not constant):
![avg force](http://blog.dotphys.net/wp-content/uploads/2008/08/avg-force.jpg)
Really what I want is the acceleration of the satellite:
![accel](http://blog.dotphys.net/wp-content/uploads/2008/08/accel.jpg)
Using the following values:
![values](http://blog.dotphys.net/wp-content/uploads/2008/08/values.jpg)
I get the average acceleration of:
![calculated accel](http://blog.dotphys.net/wp-content/uploads/2008/08/calculated-accel.jpg)
[calculated on google](http://www.google.com/search?hl=en&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=8cT&q=(6.67e-11*6e24%2F1000)*(1%2F6.4e6+-+1%2F6.7e6)&btnG=Search)
I converted this to the common acceleration unit of “g’s” so that it could be compared to manned rockets. I am pretty sure the Apollo Saturn V rockets had the highest acceleration. [Wikipedia](http://en.wikipedia.org/wiki/Saturn_V) lists it as 4 g’s. I imagine that an unmanned rocket could have a greater acceleration, but probably not this high (plus this is the average – if it were a sling shot rubber band type, the initial acceleration would be even higher). Clearly this device could not be used for humans.
How long would it have to be for it to be used with humans? Say an average acceleration of 4 g’s? To do this, the catapult would have to launch the capsule (space craft) over a distance of 71 kilometers (44 miles). Funny.
I was going to calculate how many rubber bands you would need to do this, but this post is ginourmous as it is. Maybe that can be another homework assignment.

10 thoughts on “How about a massive catapult to replace the space shuttle

  1. Obviously it wouldn’t be a mechanical catapult or slingshot, but how about something that is driven my electromagnets like a maglev type launcher with no friction.

  2. Jack,
    Actually, it doesn’t matter what exactly gets it moving. It could be an electromagnet, a sling shot, rockets, or a whole bunch of magical elves, that would still be the acceleration.

  3. Somewhere we approach the concept of a catapult tall enough to drop the satellite into orbit. This scan be accomplished. What of a catapult that is 2 miles tall? 30 miles tall? 62 miles tall?

    And the rail, someone recently tested a circular rail with a track switch leading to a ramp allowing the satellite to be accelerated gradually, like over days, and then launched up by the rail.

  4. Actually skyhooks that catapult objects to orbit are already well studied. Check out “The Hypersonic Skyhook” by Robert Zubrin, in “Islands in teh Sky” ed. Stanley Schmidt and Robert Zubrin ISBN 0471135615

  5. rhett,
    I realize that would still be the acceleration, but the author was implying, perhaps jokingly, building a large mechanical catapult. I was suggesting that using a electromagnetic maglev strategy, would be more practical and while it would still be quite large, it would be more feasible to build.

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