If you happen to be in the Bryan/College Station area tomorrow, you might consider checking out the Texas A&M Physics Festival. It’s sort of an open house with a ridiculous number of top-notch physics demonstrations as well as some very interesting talks. It’s free! I’ll be there helping out with some of the optics demos.

While we’re here, we may as well do some physics. This problem is from Halliday, Resnick, and Walker 14.62. It asks (fitting well with the space stealth theme from earlier):

One way to attack a satellite in Earth orbit is to launch a swarm of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit 500km above the Earth’s surface collides with a pellet having a mass of 4g. a) What is the kinetic energy of the pellet in the reference frame of the satellite just before the collision? b) What is the ratio os this kinetic energy to the kinetic energy of a 4g rifle bullet from a modern army rifle with a muzzle speed of 950m/s?

Well, we could “cheat” and use the virial theorem, but since at the intro level the students don’t know that yet we’ll do it the long (but still not difficult) way.

To find the kinetic energy we need to know the velocity of the pellet. We don’t know the velocity, but we do know that the pellet is in a circular orbit. We know that for any object to be in uniform circular motion, the force on that object has to be equal to m v^2 / r, where r is the radius of the circle.

But we also know the equation describing the force of gravity, as given by Newton. That force has to be providing the requisite pull to keep the object in uniform circular motion:

With M the mass of the Earth, and m the mass of the pellet. Solve for v:

Now r is not 500 km. r is the radius of the orbit, which is the radius of the Earth plus 500 km. It works out to about 7.6 km/s. But we can’t just get the kinetic energy from that either – we’re asked for the energy in the reference frame of the satellite, which is *also* moving at 7.6 km/s in the opposite direction. The total closing speed is 15.2 km/s. The satellite “sees” itself as at rest with a projectile incoming at that closing speed. Kinetic energy is 1/2 times the mass of the projectile times the square of its closing speed, for a total of some 462 kJ. This is a pretty hefty quantity of energy. The rifle bullet postulated in the problem only has 1805 J of kinetic energy, so the orbital pellet hits with something like 256 times more energy.

*Fig 1: Aluminum block hit by 7g projectile at ~7km/s
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This is why proposals for space warfare rarely involve the “explosives on rockets” paradigm which is so dominant in terrestrial warfare. At the speeds involved, adding an explosive is just an extra complication that doesn’t add anything of consequence to the destructive energy available just by virtue of the orbital velocity.