One way to attack a satellite in Earth orbit is to launch a swarm of pellets in

Question

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 720 km above Earth's surface collides with a pellet having mass 5.0 g. (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 of this kinetic energy to the kinetic energy of a 5.0 g bullet from a modern army rifle with a muzzle speed of 920 m/s?

Explanation / Answer

The pellet and the satellite must travel at the same speed to be in orbit at the same altitude, so relative to the satellite, the pellet approaches at twice it's speed.

Taking the radius of earth to be 6400km, then r = 6400 + 720 km = 7.12 *10^6 m.

So the relative speed is approximately

v = 2* sqrt(9.8*7.12 *10^6) m/s = 16706 m/s

(would actually be less if the value of g was compensated for altitude)

So the pellet's relative KE is
KE = 0.5(0.005 kg )(16706 m/s )^2 J = 697.72 kJ

(b)
The bullet speed is given as 920 m/s
So the bullet's KE is
KE = 0.5 (0.005 kg )( 920 m/s )^2 = 2116 J

The ratio is 697720 / 2116 = about 329.7 : 1

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