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 660 km above Earth's surface collides with a pellet having mass 5.3 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 5.3 g bullet from a modern army rifle with a muzzle speed of 1100 m/s?

Please help:)

Explanation / Answer

here,

mass of pellet mp = 5.3g = 0.0053 kg
mass of earth, Me = 5.972 * 10^24 kg
Radius of earth, Re = 6341 Km
height of orbit, = 660 Km

Total height from surface of earth,D = 6341+660
D = 7001 Kkm = 7001000 m

PART A:

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.

Vrel = 2 * sqrt(g*D)
Vrel = 2*sqrt(9.8 * 7001000)
Vrel = 16566.2 m/s

So Kinetic energy will be :
KEp = 0.5 * m * Vrel^2
KEp = 0.5 * 0.0053 * (16566.2)^2
KEp = 727263.303 J or 727.26 kJ

PART B:
mass of bullet = 5.3g = 0.0053 Kg
Velocity, v = 1100 m/s

KEb = 0.5 * m * v^2
KEb = 0.5 * 0.0053 * 1100^2
KEb = 3206.5 J

RAtio :
KEp/KEb = 727263.303 / 3206.5
KEp/KEb = 226.809

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