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 770 km above Earth's surface collides with a pellet having mass 6.4 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 6.4 g bullet from a modern army rifle with a muzzle speed of 550 m/s?

Answer 1

To solve the problem it is necessary to take into account the concepts related to Kinetic Energy, Gravitational Force and Centripetal Force

For the given case the energy, we apply conservation of it, in which the kinetic energy is equal to the gravitational energy, that is to say

Centripetal Force  = Gravitational Force

$\frac{mv^2}{r}= \frac{GMm}{r}$

Re-arrange for v,

$v=\sqrt{\frac{GM}{r}}$

Where,

G= Gravitational Coefficient

r= Radius of Satellite

M = Mass of earth

V = Velocity

$v=\sqrt{\frac{2GM}{r}}$

$v=\sqrt{\frac{(6.67 x 10^-11)(5.98 x 10^24)}{(6370+770)*1000}}$

$v = 7474.19m/s$

Puesto que la velocidad relativa es dos veces -se mueve con la misma magnitud pero en sentido contrario- entonces,

$v_p = 2v_s$

$v_p = 2*10570.1 = 14948.38m/s$

$KE_{pellet} = \frac{1}{2}mv^2$

$KE_{pellet} = \frac{1}{2}(6.4*10^{-3})*14948.38^2$

$KE_{pellet}= 715053.9J$

PART B) Once the speed is obtained, it is possible to compare the kinetic energies between the bullet and the satellite, like this:

$KE_b = \frac{1}{2}mv^2$

$KE_b = \frac{1}{2}(6.4*10^{-3})(550)^2$

$KE_b = 968J$

In this way the ratio is:

$\frac{KE_s}{KE_b} = \frac{715053.9}{968}$

$\frac{KE_s}{KE_b} = 738.7:1$

Therefore the ratio of this kinetic energy to the kinetic energy of a 6.4 g bullet from a modern army rifle with a muzzle speed of 1100 m/s is 739 to 1.