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 810 km above Earth's surface collides with a pellet having mass 3.7 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 3.7 g bullet from a modern army rifle with a muzzle speed of 1200 m/s?

Answer 1

Answer:

411087.52089 J

$\frac{K_r}{K_b}=154.31213$

Explanation:

R = Radius of Earth = 6370000 m

h = Altitude of satellite = 810 km

r = R+h = 63700000+810000 m

m = Mass of bullet = 3.7 g

Velocity of bullet = 1200 m/s

The relative velocity between the pellets and satellite is 2v

Now, the square of velocity is proportional to the kinetic energy

$K\propto v^2$

$\\\Rightarrow 4K\propto (2v)^2\\\Rightarrow 4K\propto 4v^2$

Kinetic energy in terms of orbital mechanics is

$K=\frac{GMm}{2r}$

In this case relative kinetic energy is

$K_r=4\frac{GMm}{2r}\\\Rightarrow K_r=2\frac{6.67\times 10^{-11}\times 5.98\times 10^{24}\times 3.7\times 10^{-3}}{(6370+810)\times 10^3}\\\Rightarrow K_r=411087.52089\ J$

The relative kinetic energy is 411087.52089 J

The ratio of kinetic energies is given by

$\frac{K_r}{K_b}=\frac{411087.52089}{\frac{1}{2}\times 3.7\times 10^{-3}\times 1200^2}\\\Rightarrow \frac{K_r}{K_b}=154.31213$

The ratio is $\frac{K_r}{K_b}=154.31213$