Question

Question 10 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 670 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 1000 m/s? (a) Number Units (b) Number Units

Answer 1

we know, orbital speed of satellite at an altitude h,

v = sqrt(G*Me/(Re + h))

= sqrt(6.67*10^-11*5.98*10^24/(6.37*10^6 + 670*10^3))

= 7527 m/s

speed of pellet with respect to satellite = 2*v

a) The kinetic energy of the pellet in the reference frame of the satellite,

KE_pellet = (1/2)*m*(2*v)^2

= (1/2)*3.7*10^-3*(2*7527)^2

= 4.19*10^5 J <<<<<<<<------------Answer

b) KE_bullet = (1/2)*m*v^2

= (1/2)*3.7*10^-3*1000^2

= 1850 J

KE_pellet/KE_bullet = 4.19*10^5/1850

= 226 <<<<<<<<------------Answer