You are in a rocket ship, in outer space. You have a nuclear reactor that suppli

Question

You are in a rocket ship, in outer space. You have a nuclear reactor that supplies a constant power , P0, and a large supply of iron pellets. The Iron pellets comprise 99/100 of your ship's mass, m. You can use the power to eject the tiny iron beads out the back of your ship with an electromagnetic "gun". You can control the rate at which you fire them and their velocity, but are limited by your power plant. (you can't fire an arbitrarily large mass at an arbitrarily large velocity.) As you fire off the beads, your ship moves in the opposite direction to conserve momentum ]. In addition, the mass of your ship decreases.

(d) what is your final velocity in part (c)? how does it compare to the answer in part (b)?

Explanation / Answer

given, mass of ship = m

Nuclear reactor power = Po

mass of pellets = 0.99m

a) energy produced by reactoir in time dT = Po*dT

velocity provided to the ejected mass dm be v

then Po*dT = 0.5*dm*v^2

v = sqroot(2Po*dT/dm)

so, from momentum conservation, velocity gained by the ship, du = ?

(m - dm)du = dm*sqroot(2*Po*dT/dm)

du = dm*sqroot(2*Po*dT/dm)/(m - dm)

so momentum of exhaust wrt rocket = dm(v - du) = dm(1 - dm/(m - dm))sqroot(2*Po*dT/dm) = dm(m - 2dm)sqroot(2*Po*dT/dm)/(m - dm)

b) consider n pellets fired in time intervals dt between 0 < t < tf

then, tf/n = dt

now, after first pellet

velocity gained by rocket = du

velocity of exhaust = v

v = sqroot(2*Po*dt/dm)

(m - dm)du = dm*sqroot(2*Po*dt/dm)

du = sqroot(2*Po*dt*dm)/(m - dm)

after second p[ellet]

v = sqroot(2*Po*dt/dm)

(m - 2*dm)du = dm*v*sqroot(2*Po*dt/dm)

du = sqroot(2*Po*dt*dm)/(m - 2*dm)

after nth pellet

du = sqroot(2*Po*dt*dm)/(m - n*dm)

total velocity of rocket = v'

v' = sqroot(2*Po*dt*dm)[1/(m - dm) + 1/(m - 2dm) + 1/(m - 3dm) + .. ( 1/(m - ndm))]

c) for optimal firing rate, let the rate be r = dm/dt

at time t

mass left of rocket = (m - rt)

so from momentum conservation

(m - rt)du = (sqroot(2*Po*dt*dm))

(m - rt)du = (sqroot(2*Po*r))dt

du/(sqroot(2*Po*r)) = dt/(m - rt)

integrating from u = 0 to u = u

t = 0 to t = t

u/(sqroot(2*Po*r)) = ln[(m)/(m - rt)]/r

u = -sqroot(2*PO*r)*ln(1 - rt/m)/r

du/dt = sqroot(2*Po*r)/m*(1 - rt/m) = 0 for maximum u

rt/m = 1

r = m/t for maximum velocity

hence

u = -sqroot(2*PO*r)*ln(1 - rt/m)/r

d) Final velocity in part c is greater than that of part b

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