# Rocket equation of motion and exhaust velocity

See attached file for full problem description.

1) (a) Consider a rocket traveling in a straight line subject to an external force Fext acting along the same line. Show the equation of motion is

mv = - mvex + Fext. (3.29)

(b) Specialize to the case of a rocket taking off vertically (from rest) in a gravitational field g, so the equation of motion becomes

mv = - mvex - mg. (3.30)

Assume that the rocket ejects mass at a constant rate, m = -k (where k is a positive constant), so that m = mo - kt. Solve equation (3.30) for v as a function of t, using separation of variables (rewriting the equation so that all terms involving v are on the left and all terms involving t on the right).

(c) Using the data the initial mass is 2x106 kg, the final mass (after 2 minutes) is about 1x106 kg, the average speed vex is about 3000m/s, and the initial velocity is zero, find the space shuttle's speed two minutes into flight, assuming (what is nearly true) that it travels vertically up during the period and that g doesn't change appreciably. Compare with the corresponding result if there were no gravity.

(d) Describe what would happen to a rocket that was designed so that the first term on the right of Equation (3.30) was smaller than the initial value of the second.

2) Integrate v(t) from part (b) in the previous problem and show that the rocket's height as a function of t is

y(t) = vext - (1/2) gt2 - (mvex/k) ln (mo/m).

Using the numbers in problem 1, estimate the space shuttle's height after two minutes.

3) (a) We know that the path of a projectile thrown from the ground is a parabola (if we ignore air resistance). In the light of the result Fext = MR. what would be the subsequent path of the CM of the pieces if the projectile exploded in midair?

(b) A shell is fired from level ground so as to hit a target 100m away. Unluckily the shell explodes prematurely and breaks into two equal pieces. The two pieces land at the same time, and one lands 100m beyond the target. Where does the other piece land?

(c) Is the same result true if they land at different times (with one piece still landing 100m beyond the target)?

See attached file for full problem description.

#### Solution Preview

According to Newton's second law:

dP/dt = F

It is convenient to rewrite this as:

dP = F dt (1)

Let's consider the change of momentum of the rocket plus fuel system and plug that into Eq. (1). We look at the rocket as it is at some time after launch. The rocket at that time contains some fuel. This is the "old state". We then look at how the momentum of that state changes:

Rocket at time t ---> Rocket at time t + dt plus some expelled burned fuel

--->

dP = (New mass of rocket) times (new velocity of rocket) + (mass of expelled fuel) times (velocity of expelled fuel) - ( old mass of rocket) times (old velocity of rocket) (2)

We define M as a function of time to be the mass of the rocket at time t. Between time

t and t + dt the mass of the rocket will increase by dM = M-dot dt. Here M-dot must, of course, be negative. The mass of the exhaust is -dM. If you insert in Eq. (2):

New mass of rocket = M + dM

new velocity of rocket = v + dv

old mass of rocket = M

old velocity of rocket = v + dv

mass of expelled fuel = -dM

velocity of expelled fuel = v - v_ex

(note that the velocity of the fuel is v_ex w.r.t. the rocket in the opposite direction in which the rocket is moving, so. w.r.t. to the ground it is v - v_ex.)

you get:

dP = M dv + v_ex ...

#### Solution Summary

A detailed solution is given.