I-A:      Goldstein exercise 1-2.

T + V = T'+ V', where T + V is the energy of a particle of mass m at the earth's surface, moving upwards with escape velocity v_e;   T'+ V' is the energy of this particle after it has escaped.  T'= 0 and V'= 0, therefore T = -V or

Plug in the numbers to get v_e = 11.2 km/s or 6.95 miles/s.

I-B:      Goldstein exercise 1-3.

In this exercise, all vectors point vertically up or down, so we don't need to bother with vector notation. We define up as positive.

We consider the rocket plus the exhaust gases emitted during time dt to be an isolated system of two "objects" to which we apply Newton's 2nd law: dp/dt = F. If gravity could be neglected, the net external force F on this system would be zero. The change in momentum of the rocket during dt would be equal and opposite to the change in momentum of the gas (which before dt was in the form of fuel, at rest relative to the rocket). In this problem, we need to set the external force F equal to the force of gravity on the system.

Assume that the rocket (including unburned fuel) has mass m at a particular instant part-way into its burn, at which instant it is moving upward with velocity v. Thus, the momentum of this system relative to the earth is mv. At a time dt later, how has the momentum of this system changed?

The rocket has burned an extra amount of fuel dm, and is now lighter by this amount, and is faster by dv. Therefore, the rocket now has momentum (m + dm)(v + dv). (Remember, as noted in the hint, that dm for the rocket is a negative number.) The total mass of the system at the end of dt is still m, so in our notation, the mass of exhaust gas generated during dt is -dm (which is a positive number!). The fuel has become hot gas streaming downward at velocity v' relative to the rocket (or -v' + v in our earth-based coordinate system, where up is positive). Therefore, the momentum of the gas relative to the earth is -dm (v - v'). Plugging everything into dp = F dt, we get

Next, we are asked to find v(m) assuming dm/dt = -k (a constant), and we are given that k = M/60 , where M is the initial mass. Thus at a time t into the burn, the rocket's mass is m = M - kt and

As usual, the arbitrary constant in the integration is determined from initial conditions: at t = 0,  v = 0, thus C = v'ln M

Now we just need to plug-in numerical values in feet and seconds, and solve for t:

Thus we get t = 59.8 s. If the rocket were really able to carry enough fuel to keep burning at dm/dt = -M/60 for 59.8 s, then the mass of burned fuel would be Mt/60 and the mass of the empty rocket would be M - Mt/60. The ratio of these two quantities is t/(60-t), about 300.

I-C:      Goldstein exercise 1-4.

The second part is more general, so let's do that first:

The first part follows directly by noting that m is now constant, and we can divide across by m.

I-D:      Goldstein's problem 1-3 also appeared in his first edition, published in 1950. In that edition, the only difference is he points out that the numerical values in the problem refer to the V-2 rocket, which the Germans used against southeastern England towards the end of World War II. At the time of writing, that was the world's most technologically advanced rocket. Its empty weight was 3 tonnes and it carried 9 tonnes of fuel and payload (explosive). Therefore, its ratio of fuel weight to empty weight was vastly less than what would be needed to reach earth escape velocity. Nevertheless, modern rockets do not have dramatically better performance than the V-2, and exploration of Mars and other planets is possible only by using multistage rockets.

(a) If the speed of a rocket (relative to the earth) were to reach and then exceed the speed v' of its exhaust gas (relative to the rocket), then viewed from the earth, its exhaust would change direction and would actually be moving towards the rocket. Is such a situation possible, or is there a flaw in this reasoning?

(b) Imagine that you are back in 1944 using V-2 technology, and wish to explore space. You use a V-2 as your first stage. The explosive warhead is replaced by the upper stages and final payload. The empty weight of each successive stage is one-tenth of the previous one, but otherwise all upper stages have the same performance as the V-2. For example, this means that the mass of the empty second stage (no payload or fuel), is 0.3 tonnes, and the total mass of its payload and fuel is 0.9 tonnes. Assume a final payload of about 11% of the total mass of the last stage. Thus, the final payload would be 132 kg for a two-stage rocket. How many stages would be needed to leave the earth travelling straight up, and what final payload could be carried?

SAMPLE SOLUTION:

(a) There were indeed erronious claims before the "space age" that a rocket could never exceed its exhaust speed. In fact, every rocket which launches a payload into earth orbit (or beyond) exceeds its exhaust speed by a large margin. In problem I-B, we solved eq. (*) for a case where v far exceeded the magnitude of v' without violating any principle of physics.

There is a more general way to think about this. Consider a coordinate system x, y, z, and another system x', y', z' moving with a speed V relative to it (along the x direction, say). Then x' = x + Vt. It is easy to show that all the laws of physics (involving 2nd derivatives of position, like forces, accelerations, etc.) are the same in both coordinate systems. It also follows that it doesn't matter which coordinate system we choose when we calculate the motion. Therefore, it is always possible to choose a coordinate system where the exhaust velocity points parallel (rather than antiparallel) to the rocket velocity.

(b) Since m_empty drops by a factor of 10 from one stage to the next, the total mass

also drops by a factor of 10. For any stage,

where T is the burning time for this stage. For every stage except the last one,

and this gives T = 39 s. Now insert into eq. (*):

or v = 5883 ft/s. Therefore, each stage (not counting the last) simply adds 5883 ft/s to the final speed of the multistage rocket, and the last stage, for which T = 38.4 s, adds 5713 ft/s. Thus 7 stages would be needed to pass escape velocity. For 7 stages, the payload would be (12 tonnes)(11%)(10^-6) or 1.32 g!

In fact, a space program based on V2-like rockets is not as unfeasible as the above numbers suggest. Aiming an interplanetary rocket straight up is rather inefficient. Also, reaching earth orbit requires only about half escape velocity, and much of the acceleration is applied at right angles to the retarding force of gravity. For a rocket orbiting eastwards, the earth's rotation helps. Allowing for all this, a V2-like multistage rocket could probably launch a satellite of at least a few kilograms.