Getting to Mars the Easy Way

posted on 19 Aug 2014 by gsgreenstein, guy
last changed 5 Apr 2015

 4.5 Average: 4.5 (2 votes) Select ratinghated itdidn't like itliked itreally liked itloved itCancel rating Your vote (click to rate)ages: 16 to 99 yrs budget: $0.00 to$0.00prep time: 0 to 20 min class time: 20 to 60 min Two different methods of getting to Mars are proposed. A simple analysis shows that one is better than the other. The reasons for the difference are explored, yielding an insight into orbital mechanics. The topic is appropriate for any high school physics class familiar with the conservation of momentum and energy. Figures with equations are included in attached lecture slides.
subjects: Astronomy, Physics

GettingToMars.pdf

Many years ago I read a science fiction story in which an astronaut, in a rocket in orbit about the earth,  decided to go to Mars. According to the story the astronaut had two options:

1. Fire the rocket engine and head to Mars.
2. Initially don’t head towards Mars: head towards the Earth! But avoid landing. Rather drop down close as possible (just above the atmosphere, maybe), fire the engine when the rocket is at this closest approach, and then zoom off toward Mars.

According to the story, the second method was better: for the same amount of fuel expended, the rocket reached a higher velocity.

The amazing thing is that I still remember this tidbit, even after all the years which have passed since reading the story. I think the reason is that I never believed it. I was sure that the author must have gotten things wrong. But recently I decided to calculate things out to see for myself. Amazingly, I found that the author had been correct.

In this writeup my first step is to show this. But there’s another step: not whether the second method is better, but why it is better. This led me on an interesting and an enjoyable journey. I thought you might like to see it, and maybe share it with your students.

FIRST STEP: COMPARE THE METHODS

This is not so easy. The full calculation would be very complicated, involving elliptical orbits, a steady drop in the rocket’s mass as its fuel is expended, and so forth. But I decided that most of these complications are merely details, and just get in the way of the essential points. So I began by setting myself the task of finding a simpler problem which would contain all the important elements of the original one.

A SIMPLE MODEL OF FIRING A ROCKET ENGINE

A rocket works by burning fuel and shooting the exhaust out the rear. It takes a certain force to shoot this exhaust off: by Newton’s 3rd law of reaction, the exhaust exerts the same force on the rocket, so accelerating it. In the process, momentum is conserved: the backwards momentum of the fuel is balanced by the forward momentum of the rocket.

So I tried to think of the simplest possible model of this process. I imagined firing the engine not steadily but in a single, short burst. If so, the stream of exhaust would be replaced by a single lump.

Let’s use the conservation of momentum to find the increase in the rocket’s velocity caused by shooting out that lump. Suppose the rocket has mass “M” and the lump mass “m,” and suppose that before firing its engine the rocket (together with its lump of fuel) is moving with velocity $V_{initial}$. Then the momentum before firing is

$$P_{initial} = (M+m) V_{initial}$$

After firing the engine, the rocket is moving at some velocity $V_{final}$. The exhaust was shot off with velocity $V_{exhaust}$ relative to the rocket: since the rocket had been moving at velocity $V_{initial}$, the lump ended up moving at velocity $U=V_{initial}-V_{exhaust}$. (If the exhaust velocity is less than $V_{initial}$ this is positive, indicating that the lump is moving in the same direction as the rocket, though more slowly: if the exhaust velocity is greater than $V_{initial}$ it is negative, indicating that the lump is moving in the other direction.) So a picture of the situation after firing the engine might look like figure 1.

So the net momentum after firing the engine is

$$P_{final}= M V_{final} + m U = M V_{final} + m (V_{initial} - V_{exhaust})$$

The law of conservation of momentum tells us that $P_{initial}= P_{final}$. Solving we get a formula for the rocket’s final velocity $V_{final}$:

$$V_{final} = V_{initial} + (m/M) V_{exhaust}$$

And finally, let’s subtract out the rocket’s initial velocity in order to find $\Delta V$, the change in the rocket’s velocity caused by firing its engine:

$$\Delta V = {m\over M} V_{exhaust}$$

where, to repeat, $m$ is the mass of the exhaust, $M$ is the mass of the rocket, and $V_{exhaust}$ is the velocity with which the rocket shoots out its fuel.

Problem for the student: Suppose we are speaking of the Space Shuttle, and its engine fires for one second. Look up the appropriate data and evaluate  $\Delta V$.

A SIMPLE MODEL OF DROPPING CLOSER TO THE EARTH

Similarly, I needed to avoid elliptical orbits, and invent a simpler model of the route taken by the rocket. Ultimately the picture came to my mind of a rocket mounted on wheels and coasting along a road.

Method 1 would have the rocket just fire its engine and head off towards Mars. In contrast, Method 2 would have the rocket take a detour and drop into a dip in the road before heading off to Mars (Figure 2). After all, dropping into that dip is just an example of maneuvering a rocket closer to the Earth!

My science fiction story was claiming that if the rocket delayed firing its engine until it was down in the dip, it would end up moving more rapidly once it had climbed back out of the dip and was on its way. I wanted to see if this was true.

Let’s analyze this more complicated scenario. The point of the dip is that the rocket speeds up as it drops into it. By how much? We can find out by invoking the principle of conservation of energy. The rocket has kinetic energy and gravitational potential energy: their sum is the rocket’s total energy, which is conserved.

Prior to entering the dip the rocket is moving with velocity $V_{initial}$ and its energy is all kinetic. So its total energy is

$$E_{initial} = {1\over2} M V_{initial}^2$$

In the dip it has both kinetic energy (greater because it is moving faster) and potential energy (negative). The total energy in the dip is

$$E_{in\thinspace dip} = {1\over2} M V_{in\thinspace dip}^2 - MgH$$

where $H$ is the depth of the dip and $g$ the acceleration of gravity.

The law of conservation of energy tells us that the total energy never changes, so that $E_{initial} = E_{in\thinspace dip}$. Writing this down and solving, we find a formula for the rocket’s velocity once it has dropped down a distance $H$:

$$V_{in\thinspace dip} = \sqrt{V_{initial}^2 + 2gH}$$

Problem for the student: carry out this derivation.
Problem for the student: suppose the rocket is moving at 60 miles per hour. How deep a dip is required for it to double its velocity?

Equation 2 above gives the rocket’s velocity in the dip: in Method 2 it is at this point, when the rocket is moving more rapidly, that it fires its engine. This increases the rocket’s velocity from $V_{in\thinspace dip}$ to

$$V_{before\thinspace the\thinspace climb\thinspace upwards} = V_{in\thinspace dip} + \Delta V$$

($\Delta V$ is given by equation 1 above). And finally, the rocket must climb up out of the dip and head off to Mars. In climbing upwards it loses velocity: we can find out how much is lost by applying the principle of conservation of energy all over again. The only differences from our previous use of this principle are

• The rocket is climbing upwards, so its velocity decreases rather than increases
• It started its upwards climb with velocity $V_{before\thinspace the\thinspace climb\thinspace upwards}$ rather than $V_{initial}$

When we go through the calculation we get the rocket’s final velocity once it has climbed up out of the well:

$$V_{final} (Method\thinspace 2) = \sqrt{ V_{initial}^2+ \Delta V^2 + 2\Delta V\sqrt{V_{initial}^2+2gH}}$$

Problem for the student: carry out this derivation.

This is the velocity with which the rocket heads off on its way to Mars according to Method 2.

$$V_{final} (Method\thinspace 1) = V_{initial} + \Delta V$$

WHICH IS BETTER?

At last we have our two answers. And now for the whole point: if we fire the rocket engine with the same force in the two methods, so that $\Delta V$ is the same, which yields the greater final velocity? The two above expressions are a little complicated, so the answer is not immediately obvious. But since the only thing we want to know is which is bigger, we can make our work easier by working not with $V_{final}$ but with $V_{final}^2$ – this gets rid of most of the square roots. If we subtract the two expressions for this we get our final result:

$$[V_{final}(Method\thinspace 2)]^2 - [V_{final}(Method\thinspace 1)]^2 = 2\Delta V\left( \sqrt{V_{initial}^2+2gH} - V_{initial} \right)$$

Even this expression is a little complicated – but the complications don’t matter, because the only question is whether the expression is positive or negative. And the answer is that it is positive (since the term involving the square root is always larger than $V_{initial}$). This proves that Method 2 is better than Method 1: for the same amount of firing of the rocket engine, Method 2 yields a greater final velocity with which the rocket heads off to Mars.

Problem for the student: if the rocket is initially in a circular orbit, the only way it can drop closer to the Earth is by firing its engine. Is it possible that this extra use of fuel will negate the advantage of using Method 2?

Problem for the student: the rocket doesn’t have to be in a circular orbit! What sort of orbit should we give it in order for it not to need to fire its engine in order to get closer to the Earth?

YES -- BUT WHY?

I felt pretty pleased when I saw that my way of looking at the problem, simple though it was, had reached the same result as the exact analysis. But after a while I realized that I was not completely satisfied. Something kept nagging at me. Eventually I realized that the real question is not why the second method is better. It is why the two methods are different at all.

For after all, you might think that the dip shouldn’t matter. Suppose the rocket simply coasts along, not bothering to fire its engine. Since the distance it drops down into the dip is the same as the distance it rises up out of it, the rocket ends up moving just as fast after emerging from the dip as before entering it. The presence of the dip has had not the slightest effect on the rocket’s velocity.

After a certain amount of stewing I managed to express this conundrum in a more productive way. We are thinking about two different ways to accelerate a rocket:

1. Spit out a mass ‘$m$’ and use momentum conservation
1. Drop down a distance ‘$H$’ and use energy conservation

Apparently the order in which we do them matters.  But why? In the long run I decided that it has to do with how the increase in velocity given by each of these depends on the initial velocity of the rocket.

Let us study this.

In the “momentum way” (i), the increase does not depend of $V_{initial}$ at all. You can see this in equation 1 which gives $\Delta V$: nowhere in this formula does $V_{initial}$ appear.

But it does appear when we consider the “energy way” (ii). You can see this from equation 2: $V_{initial}$ does appear in it.

Split the rocket’s motion into two pieces: first it drops down into the dip, and then it rises up out of it. For each piece we can calculate the change in velocity. We do this using our old tried and true conservation of energy: write down the kinetic and potential energies and require them to be equal. Finally, subtract the initial velocity in order to find $\Delta V$, the change in velocity in each piece. The results are as follows:

Dropping down We get

$$\Delta V_{dropping}= \sqrt{V_{initial}^2+2gH} - V_{initial}$$

Figure 3: The gain in velocity while dropping into a well, as a function of the initial velocity. All velocities are presented in units of the escape velocity.

Figure 4: The loss of velocity while climbing out of a well, as a function of initial velocity.

Graphing this formula we see how the change in velocity depends on the initial velocity (figure 3).

Problem for the student. Use the conservation of energy to show that escape velocity is  $\sqrt{2gH}$

Rising up we get

$$\Delta V_{rising} = \sqrt{ V_{initial}^2 - 2gH} - V_{initial}$$

We graph this formula in figure 4.

Notice that both these graphs slope downwards. This means that the faster the rocket is going before dropping down, the less effect dropping down has on its velocity. And similarly for rising up: the faster the rocket is going before starting its upwards climb, the less effect that upwards climb has.

We see --

• If we want the biggest gain in velocity from dropping down, it is smart to start out moving as slowly as possible. So don’t fire the rocket engine before dropping down.
• If we want the smallest loss in velocity from rising up, it is smart to be moving as rapidly as possible before commencing the upwards climb. So make sure you fire the rocket engine before starting upwards.

Finally, since the $\Delta V$ we get from firing the engine does not depend on the rocket’s velocity, we don’t pay any price for delaying this firing until the rocket is down in the dip. So now we understand why method (2) is better.

STILL DEEPER

This insight depends on the fact that the effect of firing a rocket’s engine does not depend on its velocity – and this depends on all sorts of physics involving momentum, Newton’s laws, and the like. But it seems to me that we don’t really need any of that physics at all. We can understand this fact based on reasoning that is almost philosophical in nature. Here’s how:

1. Suppose an observer sees a rocket at rest: it spits out a lump with relative velocity ‘u’ and ends up moving at some final velocity.
2. This observer also sees a second rocket, not at rest.  it spits out a lump with the same mass and relative velocity ‘u’ and ends up moving at some final velocity. What is the change in velocity for this rocket?

Previously I had answered this question by using physics. But there’s a way to answer this question without resorting to any physics at all! Imagine a second observer who is moving relative to first -- at just the same velocity as the second rocket. This observer sees the second rocket to be a perfect clone of the first observer’s first rocket. Therefore, what the second observer sees the second rocket do must be what the first observer saw the first rocket do. And this means that $\Delta V$ is the same for both: it is independent of velocity.

It’s a wonderfully simple argument. You might worry that it could also be used to reach the erroneous conclusion that the effect of dropping into a dip is also independent of velocity. But that’s not true, for the second observer would not see exactly the same situation as the first: the dip would be stationary for one observer and moving for the other.

In the deepest analysis it all depends on the denial of the existence of absolute space. A believer in absolute space would argue that there is a difference between those two observers: one really is moving and the other really isn’t, so there could very well be a difference between what the two observe. Only if there is no such thing as absolute space can we understand my science fiction author’s conundrum.