Lunar Elevators

I recently saw a cool exhibit about space exploration at the American Museum of Natural History.

As part of the exhibit, it was proposed that a lunar elevator be built that would greatly ease take-offs from the moon.  In particular, one could imagine a long cable connecting the surface of the moon to a top platform thousands of miles in space which seemingly floats without anything holding it up.

How could you do this? Imagine being at a point somewhere between the moon and earth. If you're on the moon, the moon's gravity is strongest and you'll be pulled towards it. If you're on earth, the opposite happens. Now as you move away from the moon and closer and closer to the earth, there exists a point at which the earth's pull and the moon's pull exactly equal each other out, and there is a zero net gravitational force.

I asked my Physics class to figure this point out today.  In particular, supposed that a particle is positioned right between the earth and moon, at distance x from the earth. What should x be in order for the gravitational forces to cancel each other out?

To solve the problem, use Newton's law of gravity and set the earth's gravitational acceleration equal to that of the moon's:

(Eq. 1)      GM / x ²  = Gm / (D-x)²

where G is Newton's gravitational constant, M  is the earth's mass, m is the moon's mass, and D is the distance between the earth and moon.  Solve this using the quadratic formula to obtain:

x = D( 1 ± / √ (m / M) / (1 - m / M )

Plugging in the following values

• D = 3.85 e +8  m
• M = 5.97 e +24 kg
• m = 7.475 e +22 kg

We get 

x ≈ 0.89 D ≈ 3.4 e +8 m

In other words, the platform should be around 89% of the way from the earth to the moon, or about 340,000 km from earth and about 45,000 km from the moon.

The analysis above was only a first approximation. A better approximation should take the centrifugal force into account. But how good of an approximation was the student calculation?  A little analysis shows that Eq. 1 above should be modified as follows:

(Eq. 2)     (2 π / T )² x = GM / x ²  - Gm / (D-x)²

where the additional variable T represents the period of the moon around the earth and is equal to 27.3 days (Hint: this is 1/12 less than a lunar month....). Unfortunately, this requires solving a quintic. However, one can solve this equation numerically using a program such as Mathematica, or Wolfram Alpha. In fact, here is a solution using Wolfram Alpha where I've plugged in

• T  = 2.36 e+6 s
• G = 6.67 e-11 m³ / (kg s²)

into Eq. 2.

We can see that there is one real solution

x ≈ 3.25 e +8 m

So the second approximation puts the elevator platform 325,000 km from earth and 60,000 km from the moon. Does it makes sense that taking into account the centrifugal force we need to put the platform closer to earth than the first approximation of 340,000? Sure, because if we put the elevator at our first approximation point, the gravity would equal out and thus the centrifugal force would simply "pull" the platform away from earth. Thus, taking centrifugal force into account requires putting the platform closer to earth, in fact, 15,000 km closer.

Finally, this is still not the best approximation. We've totally ignored all the cables and other material hanging between the platform and the surface of the moon. These cables etc. are required to make the elevator operational so that personell can be transported between the platform and lunar surface. Solving the third approximation problem requires deciding on a design, and the answer of where the platform should be located will depend on which design is used. Additionally, even if we settle on a particular design, computing the location of platform is non-trivial as it requires numerically intebrating a solution to Eq. 2 over the entire length of the elevator.

However, we can make a simple deduction: Each mass below the solution to Eq. 2 will cause a pull towards the moon on the entire elevator which will need to be counterbalanced by making the platform a little higher. I.e., we can be certain that any elevator design will need to have it's platform located closer to the earth than 325,000 km.

Using Emmy Noether's Ideas to Prove All Numbers Interesting

A great article about Emmy Noether recently appeared in the New York Times. My colleague Jon Rosenshine noticed it too. We wondered whether her ideas were accessible to eighth and ninth graders. So I did a little research and discovered that Noether is responsible for Noetherian induction, which is a proof technique used in algebra. Though it is a very powerful technique in higher mathematics, in it's most simple form it can be used to prove facts encountered in middle and high school. For example, one can use Noetherian induction to prove that every positive number greater than 1 is either prime, or can be decomposed as a product of prime numbers.

To illustrate how Noetherian induction works, let's instead prove the following pseudo-theorem:

All whole numbers are interesting.

This is a rather odd statement. But let's see if it makes any sense. Start with 0. Certainly 0 is an interesting number. It represents the absence of something. No other number has this property which shows that it is interesting. 1 is also quite interesting for many reasons. For example, it is the only number that cannot be expressed as a sum of smaller positive numbers, and also it is the only number that when multiplying other numbers does not change them (e.g. 1*3 = 3, 1*9=9, and so on). 2 is also very interesting. It is the smallest prime number, though other reasons make it interesting too. 3 is interesting as it is the smallest odd prime number. 4 is interesting, as it is the only positive number that is the product and sum of the same number ( 2+2 = 2*2 = 4 ). I could go on but proving individually that every number is interesting will make this blog infinitely long, and that's a problem. So instead, let's use the Noether's powerful induction technique to prove it.

The proof goes as follows:

Consider the set of all non-interesting numbers, and call that set S. Noether's induction principle, applied to whole numbers says that any non-empty set of whole numbers must contain a smallest element. If S were non-empty, apply Noether’s induction principle to find a minimal element, call it u (u for un-interesting). But don’t you think that u -the smallest un-interesting number is pretty darn interesting? After all, it is the smallest amongst all un-interesting numbers, which makes it quite unique, and therefore interesting. Thus we have arrived at a contradiction since we have found a number that is both interesting and un-interesting and our original supposition that S is non-empty must be false. I.e., all numbers must be interesting.

What makes this is a pseudo-proof is that there is no precise notion of what interesting means and the proof is appealing to the reader’s emotion, but this basic proof method can be used to give real proofs of real theorems such as the prime number theorem mentioned above (that theorem is called the fundamental theorem of arithmetic).