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)2where 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 mIn 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)2where 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)
We can see that there is one real solution
x ≈ 3.25 e +8 mSo 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.
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