
Space Elevators Could Totally Work—If Earth Days Were Much Shorter
But maybe a more germane question is, why do physicists ask nutty questions like this? It's never going to happen—just move on, right? Well, here's the deal. Thinking about counterfactual scenarios gives us insight into how things work here in reality. Plus, it's fun! Need I say more? OK, it might even help us build a working space elevator.
Oh, you don't know what a space elevator is? It's a sci-fi staple, a tether from Earth up to an orbiting space station in geosynchronous orbit. A cable-climbing car would ride up and down, just like a regular elevator. Basically it's a way of getting out to space as easily and routinely as you ride an elevator to your office in the morning—no rockets required.
Let's start off with some basic questions and build up to some more complicated physics. What Is a Day?
You can't get more basic than that. But the answer isn't simple. If you said a day is 24 hours, you're right—and wrong. If you're standing outside, the time at which the sun is at its highest point in the sky is called local noon. If you stand there until the next local noon, the amount of elapsed time is defined as 24 hours. So an hour is 1/24th of the time between two local noons.
But wait! This isn't the same as a complete revolution of the Earth. If you measured the time of a complete revolution, you'd find that it's not exactly 24 hours. The reason is that the Earth is doing two things at once: It's spinning on its axis, which causes the sun to appear to move across the sky. But it's also orbiting the sun over the course of a year, which means a complete rotation will not result in the sun being in the same position in the sky.
There are actually two different types of days. The solar day is the one you are thinking of, and it's the one described above. The other type is called a sidereal day . Here's a totally not-to-scale diagram that will help you understand the difference:
In position 1, there is a stick marking a location. That stick is pointing toward the sun so that this would be local noon. As the Earth moves to position 2, it makes one complete rotation. However it's not yet local noon, because the relative position of the sun has changed due to the Earth's orbital motion. This is called a sidereal day.
Finally, the Earth moves a little bit beyond one complete revolution, so the stick again points towards the sun for a second local noon. The sidereal day is just a little bit shorter—approximately 23 hours and 56 minutes.
Why does that matter? Well, if we are going to make a day half as long, we need to decide which one to divide by 2. Just for simplicity, let's say the solar day is 12 hours instead of 24 hours, but the orbit around the sun (and the length of a year) is the same. You'd Feel Lighter at the Equator
There are many things that would change with a 12-hour day. Like, how long would you sleep? Would we still work 40 hours a week? Would a week still be seven days (and still named after objects in the sky?). But let's focus on some of the physics stuff.
Here's the fun part. If you stood on a scale at the north pole and then did the same thing at the equator, the scale would give a higher value at the north pole. Actually, it's true for both a 24-hour day and a 12-hour day—but it's more noticeable with a shorter day. Let's start at the north pole. Here is a force diagram for a normal human standing on a scale:
There are two forces acting on the person. First, there is the downward pulling gravitational force due to the interaction with the Earth. (This is the mass, m, multiplied by the gravitational field, g.) Second, there is the upward-pushing force from the scale (we call this a normal force since it's perpendicular to the ground). The reading on the scale is actually the magnitude of the normal force and not the weight. Newton's second law states that the net force on an object is equal to the product of the mass and acceleration. For a person at the north pole, the acceleration would be zero (they are just standing there). That means that the normal force is equal in magnitude to the gravitational force.
What if you are instead standing on the equator? Here's a force diagram for that.
Isn't it the same except sideways? No, it's different. Notice that in this case the normal force isn't as strong as the gravitational force (the arrow is shorter). This is because a person standing at the equator is not stationary. They're moving in a circular path as the Earth rotates. When an object moves in a circle, it has an acceleration toward the center. This centripetal acceleration has a magnitude that increases with the angular velocity (ω) as well as the radius of the circular path (r).
The sum of the two forces (gravity and the scale) must equal the mass multiplied by the acceleration. This means that the force of the scale will be:
Why is the north pole different? Yes, you are still rotating, but you are ON the axis of rotation, so the radius (your distance from the axis) is zero, and that gives you a zero acceleration. If you use an angular velocity for a 24-hour day, your effective weight at the equator is 99.7 percent of the value at the north pole. With a 12-hour day (which means the Earth is spinning twice as fast and your angular velocity is twice as high), the scale would read a value that's 98.6 percent of the actual gravitational force. The faster you spin, the lighter you are.
Would you notice that in real life? I think that if you flew straight from the north pole to the equator, you might feel a change in effective weight of over 1 percent. With this lower weight, you could jump just a little bit higher and walk around with a lighter step. Space Elevators
Let's think about orbits for a moment. If you put an object near the Earth, there will be a downward-pulling gravitational force. As you get farther away from the surface of the Earth, this gravitational force gets weaker. However, if you have an object in space that's initially at rest, the gravitational force will cause it to fall down and crash. But wait! If we use the same circular motion trick for the effective weight we can make the object move in a circle such that the mass multiplied by the centripetal acceleration is equal to the gravitational force. It would be the same as standing on a scale with an effective weight of zero. We call this a circular orbit.
The rate that an object orbits depends on the distance from the center of the Earth (r). We can calculate that as:
Here G is the universal gravitational constant and M is the mass of the Earth. If you put in a value of r that is 400 kilometers above the surface of the Earth, you get an angular velocity that would take the object 92 minutes to complete an orbit. Note: This is pretty much what the international space station (ISS) does.
Wouldn't it be cool if the International Space System had a cable running down to the Earth? Unfortunately, the dangling cable would be whipping around the Earth so fast, you wouldn't be able to embark or disembark.
Well, it's possible to fix this problem. Suppose you move the space station up to a distance of 36,000 kilometers instead of 400 kilometers? In that case, the angular velocity of the ISS would be the same as the rotation rate of the Earth. As seen from the surface of the Earth, the ISS would remain in the same spot in the sky because they would both take 24 hours to rotate. We call this a geostationary orbit—but it has to be directly over the equator so that the direction of the rotations are the same.
With an object in geostationary orbit, you could run a cable down to the Earth. Boom —there's your space elevator. But wait! There are some problems. Can you imagine a cable that's 36,000 kilometers long? That's a LOT of cable. It's so much that you'd also have to counterbalance the weight of the cable with some big mass a little past the geostationary level. This system would require a tension in the material that exceeds the maximum value for the strongest steel cables. It could only work with something like a carbon nanotube cable—which we don't have (yet).
OK, but what if we make the Earth spin twice as fast with a 12-hour day? In that case, a geostationary orbit would have a larger angular velocity (to match the faster Earth). If you crunch the numbers, the geostationary distance would be only 20,000 kilometers, or around 45 percent shorter.
What if the Earth rotated so fast that the ISS was in a geostationary orbit just 400 kilometers above the surface? That might make the space elevator possible. Of course now we are going to have a MUCH shorter day of only 92 minutes. That's not worth it. Can you imagine having to get out bed every 92 minutes? I might even get dizzy. It's too bad because I really want a space elevator.
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