# What forces should I consider for my hotel room on the Space Elevator?

I asked a similar question (in a SpaceEx manner) on Space Exploration, but here I've made it a bit more ... worldbuilding.

We have a nice Space Elevator, and I'm in charge of the luxury guest accommodation for the six-day round trip.

Setting aside the design considerations, I am interested in what the experience would be physically.

Standard conceptual 50,000km space elevator attached to a counterweight; accelerating at low altitudes, then a constant velocity, and then a deceleration.

What effects will my guests feel (1) accelerating upwards at low altitudes, (2) constant velocity at very high altitudes, (3) deceleration, but 'looking up' at the Earth, and (4) stationary on the counterweight.

No numbers needed, just "then they fall to the ceiling..." You see, I have to design the suite right, with seatbelts, beds, showers, etc. for the most discerning guest!

The apparent gravity felt from the Earth as one travels up a space elevator is given by:

$g_{apparent} = -G {{M}\over{r^2}}+\omega^2r$

Where $G$ is the gravitational constant, $M$ is the mass of the Earth, $r$ is the distance from that point to Earth's center, and $\omega$ is Earth's rotation speed.

I plotted this as a function of distance from Earth's surface. Down at the surface gravity is normal, shown here as -9.8 $m/s^2$.

At around 35,000 km, the zero gravity point is hit and begins to reverse (this would be geosynchronous orbit if they weren't attached to the Earth). So, as others have noted, there will be a small amount of apparent gravity at 50,000 km, but it's in the opposite direction of Earth's.

It seems then, that one good design for a hotel room is one which can flip its orientation. As the elevator begins it departure from Earth it can begin slowly and then continue to increase its acceleration to make up for lost gravity, or at least taper the effect to ease guests into zero gravity.

Flipping the orientation of the hotel room at around 10 km above the ground while beginning to decelerate will provide a zero gravity coast through the natural zero gravity point. Additionally this will help reduce the vertigo felt by people at lower altitudes. Finally settling at around 2% Earth gravity with a fantastic view out of the skylights of the Earth below.

• If I am at the counterweight, would my guests not feel significant gravitational "pull" outward, much like a rotating space station? – Mikey Dec 31 '16 at 22:11
• @Mikey They might feel a slight outward pull. The above plot is essentially the centrifugal force (rotating space station force) minus the opposing gravitational force at that same radius from the Earth's center. The Earth's rotation is relatively slow, so even at 50,000km the centrifugal force is only 0.027g. However, lucky for us, Earth's gravity at that point is only 0.013g, so the net force is outward. But only a physicist would call 0.01g significant, for your guests they'll just notice that things seem to settle on one side of the room given enough time. – Samuel Jan 3 '17 at 17:54

I did the math for extra credit.

1. is easy: just like in an elevator at its lowest point, your distinguished guests will feel heavier. How much heavier depends on how fast it is accelerating upwards. Specifically, they will feel 1+(a/9.8) times as heavy, where a is acceleration in m/s^2.
2. is also easy: if they aren't accelerating, it is just like standing still. As they get farther away from Earth, they will slowly start feeling lighter. Specifically, they will feel 6000000^2/(6000000+e)^2 times as heavy, where e is the elevation above the surface of the Earth in m.
3. is the complicated part: it depends on how fast they are decelerating. If they are decelerating slowly (less than 1G), they will feel lighter, but still right-side up, like when an elevator is at its highest point. With such high altitudes this will barely be noticeable. At 1G, they will be perfectly weightless. Above 1G, they will feel flipped upside down and be pulled slightly towards the ceiling of the elevator. Specifically, they will feel an acceleration of a-9.8*g(e), where a is the acceleration and g(e) is the weight from step 2. You need to be careful to limit the accelerations to about 5G in the first step and about 2G in the third. You don't want your esteemed passengers dying on you (too much paperwork).
4. is again simple: we plug in the elevation to the equation from step 2 to find that gravity at 50000000m is about 1.1% of surface gravity, which is not quite negligible. If they are very careful, they should be able to stand on the earthward wall of the station. The slightest push will send them careening out into the same weightless environment they have on space stations.
• #4 - Is it as easy as this? I imagined there would be some kind of pull, but I suppose that is on the counterweight, not on my guests chilling out at the cocktail bar on the counterweight. They're indeed experiencing the same sensation as on space stations? – Mikey Jun 2 '15 at 8:23
• not perfectly so: items would slowly drift to one wall. A brick dropped down a 100 meter shaft would take 44 seconds to reach the bottom but by then it would be traveling at about 4.4m/s and would hit you like a brick dropped from about 1.25 meters above your head at ground level. – Murphy Jun 2 '15 at 10:31
• Not quite right, you left out the math that makes space elevators work. If the counterweight experiences net acceleration towards the planet (which would be the case if "they should be able to stand on the earthward wall of the station") then the entire system will fall out of the sky. The apparent acceleration would be in the opposite direction. – Samuel Jun 2 '15 at 20:27
• My mistake - I didn't account for rotation. This answer should be correct for a very tall tower supporting its own weight, but not a counterweight supported by centrifugal force. – evankh Jun 3 '15 at 0:54