Releasing the object from Space Elevator

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Discussion Overview

The discussion revolves around the trajectory of an object released from a Space Elevator cabin at an altitude of approximately 100 km. Participants explore the effects of Coriolis force and the object's motion relative to the Earth's rotation, considering various scenarios and conditions that may influence the object's path as it falls towards the Earth.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants propose that the object will not fall straight down due to Coriolis and centrifugal forces, suggesting it will curve as it descends.
  • Others argue that the object will have a linear velocity greater than the cable anchor's velocity, leading to the conclusion that it will land east of the cable anchor.
  • A participant questions whether the angle of the cable's lean changes during ascent or descent of the cabin.
  • There is a suggestion that to avoid the object hitting the tether, it should be released on the appropriate side to ensure it curves away from the cable.
  • One participant discusses the relationship between altitude and the object's motion, indicating that objects released below geostationary orbit will fall downwards and forward, while those above will move upwards and backwards.

Areas of Agreement / Disagreement

Participants express differing views on the trajectory of the object, with no consensus reached on whether it will fall vertically or along a curved path. The discussion remains unresolved regarding the exact nature of the object's descent and the implications of the cable's lean.

Contextual Notes

There are unresolved assumptions regarding the effects of wind on the object's trajectory and the influence of the cable's lean on the object's motion. The discussion also highlights the complexities of the dynamics involved in the scenario.

Eagle9
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Let’s imagine that the Space Elevator is built and we ascend to space with its cabin. The cabin stops temporarily at the altitude of 100 km (or a bit more) and then we release some object (sphere, hemisphere) downwards, the question is: what trajectory will it (object) follow when falling down?
In Wikipedia is written that:
“As the car climbs, the elevator takes on a 1 degree lean, due to the top of the elevator traveling faster than the bottom around the Earth (Coriolis force). This diagram is not to scale” (under the picture).
http://en.wikipedia.org/wiki/Space_Elevator#Climbers
So, when the cabin ascends the cable will be bent due to Coriolis force as shown on the picture.
And I would like to know how exactly the object will fall that we will release from the cabin? Will it fall vertically (straight line) and hence continuously recede from east-going cable or will it fall in a curved line due to the same Coriolis force? I have read that if you release the object (never mind from Space Elevator or other) it will fall exactly vertically on the poles but not on the equator (the Space Elevator will be built on the equator). So, which trajectory will the object “choose”? There are two options:
1. Object falls vertically (red spheres)
2. Object fall along the cable (green spheres) and therefore the distance between cable and object remains the same (let’s forget about atmosphere’s resistance for a while).
w6uly0.jpg

I want to find the solution for the following problem: how to release some object from Space Elevator so that it not to cross/cut Space Elevator’s cable during falling down :smile:
 
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You should throw this into the Classical or General physics forums. I bet you'd get more help there.
 
From a reference frame attached to the surface of the Earth (i.e. a non-inertial rotating reference frame), the object will not fall 'straight' down, but will instead appear to curve due to the coriolis and centrifugal forces. Depending on the nature of the object (e.g. density) the effects of wind could be highly non-negligible.

To avoid having the falling object hit the tether, one would simply drop the object on the appropriate side so that it curves away.
 
Even though there is a 1 degree lean, I think the object will have a linear velocity greater than 25000 miles per 24 hours.
The cable anchor at Earth's surface will have a velocity of 25000 miles per 24 hours.
Therefore, I think the object will strike the suface to the east of the cable anchor.
I think one would arrive at the same conclusion when considering the problem from the standpoint that the Earth is not flat and using angular momentum, energy, etc.
 
zhermes
From a reference frame attached to the surface of the Earth (i.e. a non-inertial rotating reference frame), the object will not fall 'straight' down, but will instead appear to curve due to the coriolis and centrifugal forces. Depending on the nature of the object (e.g. density) the effects of wind could be highly non-negligible.
So, as I understand from your post and my little knowledge from celestial mechanics the objects released from altitude below Geostationary orbit will fall downwards and forward (to East), at Geostationary orbit-stay there motionlessly, above Geostationary orbit-will go upwards and backwards (to West), right? :rolleyes:
http://img39.imageshack.us/img39/3576/graphic1oa.jpg
To avoid having the falling object hit the tether, one would simply drop the object on the appropriate side so that it curves away.
That is the object should be pushed northwards or southwards?

Fun Value
Even though there is a 1 degree lean
By the way I would like to clarify this issue: when the cabin ascends along the the elevator takes on a 1 degree lean. Does this angle changes when the cabin ascends/descends? :smile:
 
Last edited by a moderator:
Eagle9 said:
In Wikipedia is written that:
“As the car climbs, the elevator takes on a 1 degree lean, due to the top of the elevator traveling faster than the bottom around the Earth (Coriolis force). This diagram is not to scale” (under the picture).
http://en.wikipedia.org/wiki/Space_Elevator#Climbers
:smile:

Roger, Eagle9. I have received and copy your transmission sent on July 7.
I think I am trying to catch up to you on this. Here is the way I see it so far, and I am trying to see it in terms of linear velocity.
At 62000 miles the counterweight is stationary above a point on the Earth's surface. Its angular velocity is 2pi radians per 24 hours which translates to a linear velocity of about 389360 miles per 24 hours or about 16,233 miles per hour (ignoring sig.figs). This velocity, by the way, is greater than the normal free satelite at geosynchronous orbit having a 26000 mile radius.
As the elevator ascends I was thinking the counterweight would be pulled down toward Earth - pulled down to an orbit where the tethered geostationary velocity is a bit less than 16,233 mph. Yet the counterweight might not lose any linear velocity (well, no loss if rocket engines could keep it going at the same speed - which I know is not part of the plan). Anyway, not counting loses to the ascending cabin, the counterweight would take up a lower orbiting altitude and somehow not lose linear velocity (or angular momentum). For a short time, it would try to keep going 16,233 mph, but at a lower altitude. For a short adjustment time it would go faster than the geosynchronous linear velocity required for the lower altitude. Due to this greater linear velocity or greater angular velocity, I thought it would move ahead (east) of its normal resting (unloaded) geostationary point above the earth.
To answer your question, I think the angle will change as the cabin ascends and gains linear velocity. But I think rocket engines are needed either at the counterweight or on the cabin to push the cabin from a linear velocity of about 25000 miles / 24 hours or about 1000 mph to about 16000 miles per hour - in order to recover the angular loss or lean.
Finally, as I said on the other space elevator thread, even though cabin or counterweight rockets seem to defeat the purpose or intent of the space elevator, I think we need as many ways to access the opportunities of space travel as we can find.
 
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