| Topic: |
Science > Physics |
| User: |
"Cyde Weys" |
| Date: |
11 Dec 2004 09:07:25 PM |
| Object: |
Another question about space elevators - gravity |
What would it feel like to ride up in a space elevator. As you get
on at ground level, obviously you still feel 9.8 m/s^2 acceleration. But
what happens as you start going up the elevator? Goes gravity decrease
according to F=Gm1m2/r^2? If that was the case, then gravity would be
less, but not negligible, at geosynchronous orbit. But if the space
elevator is in orbit, shouldn't you feel no gravitational acceleration at
its center of mass? These seem to be contradictory. Which is true? How
would one go about finding an equation for the gravitational acceleration a
passenger feels along a space elevator, such that it equals zero at
geosynchronous orbit, if that is indeed the case? And what would you feel
past geosynchronous orbit? Would you feel like you're being slingshotted
away from the Earth?
--
~ Cyde Weys ~
Sing me that sweet headcrab elegy.
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| User: "Sam Wormley" |
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| Title: Re: Another question about space elevators - gravity |
11 Dec 2004 10:09:19 PM |
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See: http://en.wikipedia.org/wiki/Space_elevator
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| User: "Sam Wormley" |
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| Title: Re: Another question about space elevators - gravity |
11 Dec 2004 10:54:20 PM |
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Sam Wormley wrote:
See: http://en.wikipedia.org/wiki/Space_elevator
Ribbon to the Stars
Pushing the space elevator closer to reality
http://www.sciencenews.org/articles/20021005/bob9.asp
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| User: "Sam Wormley" |
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| Title: Re: Another question about space elevators - gravity |
11 Dec 2004 11:06:30 PM |
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Sam Wormley wrote:
See: http://en.wikipedia.org/wiki/Space_elevator
http://www.isr.us/Downloads/niac_pdf/chapter1.html
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| User: "Lewis Mammel" |
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| Title: Re: Another question about space elevators - gravity |
11 Dec 2004 10:14:29 PM |
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Cyde Weys wrote:
What would it feel like to ride up in a space elevator. As you get
on at ground level, obviously you still feel 9.8 m/s^2 acceleration. But
what happens as you start going up the elevator? Goes gravity decrease
according to F=Gm1m2/r^2? If that was the case, then gravity would be
less, but not negligible, at geosynchronous orbit. But if the space
elevator is in orbit, shouldn't you feel no gravitational acceleration at
its center of mass? These seem to be contradictory. Which is true? How
would one go about finding an equation for the gravitational acceleration a
passenger feels along a space elevator, such that it equals zero at
geosynchronous orbit, if that is indeed the case? And what would you feel
past geosynchronous orbit? Would you feel like you're being slingshotted
away from the Earth?
If R is the radius of the earth, the downward force you feel moving
along a line straight up from the earths surface, at constant speed,
is
g/(r/R)^2 - omega^2 r/R
where omega is 2pi/day ( Siderial day, to be exact. ) That of course
is the centrifugal force due to the rotating frame of reference.
It's what makes the Missisippi River flow, by the way. The Mississippi
River flows against the force of gravity, strictly speaking.
At the geosynchronous altitude, gravity and centrifugal force
cancel each other.
These are "g-forces" or "body forces" which effect every
element of mass equally. You experience them as forces
only because of the reaction forces which constrain you
to hold your position in the noninertial frame of reference.
There is an additional sideways "g-force" of "2 omega cross v", where
v is your upward velocity. The earth is rotating to the east, and if
you shoot a projectile straight up, it will fall behind to the west
as it rises, because it would have to move faster towards the east
than its initial eastward speed, if it were to remain directly
above its launch point, and because the launch pad and the rest of
the earth frame rotates (i.e. tilts, as opposed to sliding ) so that
the launch velocity appears to veer in the opposite direction of the
tilt. These two reasons are actually the source of the factor of 2
in the expression for the coriolis force, which is said to be
acting on the moving body and causing it to veer.
The rising elevator would be "thrown to the west" by the coriolis
force just as you are "thrown against the door" by the centrifugal
force when your car goes around a curve.
At an upward speed of 500 m/sec, the coriolis force is about
0.07 g .
Beyond the geosynchronous point, the centrifugal force is greater
than the gravitational force, so you would feel you were being
flung away, as long that is, as you were prevented from doing
so by a restraining force.
Lew Mammel, Jr.
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| User: "Cyde Weys" |
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| Title: Re: Another question about space elevators - gravity |
12 Dec 2004 03:12:35 AM |
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Lewis Mammel <l.mammel@worldnet.att.net> wrote in
news:41BBC655.D28AC8D8@worldnet.att.net:
Cyde Weys wrote:
What would it feel like to ride up in a space elevator. As
you get
on at ground level, obviously you still feel 9.8 m/s^2 acceleration.
But what happens as you start going up the elevator? Goes gravity
decrease according to F=Gm1m2/r^2? If that was the case, then
gravity would be less, but not negligible, at geosynchronous orbit.
But if the space elevator is in orbit, shouldn't you feel no
gravitational acceleration at its center of mass? These seem to be
contradictory. Which is true? How would one go about finding an
equation for the gravitational acceleration a passenger feels along a
space elevator, such that it equals zero at geosynchronous orbit, if
that is indeed the case? And what would you feel past geosynchronous
orbit? Would you feel like you're being slingshotted away from the
Earth?
If R is the radius of the earth, the downward force you feel moving
along a line straight up from the earths surface, at constant speed,
is
g/(r/R)^2 - omega^2 r/R
where omega is 2pi/day ( Siderial day, to be exact. )
Is a Cydereal day 24 hours, or is it the 23 hours, 56 minutes, xx
seconds day? (Hehe, get it, Cydereal instead of sidereal? Gosh, I kill
myself)
More seriously, in that equation, is r the distance from the Earth's
surface, or from the center of mass of the Earth?
That of course
is the centrifugal force due to the rotating frame of reference.
It's what makes the Missisippi River flow, by the way. The Mississippi
River flows against the force of gravity, strictly speaking.
I'd never heard of that. Is the top of the Mississippi closer to the
center of mass of the Earth or something? I'm guessing this only works
for north-south rivers, not east-west rivers? Come to think of it, all
the large rivers are north-south .... hrmmm ...
At the geosynchronous altitude, gravity and centrifugal force
cancel each other.
Wow, that's really neato. It's like another definition of
geosynchronous orbit. Awesome.
These are "g-forces" or "body forces" which effect every
element of mass equally. You experience them as forces
only because of the reaction forces which constrain you
to hold your position in the noninertial frame of reference.
Gotcha. I have a question though ... why is the space elevator held
strictly vertical? You'd think the parts past geosynchronous orbit
would tend to want to slow down and the parts further in would want to
speed up. Is the only thing keeping it together the tensile strength of
the material? And if so, wouldn't there still be a little bit of shape
changing (trailing past geosynch)?
Another question - if the Earth wasn't spinning, would we feel a
different acceleration at the Earth's surface? Would we feel more
attracted to the Earth? It seems like the centripetal acceleration is
working in the opposite direction of gravity, so killing the rotation
would remove a force that works opposite to gravity.
There is an additional sideways "g-force" of "2 omega cross v", where
You sure that's not a cydeweys g-force?
v is your upward velocity. The earth is rotating to the east, and if
you shoot a projectile straight up, it will fall behind to the west
as it rises, because it would have to move faster towards the east
than its initial eastward speed, if it were to remain directly
above its launch point, and because the launch pad and the rest of
the earth frame rotates (i.e. tilts, as opposed to sliding ) so that
the launch velocity appears to veer in the opposite direction of the
tilt. These two reasons are actually the source of the factor of 2
in the expression for the coriolis force, which is said to be
acting on the moving body and causing it to veer.
The rising elevator would be "thrown to the west" by the coriolis
force just as you are "thrown against the door" by the centrifugal
force when your car goes around a curve.
At an upward speed of 500 m/sec, the coriolis force is about
0.07 g .
Beyond the geosynchronous point, the centrifugal force is greater
than the gravitational force, so you would feel you were being
flung away, as long that is, as you were prevented from doing
so by a restraining force.
So once you reached geosynch orbit in the space elevator, you'd become
weightless, and past that point, the floor would become the ceiling and
vice versa? Awesome stuff!
--
~ Cyde Weys ~
Sing me that sweet headcrab elegy.
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| User: "Lewis Mammel" |
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| Title: Re: Another question about space elevators - gravity |
12 Dec 2004 05:04:01 PM |
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Cyde Weys wrote:
Lewis Mammel <l.mammel@worldnet.att.net> wrote in
news:41BBC655.D28AC8D8@worldnet.att.net:
If R is the radius of the earth, the downward force you feel moving
along a line straight up from the earths surface, at constant speed,
is
g/(r/R)^2 - omega^2 r/R
where omega is 2pi/day ( Siderial day, to be exact. )
Is a Cydereal day 24 hours, or is it the 23 hours, 56 minutes, xx
seconds day? (Hehe, get it, Cydereal instead of sidereal? Gosh, I kill
myself)
More seriously, in that equation, is r the distance from the Earth's
surface, or from the center of mass of the Earth?
From the center of course, except I misstated it. It s/b
g/(r/R)^2 - omega^2 R (r/R)
.... so at r=R we have earth surface conditions.
These are "g-forces" or "body forces" which effect every
element of mass equally. You experience them as forces
only because of the reaction forces which constrain you
to hold your position in the noninertial frame of reference.
Gotcha. I have a question though ... why is the space elevator held
strictly vertical? You'd think the parts past geosynchronous orbit
would tend to want to slow down and the parts further in would want to
speed up. Is the only thing keeping it together the tensile strength of
the material?
Not the tensile strength, the tension. The gradient of the tension
must exactly balance the gravity-plus-centrifugal force.
And if so, wouldn't there still be a little bit of shape
changing (trailing past geosynch)?
Not ideally. In actuality there would be issues with all kinds
of tension waves and what not. It's a theoretically stable
configuration against "small" displacements, but what's
"small" for such a monstrously extended object ?
Another question - if the Earth wasn't spinning, would we feel a
different acceleration at the Earth's surface? Would we feel more
attracted to the Earth? It seems like the centripetal acceleration is
working in the opposite direction of gravity, so killing the rotation
would remove a force that works opposite to gravity.
"Centrifugal force" acts oppositely to gravity. The centripetal
acceleration is the acceleration towards the center of circular
motion. So we have: ( in vector terms )
F_gravity + F_ground = mass * a_centripetal
and F_ground acts upward, but is lesser in magnitude than F_gravity .
In the rotating frame, we claim we are not accelerating, but
we have centrifugal force acting:
F_gravity + F_centrifugal + F_ground = 0
so in vector terms, F_centrifugal = - mass * a_centripetal
Note g can be thought of as an acceleration, or a force per unit mass.
m/sec2 = newton/kg
The magnitude of the centrifugal force ( per unit mass ) is
about 0.0034 g in the earth frame of reference near the surface.
Lew Mammel, Jr.
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