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A math question about acceleration over a distance
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STJensen
Dec 31, 2011, 10:49:23 PM12/31/11
to
Let us say that you had a 62,000-mile-long Earth-anchored space
elevator and let us say it has an electro-magnetic repulsion
accelerator along its entire length. If you were to accelerate a
human so that he experienced only 2g (twice the force of gravity)
during the entire length of the space elevator, what velocity would
that human be expelled from the space elevator and how long would it
take the human to travel the entire length?
Where I'm having problems with the above question is the need for the
human to be constantly increasing in speed during the entire length of
the space elevator. As soon as the human goes at the same speed, the
sensation of two gravities would stop.
32 feet (9.8 meters) per second squared = 1 gravity
So 64 feet per second squared = 2 gravities?
But that's for the first second. For the second second, the human
would have to be going at 128 feet per second squared to continue to
feel the force of two gravities, right?
If the above question isn't hard enough, there are two other issues to
deal with: 1) when the human is on the Earth's surface, there is also
the Earth's gravity but as the human goes up the space elevator, that
gravitational force decreases. This would mean a slower take-off
speed so the human only experiences 2gs at all times. 2) once the
human passes the 22,000 mile mark, centrifugal force begins to push
the human away from the Earth so the further away from the 22,000-mile
point, the more outward force is being felt by the human and his
acceleration would have to be increased even more so that the human
would feel 2gs in the opposite direction he is traveling.
So without the impact of Earth's gravity and then centrifugal force
past 22,000 miles, how fast would a human be traveling at the end of
62,000 miles and how long would he take to travel that 62,000 miles?
Now if someone knows how Earth's gravity and post-22,000-mile-point
centrifugal force would change the previous answer, I would love to
know that too.
Lastly, is there an equation for calculating the above? If so, I
would appreciate knowing it since I could then enter any length and
see what the final velocity would be.
Thanks in advance!
Scott Jensen
elevator and let us say it has an electro-magnetic repulsion
accelerator along its entire length. If you were to accelerate a
human so that he experienced only 2g (twice the force of gravity)
during the entire length of the space elevator, what velocity would
that human be expelled from the space elevator and how long would it
take the human to travel the entire length?
Where I'm having problems with the above question is the need for the
human to be constantly increasing in speed during the entire length of
the space elevator. As soon as the human goes at the same speed, the
sensation of two gravities would stop.
32 feet (9.8 meters) per second squared = 1 gravity
So 64 feet per second squared = 2 gravities?
But that's for the first second. For the second second, the human
would have to be going at 128 feet per second squared to continue to
feel the force of two gravities, right?
If the above question isn't hard enough, there are two other issues to
deal with: 1) when the human is on the Earth's surface, there is also
the Earth's gravity but as the human goes up the space elevator, that
gravitational force decreases. This would mean a slower take-off
speed so the human only experiences 2gs at all times. 2) once the
human passes the 22,000 mile mark, centrifugal force begins to push
the human away from the Earth so the further away from the 22,000-mile
point, the more outward force is being felt by the human and his
acceleration would have to be increased even more so that the human
would feel 2gs in the opposite direction he is traveling.
So without the impact of Earth's gravity and then centrifugal force
past 22,000 miles, how fast would a human be traveling at the end of
62,000 miles and how long would he take to travel that 62,000 miles?
Now if someone knows how Earth's gravity and post-22,000-mile-point
centrifugal force would change the previous answer, I would love to
know that too.
Lastly, is there an equation for calculating the above? If so, I
would appreciate knowing it since I could then enter any length and
see what the final velocity would be.
Thanks in advance!
Scott Jensen
James Waldby
Jan 1, 2012, 10:36:22 AM1/1/12
to
On Sat, 31 Dec 2011 22:49:23 -0500, STJensen wrote:
> Let us say that you had a 62,000-mile-long Earth-anchored space elevator
> and let us say it has an electro-magnetic repulsion accelerator along
> its entire length. If you were to accelerate a human so that he
> experienced only 2g (twice the force of gravity) during the entire
> length of the space elevator, what velocity would that human be expelled
> from the space elevator and how long would it take the human to travel
> the entire length?
I've snipped some paragraphs where you confuse velocity and acceleration
> Let us say that you had a 62,000-mile-long Earth-anchored space elevator
> and let us say it has an electro-magnetic repulsion accelerator along
> its entire length. If you were to accelerate a human so that he
> experienced only 2g (twice the force of gravity) during the entire
> length of the space elevator, what velocity would that human be expelled
> from the space elevator and how long would it take the human to travel
> the entire length?
> If the above question isn't hard enough, there are two other issues to
> deal with:
think your take on centrifugal force treats it as a source of free
energy, which so far as I know it isn't.]
> Lastly, is there an equation for calculating the above? If so, I would
> appreciate knowing it since I could then enter any length and see what
> the final velocity would be.
Ignoring centrifugal force, a(t) is 2*g0 - g0*(re/(re+h(t)))^2.
[See 1] For h(t) you need the value of the integral of v(t) dt.
In short, v(t) depends on a(t), which depends on h(t), which
depends on v(t).
I don't know whether the integrals have closed forms, but a
simple program can give adequately-accurate results as shown by
testing the following Python program with a range of values for
dt. I tried several dt in the range 0.001 to 0.1 and for each
got the same results within 4 decimal places.
>>> g0=9.80665 # surface gravity, m/s^2
>>> re=6371000. # mean earth radius, m
>>> dt=0.05 # time increment, sec
>>> hmax=100000000. # elevator height, m
>>> v=0; h=0; t=0
>>> while h<hmax:
... h += v*dt
... p = re/(re+h)
... v += (2*g0 - g0*p*p)*dt
... t += dt
...
>>> print "At t=%f, v=%f m/s" % (t,v)
At t=3548.900000, v=61686.788167 m/s
Anyway, suppose this 100,000-km elevator goes straight up (which
probably is a bad assumption, but no other is handy). Rotational
speed of a stationary object at Earth surface is (2*pi*re)/(24*3600)
m/s or 463 m/second. At the top, (2*pi*(re+h))/(24*3600) is about
7700 m/sec. This change in velocity over about an hour amounts to an
0.2 g acceleration the program doesn't account for, because it only
looks at radial velocity. That may introduce about 10% error. The
program also ignores centrifugal force.
[1] The formula for gravity-versus-altitude is from wikipedia at
<http://en.wikipedia.org/wiki/Gravity_of_Earth#Altitude>, g0 is
surface gravity, and re = 6371 km = mean radius of Earth.
--
jiw
William Elliot
Jan 1, 2012, 10:36:29 AM1/1/12
to
On Sat, 31 Dec 2011, STJensen wrote:
> Let us say that you had a 62,000-mile-long Earth-anchored space
> elevator and let us say it has an electro-magnetic repulsion
> accelerator along its entire length. If you were to accelerate a
> human so that he experienced only 2g (twice the force of gravity)
> during the entire length of the space elevator, what velocity would
> that human be expelled from the space elevator and how long would it
> take the human to travel the entire length?
>
If you disregard angular momentum by assuming the Earth doesn't rotate,
> Let us say that you had a 62,000-mile-long Earth-anchored space
> elevator and let us say it has an electro-magnetic repulsion
> accelerator along its entire length. If you were to accelerate a
> human so that he experienced only 2g (twice the force of gravity)
> during the entire length of the space elevator, what velocity would
> that human be expelled from the space elevator and how long would it
> take the human to travel the entire length?
>
air resistance and weakening of gravity by distance from center of Earth,
the problem becomes:
If a person accelerates at one g for 62,000 miles, what is the
terminal speed and how long does it take to reach terminal speed?
d = distance, v = velocity, a = acceleration
Set d(0) = 0 = v(0), a = g.
v = integral(0,t) a dt = at
d = integral(0,t) v dt = integral gt dt = gt^2 / 2
2d = gt^2; t = sqr 2d/g (= 76 min)
v = at (= 28 miles/sec, if I've done the calculations correctly.)
--
If you disregard angular momentum by assuming the Earth doesn't rotate,
and air resistance, then the time will be shorter, the terminal speed
larger and the calculation more complex.
r = radius of Earth; s = distance above earth
Set s(0) = 0 = v(0).
Force of gravity at s is gr^2 /(r + d)^2.
(Is there a physicist in the house?)
Net acceleration at s is 2g - gr^2 /(r + d)^2.
2g - gr^2 /(r + s)^2 = dv/dt = d^2 s/dt^2 = s"
2g(r + s)^2 - gr^2 = (r + s)^2 s"
gr^2 + 4grs + 2gs^2 = (r + s)^2 s"
If I set it up right, that's a hard to solve differential equation.
----
David Bernier
Jan 1, 2012, 11:18:04 AM1/1/12
to
On Dec 31 2011, 10:49 pm, STJensen <recreationalpo...@gmail.com>
wrote:
Gravity attracts downwards, in the amount mg .
The net force is then boost - mg.
F ma, so a F/m (boost - mg)/m boost/m - g.
The mg , gravity, depends on elevation, so calling
g g_local is better.
Then acceleration boost/m - g_local .
A boost per kilogram of 9.8 Newtons/kg at earth's
surface means acceleration 0.
I think this boost per kilo would be called '1g' ...
A boost (rocket force) per kilo of 19.6 Newtons/kg
would seem like 2g, but produce an acceleration of
boost/m - g_local 19.6 Newtons/kg - g_local,
so at "inifinity" acceleration ~ 19.6 Newtons/kg 19.6 m/sec/sec .
With a space elevator that's anchored to Earth,
things that don't move relative to earth woul be
Ok. But the human in the space-elevator, being
in motion relative to earth, would be subject
to a Coriolis force:
< http://en.wikipedia.org/wiki/Coriolis_effect > .
62,000 miles is about 7.5 earth diameters.
Dave
> Where I'm having problems with the above question is the need for the
> human to be constantly increasing in speed during the entire length of
> the space elevator. As soon as the human goes at the same speed, the
> sensation of two gravities would stop.
>
> 32 feet (9.8 meters) per second squared 1 gravity
>
> So 64 feet per second squared 2 gravities?
wrote:
> Let us say that you had a 62,000-mile-long Earth-anchored space
> elevator and let us say it has an electro-magnetic repulsion
> accelerator along its entire length. If you were to accelerate a
> human so that he experienced only 2g (twice the force of gravity)
> during the entire length of the space elevator, what velocity would
> that human be expelled from the space elevator and how long would it
> take the human to travel the entire length?
The boost force accelerates upwards, aways from earth.
> elevator and let us say it has an electro-magnetic repulsion
> accelerator along its entire length. If you were to accelerate a
> human so that he experienced only 2g (twice the force of gravity)
> during the entire length of the space elevator, what velocity would
> that human be expelled from the space elevator and how long would it
> take the human to travel the entire length?
Gravity attracts downwards, in the amount mg .
The net force is then boost - mg.
F ma, so a F/m (boost - mg)/m boost/m - g.
The mg , gravity, depends on elevation, so calling
g g_local is better.
Then acceleration boost/m - g_local .
A boost per kilogram of 9.8 Newtons/kg at earth's
surface means acceleration 0.
I think this boost per kilo would be called '1g' ...
A boost (rocket force) per kilo of 19.6 Newtons/kg
would seem like 2g, but produce an acceleration of
boost/m - g_local 19.6 Newtons/kg - g_local,
so at "inifinity" acceleration ~ 19.6 Newtons/kg 19.6 m/sec/sec .
With a space elevator that's anchored to Earth,
things that don't move relative to earth woul be
Ok. But the human in the space-elevator, being
in motion relative to earth, would be subject
to a Coriolis force:
< http://en.wikipedia.org/wiki/Coriolis_effect > .
62,000 miles is about 7.5 earth diameters.
Dave
> Where I'm having problems with the above question is the need for the
> human to be constantly increasing in speed during the entire length of
> the space elevator. As soon as the human goes at the same speed, the
> sensation of two gravities would stop.
>
>
> So 64 feet per second squared 2 gravities?
Barry Schwarz
Jan 1, 2012, 10:42:30 PM1/1/12
to
On Sat, 31 Dec 2011 22:49:23 EST, STJensen
<recreati...@gmail.com> wrote:
>Let us say that you had a 62,000-mile-long Earth-anchored space
>elevator and let us say it has an electro-magnetic repulsion
>accelerator along its entire length. If you were to accelerate a
>human so that he experienced only 2g (twice the force of gravity)
>during the entire length of the space elevator, what velocity would
>that human be expelled from the space elevator and how long would it
>take the human to travel the entire length?
>
>Where I'm having problems with the above question is the need for the
>human to be constantly increasing in speed during the entire length of
>the space elevator. As soon as the human goes at the same speed, the
>sensation of two gravities would stop.
>
>32 feet (9.8 meters) per second squared = 1 gravity
>
>So 64 feet per second squared = 2 gravities?
You used the phrase "experienced only 2g". A person standing in an
<recreati...@gmail.com> wrote:
>Let us say that you had a 62,000-mile-long Earth-anchored space
>elevator and let us say it has an electro-magnetic repulsion
>accelerator along its entire length. If you were to accelerate a
>human so that he experienced only 2g (twice the force of gravity)
>during the entire length of the space elevator, what velocity would
>that human be expelled from the space elevator and how long would it
>take the human to travel the entire length?
>
>Where I'm having problems with the above question is the need for the
>human to be constantly increasing in speed during the entire length of
>the space elevator. As soon as the human goes at the same speed, the
>sensation of two gravities would stop.
>
>32 feet (9.8 meters) per second squared = 1 gravity
>
>So 64 feet per second squared = 2 gravities?
unmoving elevator is experiencing 1g but his acceleration is 0
ft/sec^2 as these problems are normally phrased. So experiencing 2g
is accelerating upward at 32 ft/sec^2.
You also need to decide how realistic you want to be (in other words,
how much simplification are you willing to accept).
Do you want the acceleration to be a constant (32 ft/sec^2 or 64
ft/sec^2 or any other fixed value)? Or do you want the acceleration
to always be related to the current value of g (as you move away from
the earth, g gets smaller).
Acceleration is a vector. Is the total acceleration 2g or only
the vertical component? If you include the Earth's rotation, then the
path traveled is a spiral which means the total length is more that
just the 62,000 mile length of the cable.
>
>But that's for the first second. For the second second, the human
>would have to be going at 128 feet per second squared to continue to
>feel the force of two gravities, right?
function. At a fixed acceleration of 64 ft/sec^2, at the end of 2
seconds the velocity will be 128 ft/sec but only at that instant.
Since the acceleration is continuous, the velocity is constantly
changing. Just before 2 seconds have elapsed, the velocity would be
slightly less than 128 ft/sec. Just after, it would be slightly more.
>
>If the above question isn't hard enough, there are two other issues to
>deal with: 1) when the human is on the Earth's surface, there is also
>the Earth's gravity but as the human goes up the space elevator, that
>gravitational force decreases. This would mean a slower take-off
>speed so the human only experiences 2gs at all times. 2) once the
rides would reduce you to a puddle. Change in speed (which is the
definition of acceleration) determines what you feel.
>human passes the 22,000 mile mark, centrifugal force begins to push
>the human away from the Earth so the further away from the 22,000-mile
>point, the more outward force is being felt by the human and his
>acceleration would have to be increased even more so that the human
>would feel 2gs in the opposite direction he is traveling.
2g".
If you mean that he feels twice as heavy as he would if the
elevator were to stop, then the acceleration would gradually decrease
until he reached synchronous orbit.
If you mean his legs always feel like they are holding up twice
his weight, or equivalently, assume he is standing on a spring-type
scale (not a balance one) and it constantly registers twice his
"normal weight", then the acceleration must gradually increase to
compensate for the lower value of g at altitude. Then the question
becomes is the scale oriented vertically (as it is on the surface) or
is it oriented to the direction of travel.
>
>So without the impact of Earth's gravity and then centrifugal force
>past 22,000 miles, how fast would a human be traveling at the end of
>62,000 miles and how long would he take to travel that 62,000 miles?
distance d, and time t:
v = at
d =.5at^2
With a set to 64 ft/sec^2 and d set to 62,000 miles, you should be
have no trouble answering both questions. (Watch the units.)
>Now if someone knows how Earth's gravity and post-22,000-mile-point
>centrifugal force would change the previous answer, I would love to
>know that too.
resulting from the differential equations
v = dd/dt
a = dv/dt
You can derive the instantaneous value of g using Newton's equation
mg = GMm/d^2
where G is the gravitational constant, M is the mass of the Earth, m
is the mass of the man, and d is the distance from the center of the
Earth.
>
>Lastly, is there an equation for calculating the above? If so, I
>would appreciate knowing it since I could then enter any length and
>see what the final velocity would be.
--
Remove del for email
STJensen
Jan 1, 2012, 10:42:56 PM1/1/12
to
Let us see if I get what William Elliot, James Waldby, and David
Bernier are saying.
Time it would take to go the entire 62,000 miles:
Elliot: 76 minutes
Waldby: 3548.9 If the time is in seconds, that would equal 59.1
minutes.
Bernier: Sorry, I couldn't find a time.
Velocity human would be traveling at end of space elevator:
Elliot: 28 miles per second
Waldby: 61,686.8 meters per second. I'm assuming you mean meters and
not miles so that would be ~ 38 miles per second
Bernier: 19.6 miles per second
If the above is correct, why so much difference between the numbers?
Scott Jensen
Bernier are saying.
Time it would take to go the entire 62,000 miles:
Elliot: 76 minutes
Waldby: 3548.9 If the time is in seconds, that would equal 59.1
minutes.
Bernier: Sorry, I couldn't find a time.
Velocity human would be traveling at end of space elevator:
Elliot: 28 miles per second
Waldby: 61,686.8 meters per second. I'm assuming you mean meters and
not miles so that would be ~ 38 miles per second
Bernier: 19.6 miles per second
If the above is correct, why so much difference between the numbers?
Scott Jensen
James Waldby
Jan 2, 2012, 8:42:33 AM1/2/12
to
19.6 miles per second; he wrote 19.6 m/sec/sec, where m is meters, and
said that that is the acceleration equal to 2*g.
Second, the numbers William Elliot supplied are for the case where
a uniform 1 g acceleration is applied for a little over 75 minutes,
causing one to arrive at the top of the elevator at a speed of about
27.5 miles per second. The numbers are from s = (1/2)*a*t^2 and
v=a*t with s = 62000 miles, a = g. If you apply a uniform 2 g
acceleration (as suggested in Barry Schwarz's post) and experience
about 3G's on takeoff and about 2G's near the top, you arrive at
the top in about 53 minutes going 39 miles per second. In my
solution, with about 2G's total all the way up it takes about 59
minutes and top velocity is about 38.3 miles per second.
--
jiw
STJensen
Jan 3, 2012, 6:17:43 PM1/3/12
to
Thanks to all for the replies!
Scott Jensen
Scott Jensen
Dr J R Stockton
Jan 3, 2012, 11:17:59 PM1/3/12
to
In sci.math message <3ebddf87-655d-4e36...@l19g2000yqc.go
oglegroups.com>, Sat, 31 Dec 2011 22:49:23, STJensen
<recreati...@gmail.com> posted:
>Let us say that you had a 62,000-mile-long Earth-anchored space
>elevator and let us say it has an electro-magnetic repulsion
>accelerator along its entire length. If you were to accelerate a
>human so that he experienced only 2g (twice the force of gravity)
>during the entire length of the space elevator, what velocity would
>that human be expelled from the space elevator and how long would it
>take the human to travel the entire length?
Since the elevator cable cannot be infinitely rigid, and will in
practice be quite flexible, one will need to consider the effect of the
sideways forces on its shape - unless the mass of a shortish length of
the cable is large in comparison with that of your human and his
accessories.
--
(c) John Stockton, near London. *@merlyn.demon.co.uk/?.?.Stockton@physics.org
Web <http://www.merlyn.demon.co.uk/> - FAQish topics, acronyms, and links.
Correct <= 4-line sig. separator as above, a line precisely "-- " (RFC5536/7)
Do not Mail News to me. Before a reply, quote with ">" or "> " (RFC5536/7)
oglegroups.com>, Sat, 31 Dec 2011 22:49:23, STJensen
<recreati...@gmail.com> posted:
>Let us say that you had a 62,000-mile-long Earth-anchored space
>elevator and let us say it has an electro-magnetic repulsion
>accelerator along its entire length. If you were to accelerate a
>human so that he experienced only 2g (twice the force of gravity)
>during the entire length of the space elevator, what velocity would
>that human be expelled from the space elevator and how long would it
>take the human to travel the entire length?
practice be quite flexible, one will need to consider the effect of the
sideways forces on its shape - unless the mass of a shortish length of
the cable is large in comparison with that of your human and his
accessories.
--
(c) John Stockton, near London. *@merlyn.demon.co.uk/?.?.Stockton@physics.org
Web <http://www.merlyn.demon.co.uk/> - FAQish topics, acronyms, and links.
Correct <= 4-line sig. separator as above, a line precisely "-- " (RFC5536/7)
Do not Mail News to me. Before a reply, quote with ">" or "> " (RFC5536/7)
K_h
Jan 8, 2012, 2:11:06 AM1/8/12
to
"Dr J R Stockton" <repl...@merlyn.demon.co.uk> wrote in message
news:BCrPJYcX...@invalid.uk.co.demon.merlyn.invalid...
> In sci.math message <3ebddf87-655d-4e36...@l19g2000yqc.go
> oglegroups.com>, Sat, 31 Dec 2011 22:49:23, STJensen
> <recreati...@gmail.com> posted:
>
>>Let us say that you had a 62,000-mile-long Earth-anchored space
>>elevator and let us say it has an electro-magnetic repulsion
>>accelerator along its entire length. If you were to accelerate a
>>human so that he experienced only 2g (twice the force of gravity)
>>during the entire length of the space elevator, what velocity would
>>that human be expelled from the space elevator and how long would it
>>take the human to travel the entire length?
>
> Since the elevator cable cannot be infinitely rigid, and will in
> practice be quite flexible, one will need to consider the effect of the
> sideways forces on its shape - unless the mass of a shortish length of
> the cable is large in comparison with that of your human and his
> accessories.
Really good point. And the Earth is rotating so the cable must rotate with the
> oglegroups.com>, Sat, 31 Dec 2011 22:49:23, STJensen
> <recreati...@gmail.com> posted:
>
>>Let us say that you had a 62,000-mile-long Earth-anchored space
>>elevator and let us say it has an electro-magnetic repulsion
>>accelerator along its entire length. If you were to accelerate a
>>human so that he experienced only 2g (twice the force of gravity)
>>during the entire length of the space elevator, what velocity would
>>that human be expelled from the space elevator and how long would it
>>take the human to travel the entire length?
>
> Since the elevator cable cannot be infinitely rigid, and will in
> practice be quite flexible, one will need to consider the effect of the
> sideways forces on its shape - unless the mass of a shortish length of
> the cable is large in comparison with that of your human and his
> accessories.
same angular speed as the Earth. So, to do this calculation properly requires
consideration of tangent speeds, tangent acceleration (tangent to the cable for
those) and so forth.
_
Dr J R Stockton
Jan 9, 2012, 8:00:12 PM1/9/12
to
In sci.math message <iYidnV_pjtnsaJXS...@giganews.com>,
Sun, 8 Jan 2012 02:11:06, K_h <KHo...@SX729.com> posted:
It should not be difficult for the rigid-cable case, where the cable is
kept under sufficient tension by having an adequate mass at its tip..
Perhaps the rigid case is best done in natural units, in which GSO is at
unit radius and the sidereal period at GSO is also unity. The variables
are then, in GSO units, the starting and finishing heights, and the
constant thrust, which ISTR was going to be twice of what is needed to
be felt to remain stationary at the starting point, a.k.a. ground level.
The acceleration outwards results from the combination of the thrust,
the inverse square gravity, and the linear centripetal pseudo-force.
Integrate once with respect to the time to get the outwards speed as a
function of time, and again to get the outwards distance as a function
of time. Then determine the time to release from the second integral,
and substitute in the first integral to get the release speed.
Unchecked.
Sun, 8 Jan 2012 02:11:06, K_h <KHo...@SX729.com> posted:
kept under sufficient tension by having an adequate mass at its tip..
Perhaps the rigid case is best done in natural units, in which GSO is at
unit radius and the sidereal period at GSO is also unity. The variables
are then, in GSO units, the starting and finishing heights, and the
constant thrust, which ISTR was going to be twice of what is needed to
be felt to remain stationary at the starting point, a.k.a. ground level.
The acceleration outwards results from the combination of the thrust,
the inverse square gravity, and the linear centripetal pseudo-force.
Integrate once with respect to the time to get the outwards speed as a
function of time, and again to get the outwards distance as a function
of time. Then determine the time to release from the second integral,
and substitute in the first integral to get the release speed.
Unchecked.
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