| Topic: |
Science > Physics |
| User: |
"Front Office" |
| Date: |
14 Nov 2006 12:27:01 PM |
| Object: |
Constant acceleration problem |
What would be the constant acceleration needed by a spaceship
for it to travel 13.7 billion light-years in 24 hours of subjective
(i.e., passenger) time?
(Clearly, the 'constant acceleration' would be experienced by
the passengers on such a spaceship, but would not be seen
as such by stationary observers.)
Thanks for any help, guidance, answers, wisecracks or whatever.
I will check back at this newsgroup in a day or so.
Bob
armistead_rap[AT]bigfoot.com
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| User: "Roland Franzius" |
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| Title: Re: Constant acceleration problem |
14 Nov 2006 12:59:09 PM |
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Front Office schrieb:
What would be the constant acceleration needed by a spaceship
for it to travel 13.7 billion light-years in 24 hours of subjective
(i.e., passenger) time?
(Clearly, the 'constant acceleration' would be experienced by
the passengers on such a spaceship, but would not be seen
as such by stationary observers.)
Thanks for any help, guidance, answers, wisecracks or whatever.
I will check back at this newsgroup in a day or so.
Bob
armistead_rap[AT]bigfoot.com
The value of the constant acceleration with respect to travellers time
in Minkowski flat space
1/a (Cosh(a/365)-1) = 13.6 10^9 y
is
a*c= c*12206.9 /year = 116043 m/s^2 =11833.1 g_earth
--
Roland Franzius
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| User: "Front Office" |
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| Title: Re: Constant acceleration problem |
16 Nov 2006 09:13:39 AM |
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Roland Franzius wrote:
Front Office schrieb:
What would be the constant acceleration needed by a spaceship
for it to travel 13.7 billion light-years in 24 hours of subjective
(i.e., passenger) time?
(Clearly, the 'constant acceleration' would be experienced by
the passengers on such a spaceship, but would not be seen
as such by stationary observers.)
Thanks for any help, guidance, answers, wisecracks or whatever.
I will check back at this newsgroup in a day or so.
Bob
armistead_rap[AT]bigfoot.com
The value of the constant acceleration with respect to travellers time
in Minkowski flat space
1/a (Cosh(a/365)-1) = 13.6 10^9 y
is
a*c= c*12206.9 /year = 116043 m/s^2 =11833.1 g_earth
This is the number that will get published. Much thanks.
Bob
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| User: "Sam Wormley" |
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| Title: Re: Constant acceleration problem |
14 Nov 2006 03:19:28 PM |
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Front Office wrote:
What would be the constant acceleration needed by a spaceship
for it to travel 13.7 billion light-years in 24 hours of subjective
(i.e., passenger) time?
Better plan on a longer trip is you want to experience it.
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| User: "Phineas T Puddleduck" |
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| Title: Re: Constant acceleration problem |
14 Nov 2006 01:06:58 PM |
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On 2006-11-14 18:27:01 +0000, Front Office <YoMo.nospam@erols.com> said:
What would be the constant acceleration needed by a spaceship
for it to travel 13.7 billion light-years in 24 hours of subjective
(i.e., passenger) time?
(Clearly, the 'constant acceleration' would be experienced by
the passengers on such a spaceship, but would not be seen
as such by stationary observers.)
Thanks for any help, guidance, answers, wisecracks or whatever.
I will check back at this newsgroup in a day or so.
Bob
armistead_rap[AT]bigfoot.com
around 6 angstroms per square hectare.
--
The biggest enemy of science is pseudo science. Oh, and Jeff Relf too...
--
Posted via a free Usenet account from http://www.teranews.com
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