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A space shuttle can theoretically do it in approximately 2195 seconds, but the burn time of the rockets is only 1854 seconds, so it would take longer in real world conditions. Here is the math I used for my calculation. This assumes that gravity is negligible because it is already in orbit, that the space shuttle boosters can somehow both accelerate and then flip around and decelerate the craft, and that the space shuttle accelerates at 29.4 m/s^2. (http://spaceflight.nasa.gov/shuttle/reference/shutref/sts/requirements.html states that the acceleration never exceeds three G's, and 1 G is 9.8 m/s^2)
the upward acceleration of the shuttle has the function a(t) = 29.4
antidifferentiate to find the Velocity function v(t) = 29.4t + C
C = 0 because we are taking the position of the shuttle relative to the ISS
therefore v(t) = 29.4t
antidifferentiate again to find the position function s(t) = 14.7t^2
once the shuttle is halfway it has to decelerate, so the total time is twice the time taken to reach the midpoint.
the ISS orbits at 350000 m(http://en.wikipedia.org/wiki/International_Space_Station) and geosynchronous orbit occurs at 35786000 m (http://en.wikipedia.org/wiki/Geosynchronous_orbit), so the shortest distance between them is 35436000 m. Half this distance is 17718000 m
set that equal to the position function and solve for t
17718000 = 14.7t^2
1205306 = t^2
t is about 1097 s
multiply by 2 to find total time
2195 seconds
Source(s):
http://en.wikipedia.org/wiki/Geosynchronous_orbit
http://en.wikipedia.org/wiki/International_Space_Station
http://spaceflight.nasa.gov/shuttle/reference/shutref/sts/requirements.html
http://en.wikipedia.org/wiki/Space_Shuttle#Fiction_and_games
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As far as what kind of ship it would take, anything that can make it to the moon will do. Actually that kind of energy pretty close to exactly what we would need. If we can make it to a geosynchronous orbit around the earth, we only need at most 5% more energy to completely escape Earth's gravity. I'm basing that on Earth's gravity's strength at the distance of a geosynchronous orbit which is only 2% what it is on the surface of the Earth. I'd bet dollars to donuts we'd need way less than 2% more energy to get to the station than we need to get to the moon.
That said, it WOULD take a lot of energy if we went straight to that space station. With current technology and fuel resources, we can't afford to go straight there. So it would probably take the same path we would take to get to the moon, and therefore take about the same amount of time, which somehow is only four days. It would probably take longer though, like a day or two longer, because as you approach the current space station you have to slow down well before you get there in order to avoid spraying the space station with debris or fuel.
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Answered Question
M$1
December 30, 2008 05:00 AM
How long would it take to travel from the International Space Station to an International Geosynchronous SS
Suppose we built a space station in geosynchronous orbit. What type of vessel would we need to travel between the two stations? Could a space refueled shuttle do it? How long would it take?
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| December 30, 2008 06:51 AM |
the upward acceleration of the shuttle has the function a(t) = 29.4
antidifferentiate to find the Velocity function v(t) = 29.4t + C
C = 0 because we are taking the position of the shuttle relative to the ISS
therefore v(t) = 29.4t
antidifferentiate again to find the position function s(t) = 14.7t^2
once the shuttle is halfway it has to decelerate, so the total time is twice the time taken to reach the midpoint.
the ISS orbits at 350000 m(http://en.wikipedia.org/wiki/International_Space_Station) and geosynchronous orbit occurs at 35786000 m (http://en.wikipedia.org/wiki/Geosynchronous_orbit), so the shortest distance between them is 35436000 m. Half this distance is 17718000 m
set that equal to the position function and solve for t
17718000 = 14.7t^2
1205306 = t^2
t is about 1097 s
multiply by 2 to find total time
2195 seconds
Source(s):
http://en.wikipedia.org/wiki/Geosynchronous_orbit
http://en.wikipedia.org/wiki/International_Space_Station
http://spaceflight.nasa.gov/shuttle/reference/shutref/sts/requirements.html
http://en.wikipedia.org/wiki/Space_Shuttle#Fiction_and_games
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Other Answers (1)
December 30, 2008 05:54 AM
It wouldn't require refueling because a geosynchronous orbit is only about seven earth radii above the earth, whereas the moon is about sixty earth radii away and it doesn't require refueling. As far as what kind of ship it would take, anything that can make it to the moon will do. Actually that kind of energy pretty close to exactly what we would need. If we can make it to a geosynchronous orbit around the earth, we only need at most 5% more energy to completely escape Earth's gravity. I'm basing that on Earth's gravity's strength at the distance of a geosynchronous orbit which is only 2% what it is on the surface of the Earth. I'd bet dollars to donuts we'd need way less than 2% more energy to get to the station than we need to get to the moon.
That said, it WOULD take a lot of energy if we went straight to that space station. With current technology and fuel resources, we can't afford to go straight there. So it would probably take the same path we would take to get to the moon, and therefore take about the same amount of time, which somehow is only four days. It would probably take longer though, like a day or two longer, because as you approach the current space station you have to slow down well before you get there in order to avoid spraying the space station with debris or fuel.
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December 30, 2008 07:01 AM
getting to the moon doesn't require refueling with a saturn V rocket, but it does require refueling with a space shuttle
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December 30, 2008 07:09 AM
You need the same basic technology to get to the moon as you need to get to geosynchronous orbit. So if you don't need to refill to get to the moon, you won't need to refill to get to that orbit either.
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I redid the problem with consideration to burn time, and the answers is 2226 seconds.
We can only accelerate for 1854 s, so we accelerate for the first half of our burn time, 927 s. Then we have to coast til the midpoint. We can still multiply by two at the end because what we are doing on either side of the midpoint is still symmetrical.
at t = 927 s, velocity v(t) = 27253 m/s, and position s(t) = 12632136 m
there are 17718000 m -12632136 m = 5085863 m left to go until the midpoint, and we are moving at 27253 m/s , so this takes 186 s. add that to 927 s, and the time to get to the midpoint is 1113 seconds. Multiply by 2 and the total time is 2226 seconds. I should probably round off to 2200 seconds, because the given altitude for the ISS only had 2 sig figs. Bottom line, approximately 2200 seconds.