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Originally posted by GOURDHEAD@May 20 2004, 03:47 PM
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Acceleration times (time squared)/2 so by the time I get to light speed in 356 days at a constant 1 g I am far beyond the Ort cloud.
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When I multiply 3600 seconds per hour by 24 hours per day by 356 days by 6.8 pounds per second and divide by 2000 pounds per ton I come up with 104578.56 tons as your propellant tonnage to reach light speed. When I multiply 3600 times 24 times 1.73 times 6.8 and divide by 2000, I come up with 3599.858832 tons of propellant to reach max velocity to Mars. What am I missing?
No one has seen the Oort cloud; it has been postulated from the somewhat random orientation of the planes of the orbits of long period comets. The cloud is assumed to be spherical, a spherical shell around the solar system. If so, it will be hard to reach any other stellar system without passing through it. I'm assuming that the spherical shell is several tenths of a light year thick and ranges from about 50000 to 76000 AUs although I've seen other values for its size. If we can't observe the Oort cloud directly, we can't know how thick intragalactic space is populated with similar objects because our current instruments can't see them either. Again, caution is advised. Remember you can't acquire data fast enough to detect and avoid objects after the ship passes about 0.4c. The radar round trip is the time required for the ship to travel 0.8c-seconds (148800 miles) then you add detection and processing time to determine whether and where trajectories will intersect and to compute and command steering instructions.
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That is a dramatic scenario and so good grist for science fiction novel; however the math does not support the conclusion
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What is the maximum rate of steering angle change the math will support. Have you constructed a safe rate of direction change as a function of ship velocity? At velocities equal to or greater than 0.5c you can't detect an object on a collision course much less avoid it.
When traveling at c, 186000 times 5280 = 982080000 feet per second. The cosine of 5 degrees = 0.999614. A five degree change in direction per second would generate (1 - 0.999614)* 982080000) feet per second velocity change = 379082.88 feet/sec. If it takes 5 seconds (the ship will have traveled 5*186000 miles) to effect the change in direction, the g force is 379082.88/5/32 = 2369.268 g's. This seems a bit much for people and equipment mountings to withstand. Science fiction writers conveniently avoid troublesome unpleasantries.
From d = at*(1 + 0.5*t) and using 32 feet/sec^2 for a, I get a tad over
1433453381934.54 miles or just under 15413.48 AUs traveled in 356 days. Due to uncertainities of our understanding of the distance to the inner radius of the Oort cloud you may not have reached it before you would have reached light speed.
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let me see; since I can get 100,000 tons to 687.960 km/sec at 1 g
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Should this value be 987.960 km/sec?
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Gourdhead and Tom2Mars
Hold your horses please
I am trying to get a three year old to bed plus trnsfer many of my text files to this computer for=e ease of reference and get some time of focus and consentration to reply. The questions your both raise are good ones so you deserve a good answer.
Adressing the oort cloud now, I will recheck my math as time permits Gourdhead may of found an error. I rejoice when a reader actully checks my math as two heads are better than one.
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The Oort cloud is an immense spherical cloud surrounding the planetary system and extending approximately 3 light years, about 30 trillion kilometers from the Sun. This vast distance is considered the edge of the Sun's orb of physical, gravitational, or dynamical influence.
Within the cloud, comets are typically tens of millions of kilometers apart. They are weakly bound to the sun, and passing stars and other forces can readily change their orbits, sending them into the inner solar system or out to interstellar space. This is especially true of comets on the outer edges of the Oort cloud. The structure of the cloud is believed to consist of a relatively dense core that lies near the ecliptic plane and gradually replenishes the outer boundaries, creating a steady state. One sixth of an estimated six trillion icy objects or comets are in the outer region with the remainder in the relatively dense core.
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Like I said I need only to take off perpindicular to the ecliptic plane to avoid the comets-well 5/6 of them any way. the remaining 1 trillion are spread so thin the likelyhood of me hiting one with a 10 meter diameter rocket is insignifigant if not zero.
I can best explain this if you can visulize my trajectory path being roughly a cylinder 10 meters in diameter and 7.5 light years long. I do not care if 10 million baseball size rocks at various velocites intersect that volume over the trip times of 4.5 years as the chance that one will be where my ship is at any one time is next to zero.
Good math does not replace good measuremnt test so the best I can do is send unmanned probes and see how many take the trip without damage.
