De orbiting from ISS

 

Our future will include a massive Space Station that is plausibly a full mile in diameter and as thick as we need making it look like a giant disc.  Simple rotation would produce one g of acceleration on the outer trim.

Getting down to surface is then a highly desirable objective and we have the capacity to carry thousands of one man life boats able to de orbit.  One man boats are likely the best option in practise as well as it can easily be used for random and normal removal as well.

This item shows us that it just may be plausible and a nice alternative to sending up insufficient transports.  After all we now have sucessful boosters able to land..

If an astronaut pushed off from the International Space Station in the direction of earth, how long would it take them to reach the ground? And assuming they had enough oxygen and a parachute, could they survive the journey?

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The astronaut and station are in orbit 350 km above the Earth and are moving at 7.697 km/sec. That is they are in an orbit with a 6721 km semi-major axis with a period of 5486.46 seconds. So, if they pushed away at 1 meter per second toward Earth, the resultant would be

sqrt(7.697^2+0.001^2) = 7.697000064960374 km/sec angled

acos(7.697/7.697000064960374)= 0.000129920747257 radians

or 26.8 arc seconds relative to the station.

Which means that in 1371.615 seconds after push off, they would be 166 meters from the station and reach perigee. They would then fly back up to the station’s altitude, but be ahead of the station by a few hundred meters 2743.23 seconds after push off. They would then reach an altitude 166 meters above the station 4114.845 seconds after push off and then come back to the station, opposite to the side they pushed off arriving at 1 meter/second 5486.46 seconds after push off.

To re-enter Earth’s atmosphere requires that the astronaut slow their speed relative to the station sufficiently to have a vacuum perigee altitude of 60 km. This is an orbital radius of 6371+60 = 6431 km.

So normalising everything from 6371 km = 1 radius we have

(6371+60)/6371 = 1.009417673834563 - perigee

(6371+350)/6371 = 1.054936430701617 - apogee

a = (1.054936430701617+1.009417673834563)/2 = 1.03217705226809 transfer

v = 7.9054*sqrt(2/1.054936430701617–1/1.03217705226809) = 7.61147905716

So the change in speed is 7.697-7.61147905716 = 0.085520942840000 km/sec delta vee. This is 307.87 km/hour (192.23 mph) speed directed against the motion of the station in its orbit.

The period of the transfer orbit is

5063.6*1.03217705226809^1.5 = 5,309.95 seconds

So, 2,654.98 seconds after the 192.23 mph push against the motion of the station, the return capsule (or astronaut) reaches vacuum perigee altitude of 60 km and begins re-entry.

So, you blast 192.23 mph against the motion of the station and 42.25 minutes later you are at 60 km altitude and begin re-entry.

Using a LOX/LNG propellant with a 3.7 km/sec exhaust speed you require a propellant fraction of

u = 1 - 1 /exp(0.085521/3.7) = 0.02285

Bell Lunar Flight Unit - designed by Bell Aerospace as an alternative to Lunar Rover for Apollo.

So, an astronaut in a spacesuit that weighs 310 pounds sitting on a flyboard that weighs 90 pounds has a 400 pound inert weight. With a 2.285% propellant weight that means propellant weight is

400/(1–0.02285)-400 = 9.353 pounds of propellant. A 100 pound propellant tank surrounded by a heat sheild, would allow an astronaut to fly from the station to the ground and land softly!

This would be a great escape system for ISS.

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 · Answer requested by 

Jim Dalrymple