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How long will it take to fall 10000km? - Answers

i'm not sure, but i think the following newton application, applies... so: 10 000 km=10 000 000 m = g/2* t^2 => 2*10 000 000/9.8=2 040 816.32=t^2 =>t= sqrt(2 040 816.33)=1 428.57 seconds = 23.80952 minutes all this disregarding atmosphere influence, which is probably huge 10,000 km is approximately the same distance as the diameter of the earth. so we're talking about falling from really far, on the one hand (it's 270 times farther then the 37 km that a man fell from to earth lately in less then 6 minutes), on the other hand, its pretty close by, as its about 1/33 from the distance from the moon. i have no idea if the figure calculated is realistic, 23 minutes seems to me, intuitively, way to short. i'd expect days at least, not minutes. anybody can tell what the deal here?



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How long will it take to fall 10000km? - Answers

https://math.answers.com/math-and-arithmetic/How_long_will_it_take_to_fall_10000km

i'm not sure, but i think the following newton application, applies... so: 10 000 km=10 000 000 m = g/2* t^2 => 2*10 000 000/9.8=2 040 816.32=t^2 =>t= sqrt(2 040 816.33)=1 428.57 seconds = 23.80952 minutes all this disregarding atmosphere influence, which is probably huge 10,000 km is approximately the same distance as the diameter of the earth. so we're talking about falling from really far, on the one hand (it's 270 times farther then the 37 km that a man fell from to earth lately in less then 6 minutes), on the other hand, its pretty close by, as its about 1/33 from the distance from the moon. i have no idea if the figure calculated is realistic, 23 minutes seems to me, intuitively, way to short. i'd expect days at least, not minutes. anybody can tell what the deal here?



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https://math.answers.com/math-and-arithmetic/How_long_will_it_take_to_fall_10000km

How long will it take to fall 10000km? - Answers

i'm not sure, but i think the following newton application, applies... so: 10 000 km=10 000 000 m = g/2* t^2 => 2*10 000 000/9.8=2 040 816.32=t^2 =>t= sqrt(2 040 816.33)=1 428.57 seconds = 23.80952 minutes all this disregarding atmosphere influence, which is probably huge 10,000 km is approximately the same distance as the diameter of the earth. so we're talking about falling from really far, on the one hand (it's 270 times farther then the 37 km that a man fell from to earth lately in less then 6 minutes), on the other hand, its pretty close by, as its about 1/33 from the distance from the moon. i have no idea if the figure calculated is realistic, 23 minutes seems to me, intuitively, way to short. i'd expect days at least, not minutes. anybody can tell what the deal here?

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      i'm not sure, but i think the following newton application, applies... so: 10 000 km=10 000 000 m = g/2* t^2 => 2*10 000 000/9.8=2 040 816.32=t^2 =>t= sqrt(2 040 816.33)=1 428.57 seconds = 23.80952 minutes all this disregarding atmosphere influence, which is probably huge 10,000 km is approximately the same distance as the diameter of the earth. so we're talking about falling from really far, on the one hand (it's 270 times farther then the 37 km that a man fell from to earth lately in less then 6 minutes), on the other hand, its pretty close by, as its about 1/33 from the distance from the moon. i have no idea if the figure calculated is realistic, 23 minutes seems to me, intuitively, way to short. i'd expect days at least, not minutes. anybody can tell what the deal here?
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