One problem for humans living in outer space is that they are apparent

kramtus51 2021-12-19 Answered
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80ms2? (b) If the space station is a waiting area for travelers going to Mars, it might be desirable to simulate the acceleration due to gravity on the Martian surface (3.70ms2). How many revolutions per minute are needed in this case?
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Expert Answer

Jacob Homer
Answered 2021-12-20 Author has 41 answers

Step 1
Given: Rspace station=Dspace station2=800m2=400m
gE=9.80ms2
gM=3.70ms2
Step 2
We know that the centripetal acceleration is given by
ar=v2R
and we also know that the linear velocity for a rotation object is given by
v=ωR
Hence, ar=(ωR)2R=ω2R2R
ar=ω2R(1)
And this is a general formula for both cases in this problem.
Step 3
a) We need to make the centripetal acceleration equals the free-fall acceleration of Earth, so, from (1);
gE=ω2R
solving for ω;
ω2=gER
ω=gER
Plug the given;
ω=9.80400rads(2)
We need to find it in rev/min, so we will use some converting factors.
Noting that 1.0rev=2πrad, hence
1.0rev2πrad=1.0(A)
and 60s=1.0min, hence
60s1.0min=1.0(B)
Step 4
Multiply (2) by the two converting factors (A) and (B).
ω=9.80400rads×1.0rev2πrad×60s1.0min
rad ans s are cancelled.
ω=1.49revmin
Step 5
b)Using the same approach above but changing the free-fall acceleration of Earth gE to the free-fall acceleration of Mars gM.
Hence, from (1) gM=ω2R
solving for ω;
ω2=gMR
ω=gMR
Plug the given;
ω=3.70400rads(3)
Multiply (3) by the two converting factors (A) and (B).

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porschomcl
Answered 2021-12-21 Author has 28 answers

A) The circumference can be used to obtain the radius of the space station
2πR=1884m R=300m
The acceleration a for a rotational motion in terms of angular velocity ω
a=g(in this case)=v2r=rω2
ω=gr=0.1807radianssec
The number of revolution required is the frequency of 2π rotation in a minute
N=ω60sec2π=0.1807606.28=1.726revmin
B) We adopt the same procedure as we did for Part A except that the acceleration in now amars=3.7msec2.
ω=gr=0.111radianssec
N=ω60sec2π=0.1807606.28=1.06revmin

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nick1337
Answered 2021-12-27 Author has 575 answers

Explanation:
In space station artificial gravity is created by rotational motion of space station
Centifugal  acceleration creates artificial gravity
ω2r=g
ω2×400=9.8
ω2=9.8/400
NSK
ω=.1565rad/s
no of turns per min ω2π per sec =1.56 rev per min.
For Mars
Same theory will apply
ω2r=g
ω2×400=3.7
ω2=3.7400
ω=.096rad/s
n=ω2π
=.0962π
=.01528 per sec
per min
=.9168 per min.

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