A space station is to provide artificial gravity to support long-term stay of as

Question

A space station is to provide artificial gravity to support long-term stay of astronauts and cosmonauts. It is designed as a large wheel, with all the compartments in the rim, which is to rotate at a speed that will provide an acceleration similar to that of terrestrial gravity for the astronauts (their feet will be on the inside of the outer wall of the space station and their heads will be pointing toward the hub). After the space station is assembled in orbit, its rotation will be started by the firing of a rocket motor fixed to the outer rim, which fires tangentially to the rim. The radius of the space station is R = 46.5 m, and the mass is M = 4.65 · 105 kg. If the thrust of the rocket motor is F = 223 N, how long should the motor fire?

Explanation / Answer

Since, the space station has to provide terrestrial gravity, so a = g =9.8 m/s/s.

We need to get the space station rotating with an angular velocity . needs to be large enough so that the centripetal acceleration felt by the people walking on the inner surface of the outer wall is equal to g (9.8 m/s/s). The centripetal acceleration (lets call it a) is given by

a = v²/r

since

v/r =

we get

a = r²

We want a = 9.8, thus:

r² = 9.8.

Since the radius is 46.5 m we get the following value for :

46.5² = 9.8
= (9.8 / 46.5) = 0.4590780.... rad/s

This is the angular velocity at which the space station needs to be rotating at to achieve a centripetal acceleration equal to g. We want to find out what angular momentum the space station acquires. Angular momentum in general is given by:

I

where I = the moment of inertia and is the angular velocity! We have not been given a value for I, so this needs to be calculated. Since I have not been given the schematics for the space station I will have to estimate a value for I. In my estimate, I will approximate the space station by a thin ring with radius of 46.5m and a mass of 4.65 · 105 kg. The moment of inertia for such a ring is given by the following formula:

I = mr² = (4.65 · 105)(46.5)² = 10054 · 10 kgm²

Now we can calculate the angular momentum of the space station:

I = (10054 · 10)(0.4590780)

4.615 · 10 kgm²/s

So the above is the angular momentum which the space station acquires.

The thrusters (rockets) blast away for t seconds exerting 223 N. t must be such that the space station acquires the angular momentum calculated above. Let us start by finding out the resulting angular acceleration caused by the thrusters.

= I
= /I

where is the torque and is the angular acceleration!

is given by

= r × F (be ware, this is the cross product, but since the force is perpendicular to the radius of the space station we just get)
= rF = (46.5)(223) = 10369.5 kgm²/s²

We already have a value for I, thus we can start calculating for the angular acceleration (alpha):

= /I = (10369.5)/(10054 · 10) = 1.03138 · 10...rad/s²

Now, since we want to achieve an angular velocity of

0.4590780..... rad/s

we need to have the rockets blasting for t seconds such that.

t =

t = / = 0.4590780 / 1.03138 · 10
= 44511.04 s = 44511.04 / (60·60) hours
= 12.36 hours

So, the motor should fire for 12.36 hours.

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