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(D) The artificial gravity experienced by the astronauts is due to the centripetal acceleration $g = \omega^2 L$,where $\omega$ is the angular velocit

Vedclass · Accepted Answer

$\left(\frac{N}{N-1}\right)^2$

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(D) The artificial gravity experienced by the astronauts is due to the centripetal acceleration $g = \omega^2 L$,where $\omega$ is the angular velocity and $L$ is the distance from the hub.Initially,the total moment of inertia of the astronauts about the central hub is $I = 2 \cdot (N \cdot m \cdot L^2) = 2Nm L^2$.When two astronauts move to the hub,they are at the axis of rotation,so their moment of inertia becomes zero. The new moment of inertia is $I' = 2 \cdot ((N-1) \cdot m \cdot L^2) = 2(N-1)m L^2$.Since there are no external torques,angular momentum is conserved: $I\omega = I'\omega'$.Substituting the values: $(2Nm L^2) \omega = (2(N-1)m L^2) \omega'$.This simplifies to $N\omega = (N-1)\omega'$,so $\omega' = \omega \cdot \frac{N}{N-1}$.The new gravitational field strength is $g' = \omega'^2 L$.The ratio is $\frac{g'}{g} = \frac{\omega'^2 L}{\omega^2 L} = \left(\frac{\omega'}{\omega}\right)^2 = \left(\frac{N}{N-1}\right)^2$.

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