Question #d4f0b

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Question #d4f0b

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Answer 1 · 2017-08-21T06:05:43

Given that speed of rotating space station $v = 12 m / s$ $v=12m"/"s$

If the radius of the orbit of rotation be $r$ $r$ m, then centrifugal accelertion which is equal in magnitude of artificial accelartion due to gravity in the space station will be $g_{a} = \frac{v^{2}}{r}$ $g_a=v^2/r$

Now the acceleration due to gravity on earth $g_{e} = 9.8 m / s^{2}$ $g_e=9.8m"/"s^2$

Now by the condition of the problem

$g_{a} = 50 % of g_{e}$ $g_a=50% "of "g_e$

$\Rightarrow \frac{v^{2}}{r} = \frac{50}{100} \times g_{e}$ $=>v^2/r=50/100xxg_e$

$\Rightarrow \frac{12^{2}}{r} = \frac{1}{2} \times 9.8$ $=>12^2/r=1/2xx9.8$

$\Rightarrow r = \frac{144 \times 2}{9.8} \approx 29.4 m$ $=>r=(144xx2)/9.8~~29.4m$

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