# Question d4f0b

Aug 21, 2017

Given that speed of rotating space station $v = 12 m \text{/} s$

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

Now the acceleration due to gravity on earth ${g}_{e} = 9.8 m \text{/} {s}^{2}$

Now by the condition of the problem

g_a=50% "of "g_e#

$\implies {v}^{2} / r = \frac{50}{100} \times {g}_{e}$

$\implies {12}^{2} / r = \frac{1}{2} \times 9.8$

$\implies r = \frac{144 \times 2}{9.8} \approx 29.4 m$