91Ó°ÊÓ

Whenever the object is in the circular track, then the kind of acceleration occurs is termed as centripetal acceleration.

Time period is defined as the time required by the object in order to make one full oscillation.

The expression for the time period is given as,

$T=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}$

Here, $\mathrm{Ó\neg }$is the angular velocity.

The centripetal acceleration is $a_c=0.70g$.

The diameter of the spaceship is $d=32\mathrm{m}$.

The expression for the centripetal acceleration is given as,

$a_c=\frac{v^2}{r}$

Substitute the values in the above equation,

role="math" localid="1646295354239" $\begin{array}{c}0.70g=\frac{v^2}{r}\\ v=\sqrt{0.70gr}…\left(i\right)\end{array}$

The expression for the time period is given as,

role="math" localid="1646295379115" $\begin{array}{c}T=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}\\ T=\frac{2\mathrm{Ï€}r}{v}\\ v=\frac{2\mathrm{Ï€}r}{T}…\left(ii\right)\end{array}$

On combining equation (i) and equation (ii), to get the value of time needed for one revolution,

$\begin{array}{c}T=\frac{2\mathrm{π}r}{\sqrt{0.70gr}}\\ T=\frac{2\mathrm{π}\left(16\mathrm{m}\right)}{\sqrt{0.70×9.81\mathrm{m}/{\mathrm{s}}^2×16\mathrm{m}}}\\ T=9.59\mathrm{s}\end{array}$

Thus, the time needed for one revolution is $9.59\mathrm{s}$.