Suppose that a space probe can withstand the stresses of a $20\mathrm{g}$acceleration. What are (a) the minimum turning radius of such a craft moving at a speed of one-tenth the speed of light? (b) How long would it take to complete a${90}^0$turn at this speed?

It is given that,

v = $\frac{1}{10}c=0.3\times {10}^8\mathrm{m}/\mathrm{s}$

here c is the velocity of light

$$\begin{gathered} a_c=20g \\ =20\times 9.8\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

distance is $\frac{\mathrm{\pi }}{2}$

This problem deals with the centripetal acceleration. Centripetal acceleration is the acceleration of a body traversing a circular path. The net force will be the centripetal force on the craft moving in a circle. It is the force towards the centre. The radius and time taken by craft using the equation of centripetal acceleration and period in circular motion can be found.

Formula:

The centripetal acceleration is given by,

$a_c=\frac{v^2}{r}$ (i)

$v\mathit{}\mathit{=}\frac{\mathit{2}\mathit{\pi }\mathit{r}}{\mathit{T}}$ (ii)

where, r is the radius, $v$ is the velocity, T is time and a is an acceleration.

Using equation (i), the radius will be,

$\begin{array}{l}r=\frac{v^2}{a_c}=\frac{(0.3\times {10}^8{)}^2}{(20x9.8)}\\ =4.6\times {10}^{12}\mathrm{m}\end{array}$

The radius of circular orbit is$4.6\times {10}^{12}\mathrm{m}$

The time taken by craft to turn 90⁰ can be calculated as follow,

Using equation (ii) the time will be,

$T=\frac{\left(2\mathrm{\pi }\right)r}{v}$

localid="1656997253730" $\begin{array}{l}T=\frac{2\times 3.14\times 4.6\times {10}^{12}\mathrm{m}}{0.3\times {10}^8\mathrm{m}/\mathrm{s}}\\ =96\times {10}^{5}s\end{array}$

Therefore, the time to make quarter turn is,

$$\begin{gathered} \frac{T}{4}=\frac{9.6\times {10}^{5}s}{4} \\ =2.4\times {10}^{5}s \end{gathered}$$

Hence, the time taken by craft to cover 90⁰of circular turn is$=2.4\times {10}^{5}s$