Q2.7 Centripetal acceleration (S)
A space probe can withstand the stresses of a 20g acceleration. What is the minimum turning radius needed if its speed is 10-3c ? How long would it take to complete a 90∘ turn?
[velocity of light c=3×108ms-1]
Hint
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Solution
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a bit more than the earth-moon distance.

Solution
According to Key Point 1.8 a body moving at uniform speed v in a circle of radius r exhibits an acceleration directed towards the centre of the circle (ie centripetal) of magnitude v2/r. Setting
![\[ \frac{v^2}{r} = 20 g \hspace{0.5cm}\mbox{\rm with}\hspace{0.5cm} v=10^{-3}c \hspace{0.5cm}\mbox{\rm gives}\hspace{0.5cm} r= 5 \times 10^{-8} \frac{c^2}{g} = 5\times 10^{5}\,km \hspace{0.5cm}\mbox{\rm (1sf)}\hspace{0.5cm} \]](mastermathpng-0.png)
The time to complete one quarter of a revolution like this is
![\[ t= \frac{1}{4} \frac{2\pi r}{v} = 3 \times 10^3 {s} \hspace{0.5cm}\mbox{\rm (1sf)}\hspace{0.5cm} \]](mastermathpng-1.png)