Question

a jet fighter flying at 300 m/s (just below the speed of sound) makes a turn of radius 1.65 km.
(a) what is its centripetal acceleration in gs?

g
(b) suppose the pilot makes an emergency turn to avoid an approaching missile, subjecting himself to a centripetal acceleration of 10 gs, while flying at 400 m/s (supersonic). what is the radius, in km, of his turn? (this must be short - lived because fighter planes can only briefly endure such large accelerations without serious damage, and the pilot will soon black out at 10 gs.)

km

Explanation:

Step1: Convert radius to meters for part (a)

Given radius \(r = 1.65\space km=1.65\times10^{3}\space m\), speed \(v = 300\space m/s\). The formula for centripetal acceleration is \(a_{c}=\frac{v^{2}}{r}\).

$$a_{c}=\frac{(300)^{2}}{1.65\times 10^{3}}$$

$$a_{c}=\frac{90000}{1650}\approx54.55\space m/s^{2}$$

Since \(g = 9.8\space m/s^{2}\), to get the value in \(g\)'s, we use \(n=\frac{a_{c}}{g}\)

$$n=\frac{54.55}{9.8}\approx 5.57$$

Step2: Use centripetal - acceleration formula for part (b)

Given \(a_{c}=10g\), \(g = 9.8\space m/s^{2}\), so \(a_{c}=10\times9.8 = 98\space m/s^{2}\), \(v = 400\space m/s\). From \(a_{c}=\frac{v^{2}}{r}\), we can solve for \(r\), \(r=\frac{v^{2}}{a_{c}}\)

$$r=\frac{(400)^{2}}{98}=\frac{160000}{98}\approx1632.65\space m$$

Convert to kilometers: \(r=\frac{1632.65}{1000}=1.63\space km\)

Answer:

(a) \(5.57g\)
(b) \(1.63\space km\)