NCERT Solutions

We need to calculate the aircraft's centripetal acceleration and compare its magnitude to the standard acceleration due to gravity.

Concept: Centripetal acceleration

The centripetal acceleration $a_c$ of an object moving in a circular path of radius $r$ at a constant speed $v$ is given by:

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

To compare this with the acceleration due to gravity $g$, we find the ratio:

$$n = \frac{a_c}{g}$$

Given:

• Radius of the loop, $r = 1.00\,\text{km} = 1000\,\text{m}$
• Speed of the aircraft, $v = 900\,\text{km}\cdot\text{h}^{-1}$
• Acceleration due to gravity, $g = 9.8\,\text{m}\cdot\text{s}^{-2}$

Solving:

1. Convert the speed to SI units ($\text{m}\cdot\text{s}^{-1}$)

$$\begin{aligned} v &= 900 \times \frac{1000\,\text{m}}{3600\,\text{s}} \\ &= 250\,\text{m}\cdot\text{s}^{-1} \end{aligned}$$

2. Calculate the centripetal acceleration

$$\begin{aligned} a_c &= \frac{(250\,\text{m}\cdot\text{s}^{-1})^2}{1000\,\text{m}} \\ &= \frac{62500\,\text{m}^2\cdot\text{s}^{-2}}{1000\,\text{m}} \\ &= 62.5\,\text{m}\cdot\text{s}^{-2} \end{aligned}$$

3. Compare with the acceleration due to gravity

$$\begin{aligned} \frac{a_c}{g} &= \frac{62.5\,\text{m}\cdot\text{s}^{-2}}{9.8\,\text{m}\cdot\text{s}^{-2}} \\ &\approx 6.38 \end{aligned}$$

Answer:

$$62.5\,\text{m}\cdot\text{s}^{-2}$$

This is approximately:

$$6.38\,g$$