Find the velocity relative to an inertial frame (in which the centre of the Earth is at rest) of a point on the Earth's equator. An aircraft is flying above the equator at \(1000 \mathrm{~km} \mathrm{~h}^{-1}\). Assuming that it flies straight and level (i.e., at a constant altitude above the surface) what is its velocity relative to the inertial frame (a) if it flies north, (b) if it flies west, and (c) if it flies east?

First, we will calculate the speed of a point on Earth's equator due to its rotation. Earth's circumference is given by \(C = 2\pi R\), where \(R\) is Earth's radius, approximately \(6371 \mathrm{~km}\). Earth takes about 24 hours to complete one rotation, so the speed of a point on the equator is: v = \frac{C}{T} = \frac{2\pi R}{24\mathrm{~h}} = \frac{2\pi(6371)~\mathrm{km}}{24\mathrm{~h}} v \approx 1670 \mathrm{~km}\mathrm{~h}^{-1} Where v is the speed of a point of the Earth's equator.

Now we will use vector addition to find the aircraft's velocity in each case. Let's denote the aircraft's speed of \(1000 \mathrm{~km} \mathrm{~h}^{-1}\) as \(v_a\). (a) If the aircraft flies north: In this case, the aircraft's velocity is orthogonal to Earth's rotational velocity, so the total velocity is the vector sum of the aircraft's northward speed and Earth's eastward speed. Using the Pythagorean theorem, we get: v_t = \sqrt{v^2 + v_a^2} = \sqrt{(1670\mathrm{~km}\mathrm{~h}^{-1})^2 + (1000\mathrm{~km}\mathrm{~h}^{-1})^2} v_t \approx 1950 \mathrm{~km}\mathrm{~h}^{-1} (b) If the aircraft flies west: In this case, the aircraft's velocity is opposite to Earth's rotational velocity. So, we will subtract Earth's equatorial speed from the aircraft's speed to get the total velocity: v_t = v - v_a = 1670\mathrm{~km}\mathrm{~h}^{-1} - 1000\mathrm{~km}\mathrm{~h}^{-1} v_t = 670 \mathrm{~km}\mathrm{~h}^{-1} (c) If the aircraft flies east: In this case, the aircraft's velocity is in the same direction as Earth's rotational velocity. We will add Earth's equatorial speed to the aircraft's speed to get the total velocity: v_t = v + v_a = 1670\mathrm{~km}\mathrm{~h}^{-1} + 1000\mathrm{~km}\mathrm{~h}^{-1} v_t = 2670 \mathrm{~km}\mathrm{~h}^{-1} In conclusion, the aircraft's velocity relative to the inertial frame (in which Earth's center is at rest) is approximately (a) \(1950 \mathrm{~km}\mathrm{~h}^{-1}\) if it flies north, (b) \(670 \mathrm{~km}\mathrm{~h}^{-1}\) if it flies west, and (c) \(2670 \mathrm{~km}\mathrm{~h}^{-1}\) if it flies east.