[YMP/CM-08002] How good are different frames as inertial frames

Problem

Estimate the following accelerations.

1. Acceleration of the Earth due to its spin;
2. Acceleration of the Earth due to its orbital motion;
3. Acceleration of the Sun due to rotation about center of our galaxy.

Solution

1. We compute the accelerations of the earth at the equator taking the radius of the earth to be \(\approx\) 6000km. The earth rotates by \(2\pi\) in 24 hours. Hence its angular velocity is \[\omega= 2\pi/(24\times3600) \text{rad/s}\] Therefore acceleration is \begin{eqnarray} a &=& \omega^2 R \nonumber\\ &=& (2\pi)^2 6400\times10^3/(24\times3600)^2 \nonumber\\ &\approx& 40\times64 \times10^5/(576\times1296\times 10^4) \text{m/s}^2\nonumber\\ &\approx& \frac{40}{60}\times\frac{65}{13}\times 10^{-2}\text{m/s}^2 \approx\frac{2}{3}\times 5 \times10^{-2}\text{m/s}^2\nonumber\\ &\approx&3.3\times10^{-3}g \end{eqnarray}
2. For orbital motion we take the distance of the Earth from sun to be \(D\sim150\times10^6\)km.Approximating the orbit to be circular the angular velocity is \begin{eqnarray} \omega&=& \frac{2\pi}{(365\times24\times 3600)}\approx \frac{2\pi}{(9000\times3600)}\nonumber\\ &=& \frac{2\pi}{9\times36} \times 10^{-5} \text{rad/s} \end{eqnarray} Therefore acceleration due to orbital motion is \begin{eqnarray} \omega^2 D &\approx& \Big(\frac{2\pi}{9\times36}\Big)^2 \times 10^{-10}\times150\times10^9 m\nonumber\\ &\approx& \frac{40 \times15}{(9\times36)^2} \text{m/s}^2 =\frac{600}{324\times324}\text{m/s}^2\nonumber\\ &\approx& 6 \times10^{-4}g \end{eqnarray}
3. With velocity of Sun = 240km/s, distance of Sun from center of galaxy to be \(8 \text{kpc}\sim24\times10^{16}\text{\,km}\) we get acceleration of the Sun \begin{eqnarray} a &=& (240\times \text{km/s})^2/(24 \times 10^{19}\text{m})\nonumber\\ &=& (2400\times10^6 \text{\,m/s}^2)/(10^{19}\text{m})\nonumber\\ &=& 2.4\times 10^{-10} \text{\,m/s}^2= 2.4\times 10^{-11}g \end{eqnarray}

• One parsec is equal to about 3.26 light-years or 30 trillion km \\ 1 kpc\(\sim 3\times10^{16}\)km.