Question: An object lying on Earth’s equator is accelerated (a) toward the center of Earth because Earth rotates, (b) toward the Sun because Earth revolves around the Sun in an almost circular orbit, and (c) toward the center of our galaxy because the Sun moves around the galactic center. For the latter, the period is $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathit{y}$and the radius isrole="math" localid="1663148507355" $\mathbf{2}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}\mathbf{ }\mathit{m}$. Calculate these three accelerations as multiples of$\mathbf{}\mathit{g}\mathbf{=}\mathbf{}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$.

The period of earth rotation,

$$\begin{gathered} T=24year \\ =86400s \end{gathered}$$

The radius of earth,$r=6.37\times {10}^{6}m$

The period of earth’s revolution around the sun,

$$\begin{gathered} T=365days \\ =3.16\times {10}^{7}s \end{gathered}$$

The radius of earth revolution,$r=1.5\times {10}^{11}m$

The period for rotation around center of galaxy,

$$\begin{gathered} T=2.5\times {10}^{8}y \\ =7.9\times {10}^{15}s \end{gathered}$$

The radius of rotation around center of galaxy,$r=2.2\times {10}^{20}m$

It is radially directed acceleration with a magnitude equal to the square of the speed of the body along the curve is divided by the total distance from the centre of the circle to the moving body. Inserting the velocity in terms of T in the formula for centripetal acceleration, we can find the acceleration of an object lying on Earth’s equator toward the center of earth, Sun and center of the galaxy toward earth’s center.

Formula:

$$\begin{gathered} V=\frac{2\pi r}{T} \\ a=\frac{V^2}{r} \end{gathered}$$

Where, V is the velocity of object,T is the time period of object and r is the radius of object

We have the equation for velocity and period as

$V=\frac{2\pi r}{T}$

Hence, the centripetal acceleration may be written as

$$\begin{gathered} a=\frac{V^2}{r} \\ a=\frac{4{\pi }^{2}r}{T^2} \end{gathered}$$

To express the result in terms of g, we divide by 9.8 m/s² .

The acceleration associated with Earth’s spin is

$$\begin{gathered} a=\frac{4{\pi }^2\times 6.37\times {10}^6 \text{m}}{{(86400 \text{s})}^2\times \left(9.8 {\text{m/s}}^2\right)} g \\ a=3.4\times {10}^{-3}g \end{gathered}$$

The acceleration due to revolution around the Sun is

$$\begin{gathered} a=\frac{4{\pi }^2\times 1.5\times {10}^{11} \text{m}}{{\left(3.16\times {10}^7 \text{s}\right)}^2\times \left(9.8 {\text{m/s}}^{\text{2}}\right)} g \\ a=6.1\times {10}^{-4}g \end{gathered}$$

The acceleration toward the center of galaxy is

$$\begin{gathered} a=\frac{4{\pi }^2\times 2.2\times {10}^{20} \text{m}}{{\left(7.9\times {10}^{15} \text{s}\right)}^2\times \left(9.8 {\text{m/s}}^{\text{2}}\right)} g \\ a=1.4\times {10}^{-11}g \end{gathered}$$

Acceleration of an object lying on earth’s equator toward the center of galaxy is$a=1.4\times {10}^{-11}g$