Using Appendix F, along with the fact that the earth spins on its axis once per day, calculate

(a) the earth’s orbital angular speed (in rad/s) due to its motion around the sun,

(b) its angular speed (in rad/s) due to its axial spin,

(c) the tangential speed of the earth around the sun (assuming a circular

orbit),

(d) the tangential speed of a point on the earth’s equator due to the planet’s axial spin, and

(e) the radial and tangential acceleration components of the point in part (d).

We know what the radius of the earth is $R_E=6.38\times {10}^{6}m$.

The earth rotation once in 1 day = 86.400s.

The orbit radius of the earth is$150x{10}^{11}$

The earth completes one orbit in y=role="math" localid="1667993094016" $3.156x{10}^{7}s$.

It$\omega$ is constant ,role="math" localid="1667993127429" $\omega =\frac{\theta }{t}$ __________(1)

$\theta =1rev=2\pi rad$androle="math" localid="1667993186939" $t=3.156x{10}^{7}s$

Put the value in equation (1)

$\omega =\frac{2\pi rad}{3.156\times {10}^{7}s}$

We know that the value of the $\pi =3.14$.

$$\begin{gathered} \omega =\frac{2x3.14rad}{3.156\times {10}^{7}s} \\ =1.99\times {10}^{-7}\frac{rad}{s} \end{gathered}$$

The motion around the sun $1.99\times {10}^{-7}\frac{rad}{s}$.

We know that,$\theta =1rev=2\pi rad$

Given that, $t=86,400s$.

We know that the angular velocity,

$\omega =\frac{\theta }{t}$

$=\frac{2\pi rad}{86,400s}$

Here we know that the value of$\pi =3.14$

role="math" localid="1667993660962" $$\begin{gathered} \omega =\frac{2\times 3.14\times \mathrm{ra}d}{86,400s} \\ =7.27\times {10}^{-5}\frac{rad}{s} \end{gathered}$$

The angular speed is $7.27\times {10}^{-5}\frac{rad}{s}$.

Given that $r=1.50x{10}^{11}m$

$\omega =1.99x{10}^{-7}\frac{rad}{s}$.

We know the formula $v=r\omega$.

Here we use the formula $v=r\omega$.

Put the value in this equation $r=150x{10}^{11}m$

And $\omega =1.99x{10}^{-7}\frac{rad}{s}$.

We get,

role="math" localid="1667994013137" $$\begin{gathered} v=\left(1.50x{10}^{11}m\right)\left(1.99x{10}^{-7}\frac{rad}{s}\right) \\ =2.98\times {10}^4\frac{m}{s} \end{gathered}$$

Here we know that the $\omega =7.27x{10}^{-5}\frac{rad}{s}$, and$r=6.38x{10}^{6}m$

We know that the formula $v=r\omega$_______(1)

Put the value$r=6.38x{10}^{6}m$ in equation (1)

Then we get,

$$\begin{gathered} v=\left(6.38x{10}^{6}m\right)\left(7.27x{10}^{-5}\frac{rad}{s}\right) \\ =464\frac{m}{s} \end{gathered}$$

the tangential speed of a point on the earth’s equator due to the planet$464\frac{m}{s}$.

We know that the value of$r=6.38\times {10}^{6}m$ and $\omega =7.27x{10}^{-5}\frac{rad}{s}$.

Here we use the formula $a_{rad}=r{w}^2$__________(1)

Put the value$\omega =7.27x{10}^{-5}\frac{rad}{s}$ and $r=6.38x{10}^{6}m$.

Then we get,

$$\begin{gathered} =\left(6.38x{10}^{6}m\right)\left(7.27x{10}^{-5}\frac{rad}{s^2}\right) \\ =0.0337\frac{m}{s^2} \end{gathered}$$

We also know that $\alpha =0$.

Since the angular velocity is constant.