The orbit of the Earth around the Sun is approximately circular, and takes one year to complete. The Earth's mass is $\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{}\mathit{k}\mathit{g}$, and the distance from the Earth to the Sun is $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathbf{}\mathit{m}$. What is $\mathbf{(}\mathit{d}\overrightarrow{\mathbf{p}}\mathbf{\mid }\mathbf{/}\mathit{d}\mathit{t}\mathbf{)}\hat{\mathbf{p}}$ of the Earth? What is $\overrightarrow{\mathbf{p}}\mathbf{(}\mathit{d}\hat{\mathbf{p}}\mathbf{/}\mathit{d}\mathit{t}\mathbf{)}$ of the Earth? What is the magnitude of the gravitational force the Sun (mass$\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{30}}\mathit{k}\mathit{g}$) exerts on the Earth? What is the direction of this force?

The orbit of the Earth around the Sun is approximately circular, and takes one year to complete. The Earth's mass is $m_E=6\times {10}^{24}\text{kg}$, and the distance from the Earth to the Sun is $R=1.5\times {10}^{11}\text{m}$.

A force is a push or pull on an object caused by the interaction of the thing with another object. Every time two things interact, a force is exerted on each of them. The two items no longer feel the force after the interaction ends.

The speed is the change in the distance with time.

$v=\frac{d}{t}$

The Earth moves above the circumference of a circle when it completes one round.

The circumference is given by$2\pi R$.

Where R is the radius of the circle (distance between the wooden horse and the center of the carousel).

So, the distance is$d=2\pi R$

Therefore,

$v=\frac{2\pi R}{t}$

Now, plug the values for R and t to get the speed of the wooden horse,

$$\begin{gathered} t=\left(1\text{year}\right)(365\text{day}/\text{year})(24\text{hr}/\text{day})(3600\text{s}/\text{h}) \\ =31.536\times {10}^6\text{s} \end{gathered}$$

Therefore,

$$\begin{gathered} v=\frac{2\pi R}{t} \\ =\frac{2\pi \left(1.5\times {10}^{11}\text{m}\right)}{31.536\times {10}^6\text{s}} \\ =2.98\times {10}^4\text{m}/\text{s} \end{gathered}$$

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two components.

The parallel rate of change of momentum${\frac{d\overrightarrow{p}}{dt}}_{\left|\right|}$and the perpendicular rate of change of momentum${\frac{d\overrightarrow{p}}{dt}}_{\perp }$are the two elements that we are concerned with.

So, the change of momentum required is given by,

$\frac{d\overrightarrow{p}}{dt}={\frac{d\overrightarrow{p}}{dt}}_{\left|\right|}+{\frac{d\overrightarrow{p}}{dt}}_{\perp }$

The parallel rate of change of momentum affects the object's speed, and since the speed is constant, the parallel rate is zero and equal to the rate of change of the size of the momentum.

$$\begin{gathered} {\frac{d\overrightarrow{p}}{dt}}_{\left|\right|}=\frac{d\left|\overrightarrow{p}\right|}{dt}\hat{p} \\ =0 \end{gathered}$$

As a result, the rate change is the direction change owing to the perpendicular rate of change.

The magnitude of the perpendicular rate change equals the rate change of the direction of the momentum and at speeds much less than the speed of light is given by

$$\begin{gathered} {\frac{d\overrightarrow{p}}{dt}}_{\perp }=\left|\overrightarrow{p}\right|\frac{d\hat{p}}{dt} \\ =\frac{m_E{v}^2}{R} \end{gathered}$$

Now, put the values for $m_E,\nu andR$to get$\left|\overrightarrow{p}\right|\frac{d\hat{p}}{dt}$

$$\begin{gathered} \left|\overrightarrow{p}\right|\frac{d\hat{p}}{dt}=\frac{m_E{v}^2}{R} \\ =\frac{\left(6\times {10}^{24}\text{kg}\right){\left(2.98\times {10}^4\text{m}/\text{s}\right)}^2}{1.5\times {10}^{11}\text{m}} \\ =3.55\times {10}^{22}\text{kg}·\text{m}/{\text{s}}^2 \end{gathered}$$

Also, there is a gravitational force between the Earth and the Sun and it is given by

$F_g=G\frac{m_E{m}_S}{R^2}$

Where G is the gravitational constant and equals $6.67\times {10}^{-11}\text{N}·{\text{m}}^2/{\text{kg}}^2{\text{,m}}_E$is the mass of the earth, and $m_s$is the mass of the Sun.

Now, put the values for$m_E,m_s,GandRtoget{F}_g$

$$\begin{gathered} F_g=G\frac{m_E{m}_S}{R^2} \\ =\left(6.67\times {10}^{-11}\text{N}·{\text{m}}^2/{\text{kg}}^2\right)\frac{\left(6\times {10}^{24}\text{kg}\right)\left(2\times {10}^{30}\text{kg}\right)}{{\left(1.5\times {10}^{11}\text{m}\right)}^2} \\ =3.55\times {10}^{22}\text{N} \end{gathered}$$

The rate of change in momentum matches the force exerted on the Earth as demonstrated by the data, which makes sense. The Earth's motion is virtually round with the period repeating itself. As a result, the Earth's velocity and momentum are tangential, and the direction of change in momentum is toward the path's centre.

Thus, the force is directed toward the path's centre.