The planets in our Solar System have orbits around the Sun that are nearly circular, and v<<c.Calculate the period T(a year-the time required to go around the Sun once) for a planet whose orbit radius is r. This is the relationship discovered by Kepler and explained by Newton. (It can be shown by advanced techniques that this result also applies to elliptical orbits if you replaceby the semi major axis, which is half the longer, major axis of the ellipse.) Use this analytical solution for circular motion to predict the Earth's orbital speed, using the data for Sun and Earth on the inside back cover of the textbook.

The planets in our Solar System have orbits around the Sun that are nearly circular, and v<<c .This is the relationship discovered by Kepler and explained by Newton.

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two parts. The parallel rate of change of momentum ${\mathbf{(}\mathbf{d}\overrightarrow{\mathbf{p}}\mathbf{/}\mathbf{dt}\mathbf{)}}_{\mathbf{\left|}\mathbf{\right|}}$and the perpendicular rate of change ofmomentum ${\mathbf{(}\mathbf{d}\overrightarrow{\mathbf{p}}\mathbf{/}\mathbf{dt}\mathbf{)}}_{\mathbf{\perp }}$are the two parts that we are concerned with.

Change in momentum is given by,

$\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{\mathbf{d}{\mathbf{t}}_{\mathbf{\left|}\mathbf{\right|}}}\mathbf{+}\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{\mathbf{d}{\mathbf{t}}_{\mathbf{\perp }}}$……………………….. (1)

The object's speed is affected by the parallel rate of change of momentum, and because the speed is constant, the parallel rate is zero and equal to the rate of change of the magnitude of the momentum.

$\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{\mathbf{d}{\mathbf{t}}_{\mathbf{\left|}\mathbf{\right|}}}\mathbf{=}\frac{\mathbf{d}\left|\overrightarrow{p}\right|}{\mathbf{d}\mathbf{t}}\stackrel{\mathbf{⏜}}{\mathbf{p}}\mathbf{=}\mathbf{0}$

The direction shift generated by the perpendicular rate of change is known as the rate change. The quantity of the perpendicular rate change matches the rate change of the direction of the momentum at speeds significantly slower than the speed of light.

$\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{{\mathbf{dt}}_{\mathbf{\perp }}}\mathbf{=}\left|\overrightarrow{p}\right|\frac{\mathbf{d}\stackrel{\mathbf{⏜}}{\mathbf{p}}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}}$………………………… (2)

It will also be equivalent to the rate of change of momentum $\mathbf{d}\overrightarrow{\mathbf{p}}\mathbf{/}\mathbf{dt}$.

Where ${\mathit{M}}_{\mathbf{E}}$denotes the Earth's mass.

Also, between the Earth and the Sun, there is a gravitational force that is perpendicular to the Earth and causes a shift in the direction of momentum, and it is given by

$\mathit{F}\mathbf{=}\mathit{G}\frac{{\mathbf{M}}_{\mathbf{E}}{\mathbf{M}}_{\mathbf{s}}}{{\mathbf{R}}^{\mathbf{2}}}$……………………. (3)

The gravitational constant is G, which equals $6.67\times {10}^{11}N.m^2/k{g}^2,M_E$is the mass of the Earth, and$M_s$is the mass of the Sun. Because both forces are equal, this may be deduced the following from equations and.

$\frac{M_E{v}^2}{R}=G\frac{M_E{M}_s}{R^2}$

$$\begin{gathered} v^2=\frac{G{M}_s}{R} \\ v=\sqrt{\frac{G{M}_s}{R}} \end{gathered}$$ …………………………… (4)

The Earth follows a nearly round course, and its speed may be determined based on the change in distance over time. As a result, it is provided by

$V=\frac{d}{T}$…………………………. (5)

Where is the amount of time it takes to complete one round. The Earth moves above the circumference of a circle when it completes one round. The circumference is calculated using

$2\mathrm{\pi R}$

The radius of the kissing circle, or the distance between the Earth and the Sun, is equal to .As a result, the Earth's journey distance is

$d=2\mathrm{\pi R}$

To get the new form, enter our expression into equation (5).

$v=\frac{2\mathrm{\pi R}}{T}$………………………. (6)

The time T can be calculated using equations (4) and (6).

v=v

$\sqrt{\frac{G{M}_s}{R}}=\frac{2\mathrm{\pi R}}{T}$

$\frac{G{M}_s}{R}={\left(\frac{2\mathrm{\pi R}}{T}\right)}^2$ ………………………….. (7)

$T=\frac{2\mathrm{\pi R}}{\sqrt{G{M}_s/R}}$

Thus, this is the required expression to find the time.

The values of R ,$M_s$and G can be inserted into equation (7) to get the value of T

$$\begin{gathered} T=\frac{2\mathrm{\pi R}}{\sqrt{G{M}_{s}R}} \\ =\frac{2\mathrm{\pi }\left(1.5\times {10}^{11}\mathrm{m}\right)}{\sqrt{\left(6.67\times {10}^{-11}N.m^2/k{g}^2\right)\left(2\times {10}^{30}kg\right)\left(1.5\times {10}^{11}m\right)}} \\ =31532932\mathrm{s} \\ =365\mathrm{days} \end{gathered}$$

Also,the values of R and T into the equation and the value of the speed of the earth will be:

role="math" localid="1656855911171" $$\begin{gathered} v=\frac{2\mathrm{\pi R}}{T} \\ =\frac{2\mathrm{\pi }\left(1.5\times {10}^{11}\mathrm{m}\right)}{31532932s} \\ =29.88\times {10}^{3}m/s \end{gathered}$$

The period T for a planet is 365 days and the speed of the Earth is$29.88\times {10}^{3}m/s$ .