91影视

The given data can be listed below as:

• The semi major axis of Mercury is $a_1=57.9脳{10}^6\text{km}$.
• The orbital period of Mercury s is T₁ = 88 Days.
• The semi major axis of Venus is $a_2=108.2脳{10}^6\text{km}$.
• The orbital period of Venus is T₂ = 224.7 Days. .
• The semi major axis of Earth is $a_3=149.6脳{10}^6\text{km}$.
• The orbital period of Earth is .T₃ = 365.2 Days.
• The semi major axis of Mars is$a_4=277.9脳{10}^6\text{km}$ .
• The orbital period of Mars is T₄ = 687 Days. .
• The semi major axis of Jupiter is$a_5=778.3脳{10}^6\text{km}$ .
• The orbital period of Jupiter is T₅ = 4331 Days. .
• The semi major axis of Saturn is $a_6=1426.7脳{10}^6\text{km}$.
• The orbital period of Saturn is T₆ = 10747 Days. .
• The semi major axis of Uranus is$a_7=2870.7脳{10}^6\text{km}$ .
• The orbital period of Uranus is .T₇ = 30589 Days.
• The semi major axis of Neptune is $a_8=4498.4脳{10}^6\text{km}$.
• The orbital period of Neptune is T₈ = 59800 Days. .

The orbital period is referred to as the time which is required for completing one particular orbit of a planet. For an example, the orbital period of Earth is about 365 days.

The equation of the Kepler鈥檚 third law is expressed as:

$T=\frac{2蟺a^{\frac{3}{2}}}{\sqrt{G{m}_{sun}}}$

Here, Tis the orbital period of a particular planet, is the semi-major axis, is the gravitational constant and is the mass of the sun.

Squaring the both sides of the above equation, it can be represented as:

$T^2=\frac{4{蟺}^2{a}^3}{G{m}_{sun}}$

鈥�(颈)

From the above equation, it can be identified that $T^2鈭�a^3$. This is the main reason the values fall close to a straight line.

Thus, as $T^2鈭�a^3$, the values fall close to a straight line.

It has been identified that the above answer has been gathered with the help of the Kepler鈥檚 third law.

Thus, the Kepler鈥檚 third law is being tested.

The equation (i) has been recalled below:

$$\begin{gathered} {\mathrm{T}}^2=\frac{4{\mathrm{蟺}}^2{\mathrm{a}}^3}{{\mathrm{Gm}}_{\mathrm{sun}}} \\ \frac{{\mathrm{T}}^2}{{\mathrm{a}}^3}=\frac{4{\mathrm{蟺}}^2}{{\mathrm{Gm}}_{\mathrm{sun}}} \end{gathered}$$

Applying on the sides of the above equation.

$$\begin{gathered} \mathrm{Log}\left(\frac{{\mathrm{T}}^2}{{\mathrm{a}}^3}\right)=\mathrm{Log}\left(\frac{4{\mathrm{蟺}}^2}{{\mathrm{Gm}}_{\mathrm{sun}}}\right) \\ {\mathrm{LogT}}^2-{\mathrm{Loga}}^3=\mathrm{Log}\left(\frac{4{\mathrm{蟺}}^2}{{\mathrm{Gm}}_{\mathrm{sun}}}\right) \\ 2\mathrm{logT}=3\mathrm{loga}+\mathrm{Log}\left(\frac{4{\mathrm{蟺}}^2}{{\mathrm{Gm}}_{\mathrm{sun}}}\right) \\ \mathrm{LogT}=\frac{3}{2}\mathrm{loga}+\frac{1}{2}\mathrm{Log}\left(\frac{4{\mathrm{蟺}}^2}{{\mathrm{Gm}}_{\mathrm{sun}}}\right) \end{gathered}$$

鈥�(颈颈)

Substitute $6.67脳{10}^{-11}\text{N}路{\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$for G and $1.99脳{10}^{30}\text{kg}$for in the above equation.

$\begin{array}{rcl}LogT& =& \frac{3}{2}\log a+\frac{1}{2}\log \frac{4脳{\left(3.14\right)}^2}{\left(6.67脳{10}^{-11}\text{N}路{\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(1.99脳{10}^{30}\text{kg}\right)}\\ & =& \frac{3}{2}\log a+\frac{1}{2}\log \frac{4脳9.8596}{\left(1.32脳{10}^{20}\text{N}路{\text{m}}^{\text{2}}\text{/kg}\right)}\\ & =& \frac{3}{2}\log a+\frac{1}{2}\log \frac{39.43}{\left(1.32脳{10}^{20}\text{N}路{\text{m}}^{\text{2}}\text{/kg}\right)}\\ LogT& =& \frac{3}{2}\log a+\frac{1}{2}\log 2.98脳{10}^{-19}\text{kg/N}路{\text{m}}^{\text{2}}\\ & & \end{array}$

According to the above equation, the graph of versus has been drawn below:

As $LogT鈭�Loga$, that is one of the main reason the data should fall close to a straight line and it has also a straight line according to the graph.

Thus, as $LogT鈭�Loga$, that is one of the main reason the data should fall close to a straight line.

The equation (ii) has been recalled below:

$LogT=\frac{3}{2}\log a+\frac{1}{2}Log\left(\frac{4{蟺}^2}{G{m}_{sun}}\right)$

However, from the above diagram, it can be identified that the slope is 3/2 or 1.5.

Thus, the slope of the graph is 1.5 .

With the help of trend line feature of excel, it can be identified that the graph of versus mainly shows the prediction of the second law of Kepler.

Thus, the graph has this slope which follows the second law of Kepler.

The graph of the -intercept has been drawn below

According to the third law of Kepler,

$$\begin{gathered} \frac{1}{2}\log \frac{4{蟺}^2}{G{m}_{sun}}=-9.2633 \\ -18.5266=\log \frac{4{蟺}^2}{G{m}_{sun}} \end{gathered}$$

Applying exponential on both sides of the above equation.

$$\begin{gathered} {10}^{-18.5266}={10}^{Log\frac{4{蟺}^2}{G{m}_{sun}}} \\ -18.5266=\frac{4{蟺}^2}{G{m}_{sun}} \\ {m}_{sun}=\frac{4{蟺}^2}{\left({10}^{-18.5266}\right)G} \\ =\frac{4{\left(3.14\right)}^2}{\left({10}^{-18.5266}\right)\left(6.67脳{10}^{-11}\right)} \end{gathered}$$

Hence, further as:

$$\begin{gathered} m=\frac{4\left(9.8586\right)}{\left(2.97脳{10}^{-19}\right)\left(6.67脳{10}^{-11}\right)} \\ =\frac{4\left(9.8586\right)}{\left(1.98脳{10}^{-29}\right)} \\ =\frac{39.43}{\left(1.98脳{10}^{-29}\right)} \\ =1.99脳{10}^{30} \end{gathered}$$

Thus, the mass of the sun is$1.99脳{10}^{30}\text{kg}$ .

The value in the appendix F is equal to the mass of the sun that is $1.99脳{10}^{30}\text{kg}$.

Thus, the calculated mass of the sun is equal to the value of the Appendix F.

The equation of the length of the semi-major axis is expressed as:

$$\begin{gathered} {\mathrm{a}}^3=\frac{{\mathrm{T}}^2{\mathrm{Gm}}_{\mathrm{sun}}}{2\mathrm{蟺}} \\ \mathrm{a}=\sqrt[3]{\frac{{\mathrm{T}}^2{\mathrm{Gm}}_{\mathrm{sun}}}{2\mathrm{蟺}}} \end{gathered}$$

Here, a is the length of the semi-major axis and T is the orbital period of Vesta鈥檚 orbit.

Substitute the values in the above equation.

$\begin{array}{rcl}\mathrm{a}& =& \sqrt[3]{\frac{{\left(114514560\text{s}\right)}^2\left(6.67脳{10}^{-11}\text{N}路{\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(1.99脳{10}^{30}\text{kg}\right)}{2\left(3.14\right)}}\\ & =& \sqrt[3]{\frac{\left(1.3脳{10}^{16}{\text{s}}^2\right)\left(1.32脳{10}^{20}\text{N}路{\text{m}}^{\text{2}}\text{/kg}\right)}{6.28}}\\ & =& \sqrt[3]{\frac{\left(1.716脳{10}^{36}\text{N}路{\text{m}}^{\text{2}}路{\text{s}}^2\text{/kg}\right)}{6.28}}\\ & =& \sqrt[3]{2.7脳{10}^{35}\text{N}路{\text{m}}^{\text{2}}路{\text{s}}^2\text{/kg}}\\ & & \end{array}$

Hence, further as:

$$\begin{gathered} a=\sqrt[3]{2.7脳{10}^{35}\text{N}路{\text{m}}^{\text{2}}路{\text{s}}^2\text{/kg}} \\ =\sqrt[3]{2.7脳{10}^{35}\text{N}路{\text{m}}^{\text{2}}路{\text{s}}^2\text{/kg}脳\frac{\text{1}\text{kg}路{\text{m/s}}^{\text{2}}}{\text{1}\text{N}}} \\ =\sqrt[3]{2.7脳{10}^{35}\text{kg}路{\text{m/s}}^{\text{2}}路{\text{m}}^{\text{2}}路{\text{s}}^2\text{/kg}} \\ =\sqrt[3]{2.7脳{10}^{35}{m}^3} \\ =3.5脳{10}^{11}\text{m} \end{gathered}$$

Thus, the length of the semi major axis is $3.5脳{10}^{11}\text{m}$.

According to the table given in the question, the orbit of Vesta is between the orbit of Mars and also Jupiter.

Thus, the orbit of Vesta is between the orbit of Mars and also Jupiter.