91影视

The given data can be listed below as follows,

This law describes that the orbital period's square of a particular planet is directly proportional to the cubes of "semi-major axes" of the planet's orbit.

The root of the product of the cube of the average orbital radius of Vulcan with the Kepler鈥檚 constant gives the orbital period of Vulcan.

From Kepler鈥檚 third law, the orbital period of Vulcan can be expressed as:

${\mathrm{T}}^2=\left(\frac{4{\mathrm{蟺}}^2}{\mathrm{GM}}\right){\mathrm{r}}^3$

Here, T is the orbital period, and the value of $\mathrm{蟺}$is $3.14$. M is the mass of the sun, which is $1.99脳{10}^{30}\text{鈥塳驳}$. Moreover, G is the gravitational constant $6.673脳{10}^{鈭�11}\text{鈥�}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}$ , and r is the orbital radius of Vulcan $2/3脳5.79脳{10}^{10}\text{鈥墂丑颈肠丑鈥塱蝉鈥塨别肠辞尘别鈥�}3.86脳{10}^{10}\text{m}$.

Substituting the values in the above equation, we get

$\begin{array}{c}{\mathrm{T}}^2=\frac{4脳{(3.14)}^2脳{(3.86脳{10}^{10}\text{鈥尘})}^3}{6.673脳{10}^{鈭�11}\text{鈥�}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}脳1.99脳{10}^{30}\text{鈥塳驳}}\\ {\mathrm{T}}^2=1.701脳{10}^{13}\\ \mathrm{T}=4134982.965\text{鈥塻}脳\frac{1\text{鈥塰辞耻谤}}{3600\text{鈥塻}}脳\frac{1\text{鈥塪补测}}{24\text{鈥塰辞耻谤}}\\ \mathrm{T}=47.85\text{鈥塪补测}\end{array}$

Thus, the orbital period of such a planet is $47.85\text{鈥塪补测}$.