91Ó°ÊÓ

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’s constant gives the orbital period of Vulcan.

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

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

Here, T is the orbital period, and the value of $\mathrm{Ï€}$is $3.14$. M is the mass of the sun, which is $1.99×{10}^{30}\text{ k²µ}$. Moreover, G is the gravitational constant $6.673×{10}^{−11}\text{ }\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}$ , and r is the orbital radius of Vulcan $2/3×5.79×{10}^{10}\text{ w³ó¾±³¦³ó i²õ b±ð³¦´Ç³¾±ð }3.86×{10}^{10}\text{m}$.

Substituting the values in the above equation, we get

$\begin{array}{c}{\mathrm{T}}^2=\frac{4×{(3.14)}^2×{(3.86×{10}^{10}\text{″¾})}^3}{6.673×{10}^{−11}\text{ }\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}×1.99×{10}^{30}\text{ k²µ}}\\ {\mathrm{T}}^2=1.701×{10}^{13}\\ \mathrm{T}=4134982.965\text{ s}×\frac{1\text{ h´Ç³Ü°ù}}{3600\text{ s}}×\frac{1\text{ d²¹²â}}{24\text{ h´Ç³Ü°ù}}\\ \mathrm{T}=47.85\text{ d²¹²â}\end{array}$

Thus, the orbital period of such a planet is $47.85\text{ d²¹²â}$.