91影视

• The orbit is$\frac{1}{9}$ the distance of the Mercury from the sun.
• The planet takes about$3.09$ making one orbit.

This law illustrates that a planet鈥檚 orbital period鈥檚 square is equal to its orbit鈥檚 鈥渟emi-major axes鈥�.

The product of the cube of the radius and the Kepler's constant, along with the time period, will provide the star's mass. Moreover, dividing the perimeter of the star by the time period gives the star's speed.

$\begin{array}{c}\mathrm{T}=\frac{2{\mathrm{蟺谤}}^{3/2}}{\sqrt{{\mathrm{Gm}}_{\mathrm{star}}}}\\ {\mathrm{m}}_{\mathrm{star}}=\frac{4{\mathrm{蟺}}^2{\mathrm{r}}^3}{{\mathrm{T}}^2\mathrm{G}}\end{array}$

Here, T is the planet's orbital period, which is$3.09\text{鈥塪补测蝉=2.67}脳{10}^{5\text{鈥�}}\text{鈥塻}$r is the orbit radius of theearth $\frac{5.79脳{10}^{10}\text{鈥尘}}{9}=6.43脳{10}^9\text{m}$. The value of G, which is the gravitational constant is $6.67脳{10}^{鈭�11}\text{鈥�}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}$.

Substituting the values in the above equation, we get,

$\begin{array}{l}{m}_{star}=\frac{{\text{4蟺}}^{\text{2}}{\left({\text{6.43脳10}}^{\text{9}}\text{鈥尘}\right)}^{\text{3}}}{{\left({\text{2.67脳10}}^{\text{5}}\text{鈥塻}\right)}^{\text{2}}\left({\text{6.67脳10}}^{\text{-11}}\text{鈥�}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)}\\ m_{star}=2.21脳{10}^{30}\text{鈥塳驳}\end{array}$

Hence, the mass of the star is $2.21脳{10}^{30}\text{鈥塳驳}$.

Apart from that, it has been observed that $\frac{{\mathrm{m}}_{\mathrm{star}}}{{\mathrm{m}}_{\mathrm{sun}}}=1.11$, hence,${\mathrm{m}}_{\mathrm{star}}=1.11{\mathrm{m}}_{\mathrm{sun}}$

$\mathrm{v}=\frac{2\mathrm{蟺谤}}{\mathrm{T}}$

Here v is the orbital velocity, r is the orbital radius of the planet,$6.43脳{10}^9\text{鈥尘}$ and T is the earth's orbital period $2.67脳{10}^5\text{鈥塻}$.

Substituting the values in the above equation, we get-

$\begin{array}{c}\mathrm{v}=\frac{2\mathrm{蟺}(6.43脳{10}^9\text{鈥尘})}{2.67脳{10}^5\text{鈥塻}}\\ \mathrm{v}=1.51脳{10}^5\text{鈥塻}\end{array}$

Thus, the planet is moving at a speed of $1.51脳{10}^5\text{鈥塻}$.