91影视

The given data can be listed below as,

• The orbital distance of the planet is,$r=10.5\text{million}\mathrm{km}脳\frac{{10}^9\mathrm{m}}{1\text{million}\mathrm{km}}=10.5脳{10}^{9}m$
• The orbital period is,$T=6.3days脳\frac{86400s}{1day}=54.43脳{10}^{4}s$

The relationship between orbital period and orbital distance is expressed as follows:

$\mathit{T}\mathbf{=}\frac{\mathbf{2}{\mathbf{蟿蟿谤}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}{\sqrt{\mathbf{G}\mathbf{m}}}$

Here, r is the orbital distance of the planet, T is the orbital period, $G=6.673x{10}^{-11}N路\frac{m^2}{k}{g}^2$is Gravitational constant and m is the mass of the star.

From the relation of orbital distance and orbital time, the mass can be expressed in the following form:

$m=\frac{4{蟺}^2{r}^3}{T^{2}G}$

Substitute all the values in the above equation.

$$\begin{gathered} m_P=\frac{4{蟺}^2{\left(10.5脳{10}^9\mathrm{m}\right)}^3}{{\left(54.43脳{10}^4\mathrm{s}\right)}^2\left(6.673脳{10}^{鈭�11}\mathrm{N}鈰�\frac{{\mathrm{m}}^2}{\mathrm{k}}{\mathrm{g}}^2\right)} \\ =2.3脳{10}^{30}\mathrm{kg} \end{gathered}$$

It is known that the mass of sun is,$M=1.99脳{10}^{30}kg$, then convert the mass of planet in terms of mass of sun.

localid="1668309639393" $$\begin{gathered} m=\left(2.3脳{10}^{30}\mathrm{kg}\right)\left(\frac{M}{1.99脳{10}^{30}\mathrm{kg}}\right) \\ =1.2M \end{gathered}$$

Thus, the mass of HD 68988 in terms of mass of the Sun is 1.2M.