91影视

Given that Pluto鈥檚 diameter, ${\mathrm{d}}_{\mathrm{P}}=2370\mathrm{km}$,

The diameter of its satellite Charon,${\mathrm{d}}_{\mathrm{C}}=1250\mathrm{km}$ .

They are often about 19700 km apart, center to center.

Assume that both Pluto and Charon have the same composition and hence the same average density

Let the origin be at center of Pluto.

Let x-axis lies along the line joining the center of Pluto and Charon.

Let p is the density of Pluto and Charon.

So mass m = pV

$$\begin{gathered} \mathrm{m}=\mathrm{p}\frac{4}{3}{\mathrm{蟿蟿谤}}^3 \\ =\frac{4}{3}\mathrm{辫蟿蟿}{\left(\frac{\mathrm{d}}{2}\right)}^3 \\ =\frac{1}{6}{\mathrm{辫蟿蟿d}}^3 \end{gathered}$$

Mass of the Pluto${\mathrm{m}}_{\mathrm{P}}=\frac{1}{6}{{\mathrm{辫蟿蟿d}}_{\mathrm{P}}}^3$

Mass of Jupiter ${\mathrm{m}}_{\mathrm{C}}=\frac{1}{6}{{\mathrm{辫蟿蟿d}}_{\mathrm{C}}}^3$

Distance between centre of Pluto and Charon is km

So ${\mathrm{x}}_{\mathrm{P}}=0$

And ${\mathrm{x}}_{\mathrm{C}}=19700\mathrm{km}$

Center of mass is given by

$\mathbf{x}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{A}}{\mathbf{x}}_{\mathbf{A}}\mathbf{+}{\mathbf{m}}_{\mathbf{B}}\mathbf{+}{\mathbf{x}}_{\mathbf{B}}\mathbf{+}{\mathbf{m}}_{\mathbf{C}}{\mathbf{x}}_{\mathbf{C}}}{{\mathbf{m}}_{\mathbf{A}}\mathbf{+}{\mathbf{m}}_{\mathbf{B}}\mathbf{+}{\mathbf{m}}_{\mathbf{C}}}$

The masses are${\mathbf{m}}_{\mathbf{i}}\mathbf{,}\mathbf{s}$ and position are${\mathbf{x}}_{\mathbf{i}}\mathbf{\text{'}}\mathbf{s}$ .

Now center of mass is$\mathrm{x}=\frac{{\mathrm{m}}_{\mathrm{P}}{\mathrm{x}}_{\mathrm{P}}+{\mathrm{m}}_{\mathrm{C}}{\mathrm{x}}_{\mathrm{C}}}{{\mathrm{m}}_{\mathrm{P}}+{\mathrm{m}}_{\mathrm{C}}}$

$$\begin{gathered} \mathrm{x}=\frac{{\mathrm{m}}_{\mathrm{C}}}{{\mathrm{m}}_{\mathrm{P}}+{\mathrm{m}}_{\mathrm{C}}}\left({\mathrm{x}}_{\mathrm{C}}\right) \\ =\frac{{\displaystyle \frac{1}{6}}{{\mathrm{辫蟿蟿d}}_{\mathrm{c}}}^3}{\frac{1}{6}{{\mathrm{辫蟿蟿d}}_{\mathrm{c}}}^3+\frac{1}{6}{{\mathrm{辫蟿蟿d}}_{\mathrm{c}}}^3}\left({\mathrm{x}}_{\mathrm{C}}\right) \\ =\frac{{\mathrm{d}}_{\mathrm{C}}^3}{{\mathrm{d}}_{\mathrm{C}}^3+{\mathrm{d}}_{\mathrm{C}}^3}\left({\mathrm{x}}_{\mathrm{C}}\right) \end{gathered}$$

So

$$\begin{gathered} \mathrm{x}=\left(\frac{1250{\mathrm{km}}^3}{2370{\mathrm{km}}^3+1250{\mathrm{km}}^3}\right)\left(19700\mathrm{m}\right) \\ =2.52脳{10}^3\mathrm{m} \end{gathered}$$

Hence, the center of mass is localid="1665052443079" $2.52脳{10}^3$m from the center of Pluto.