Pluto’s diameter is approximately 2370 km, and the diameter of its satellite Charon is 1250 km. Although the distance varies, they are often about 19700 km apart, center to center. Assuming that both Pluto and Charon have the same composition and hence the same average density, find the location of the center of mass of this system relative to the center of Pluto.

Given that Pluto’s 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

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

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

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

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$\mathrm{x}=\frac{{\mathrm{m}}_{\mathrm{P}}{\mathrm{x}}_{\mathrm{P}}+{\mathrm{m}}_{\mathrm{C}}{\mathrm{x}}_{\mathrm{C}}}{{\mathrm{m}}_{\mathrm{P}}+{\mathrm{m}}_{\mathrm{C}}}$

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

So

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

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