In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. Pluto already was known to have a large satellite Charon, orbiting at 19,600 km with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites withoutusing the mass of Pluto.

• The first satellite is orbiting Pluto at about $48000\text{\hspace{0.17em}km}$.
• The second satellite is orbiting Pluto at about $64000\text{\hspace{0.17em}km}$.
• Charon is orbiting Pluto $19600\text{\hspace{0.17em}km}$.
• The orbital period of Charon is $6.39\text{\hspace{0.17em}days}$.

Newton’s second law of motion illustrates that the rate of change of the momentum of a body is mainly equal to the direction and the magnitude of the force.

The attraction law illustrates the force exerted is equal to the masse’s product and divided by the distance’s square amongst them.

The law of motion helps identify the circular motion of the small satellites, and the law of attraction helps identify the orbital period of the small satellites.

From Newton’s second law of motion, the force required for the small satellites to orbit Pluto is expressed as:

$\begin{array}{c}{\mathrm{F}}_{\mathrm{net}}=\mathrm{ma}\\ =\mathrm{m}\frac{{\mathrm{v}}^2}{\mathrm{r}}\end{array}$

Here, m is the mass of a satellite, and a is the satellite's acceleration. Moreover, v is the satellite's velocity, and r is the radius of the satellite.

Hence, equation i) can also be written as:

$\begin{array}{c}\mathrm{G}\frac{{\mathrm{mm}}_{\mathrm{p}}}{{\mathrm{r}}^2}=\mathrm{m}\frac{{\mathrm{v}}^2}{\mathrm{r}}\\ {\mathrm{Gm}}_{\mathrm{p}}={\mathrm{rv}}^2=\text{constant}\end{array}$ …..ii)

Where G is the gravitational constant and is the mass of Pluto.

Hence, from equations i) and ii), it can be concluded as:

$\begin{array}{c}{\mathrm{r}}_1{\mathrm{v}}_1^2={\mathrm{r}}_2{\mathrm{v}}_2^2\\ {\mathrm{r}}_1{\left(\frac{2{\mathrm{\pi r}}_1}{{\mathrm{T}}_1}\right)}^2={\mathrm{r}}_2{\left(\frac{2{\mathrm{\pi r}}_2}{{\mathrm{T}}_2}\right)}^2\\ \frac{{\mathrm{r}}_1^3}{{\mathrm{T}}_1^2}=\frac{{\mathrm{r}}_2^3}{{\mathrm{T}}_2^2}\\ \end{array}$ ….iii)

Here $r_1$ is the radius of Charon satellites and $r_2$the radius of the first small satellite. Moreover, it role="math" localid="1655716360679" $T_{1}and{T}_2$is the orbital period of the Charon and the first small satellite, respectively.

Hence, the equation iii) can be expressed as:

$T_2=T_1{\left(\frac{r_2}{r_1}\right)}^{3/2}$ …..iv)

Substituting the values in equation iv), we get-

$\begin{array}{l}{\text{T}}_{\text{2}}\text{=}\left(\text{6.39\hspace{0.17em}days}\right){\left(\frac{\text{48000}}{\text{19600}}\right)}^{\text{3/2}}\\ {\text{T}}_{\text{2}}\text{=24.5\hspace{0.17em}days}\end{array}$

Hence, the orbital period of the first satellite is$24.5\text{\hspace{0.17em}days}$ .

In the same way, the orbital period of the second satellite is-

$\begin{array}{c}{\text{T}}_{\text{3}}\text{=}\left(\text{6.39\hspace{0.17em}days}\right){\left(\frac{\text{64000\hspace{0.17em}km}}{\text{19600\hspace{0.17em}km}}\right)}^{\text{3/2}}\\ {\text{T}}_{\text{3}}\text{=37.7\hspace{0.17em}days}\end{array}$

Hence, the orbital period of the second satellite is$37.7\text{\hspace{0.17em}days}$ .