# The pluto orbit around the sun with a period of 247.9 years, its perihelion is 4.45 10 12 m ....

## Question:

The pluto orbit around the sun with a period of 247.9 years, its perihelion is {eq}4.45\times10^{12}\;m {/eq}. What is pluto`s aphelion?

## Kepler's Third law of Planetary Motion:

Kepler's third law of planetary motion states that the square of the period of a planet is proportional to the cube of its semi-major axis. In this problem, we will utilize the idea that the sum of the perihelion and aphelion of a planet is equal to its major axis.

The following variables are given in the problem.

Given:

{eq}T = 247.9 \ \rm years{/eq}

{eq}\rm Perihelion = 4.45 \times 10^{12} \ \rm m{/eq}

If we know the major axis of Pluto, we can get its aphelion. Thus, we need to use Kepler's third law of planetary motion shown below.

$$\frac{T^2}{r^3} = \frac{4 \pi^2}{GM}$$

Where:

{eq}T{/eq} - period

{eq}r{/eq} - semi-major axis

{eq}G = 6.67408 \times 10^{-11} \ \rm m^{3} kg^{-1} s^{-2}{/eq} (Gravitational Constant)

{eq}M = 1.9891 \times 10^{30} \ \rm kg{/eq} (Mass of the Sun)

From the above equation, we can get the semi-major axis.

\begin{align} \frac{T^2}{r^3} &= \frac{4 \pi^2}{GM}\\ \\ r^3 &= \frac{T^2GM}{4 \pi^2}\\ \\ r &= (\frac{T^2GM}{4 \pi^2})^{1/3}\\ \end{align}

Before we substitute the given variables we need to express the period in the unit of second.

$$T = 247.9 \ \rm years \ \text{or} \ 7.81777 \times 10^9 \ \rm s$$

\begin{align} r &= (\frac{T^2GM}{4 \pi^2})^{1/3}\\ r &= (\frac{(7.81777 \times 10^9 \ \rm s)^2(6.67408 \times 10^{-11} \ \rm m^{3} kg^{-1} s^{-2})(1.9891 \times 10^{30} \ \rm kg)}{4 \pi^2})^{1/3}\\ r &\approx{5.901 \times 10^{12} \ \rm m}\\ \end{align}

The major axis is twice the value of {eq}r{/eq}.The aphelion can now be computed.

\begin{align} \rm Aphelion &= 2r - \rm Perihelion\\ &= 2(5.901 \times 10^{12} \ \rm m) - (4.45 \times 10^{12} \ \rm m)\\ &= 7.352 \times 10^{12} \ \rm m\\ \end{align}