The dwarf planet Pluto has an elliptical orbit with a semi-major axis of $\mathbf{5}.\mathbf{91}\times {\mathbf{10}}^{12}\mathbf{m}$and eccentricity $\mathbf{0}.\mathbf{249}$.

(a) Calculate Pluto’s orbital period. Express your answer in seconds and in earth years.

(b) During Pluto’s orbit around the sun, what are its closest and farthest distances from the sun?

• The semi-major axis of Pluto is $5.91\times {10}^{12}\text{\hspace{0.17em}m}$.
• The eccentricity of Pluto is $0.249$.

The law states that the square of the orbital period of a planet is mainly proportional to the cube of the "semi-major axes" of the orbit of the particular planet.

The product of the cube of the orbital radius of Pluto and the Kepler’s constant gives the orbital period of Pluto.

${\mathrm{T}}^2=\left(\frac{4{\mathrm{\pi }}^2}{\mathrm{GM}}\right){\mathrm{r}}^3$

Where the value $\mathrm{\pi }$ is3.14 andT is the orbital period. Apart from that, M is the mass of the sun which is $1.99\times {10}^{30}\text{\hspace{0.17em}kg}$. Furthermore, G is the gravitational constant $6.673\times {10}^{-11}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}$, and r is Pluto's orbital radius $5.91\times {10}^{12}\text{\hspace{0.17em}m}$.

Substituting the values in the above equation, we get,

$\begin{array}{c}{\mathrm{T}}^2=\frac{4\times {(3.14)}^2\times {(5.91\times {10}^{12}\text{\hspace{0.17em}m})}^3}{6.673\times {10}^{-11}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\times 1.99\times {10}^{30}\text{\hspace{0.17em}kg}}\\ {\mathrm{T}}^2=6.13\times {10}^{19}\\ \mathrm{T}=7829431652\text{\hspace{0.17em}s}\end{array}$

Thus, the orbital period of Pluto is about $7829431652\text{\hspace{0.17em}s}$.

b)

From the Kepler’s third law, the equation of the closest distance of the Pluto from the sun is-$R=a(1-e)$

Here, R is the closest distance, a is the semi-major axis and e is the eccentricity.

Substituting the values in the above equation, we get-

$\begin{array}{c}\mathrm{R}=5.91\times {10}^{12}\text{\hspace{0.17em}m}\times (1-0.249)\\ =4.4\times {10}^{12}\text{\hspace{0.17em}m}\end{array}$

From the Kepler’s third law, the equation of the farthest distance of the Pluto from the sun is-

$r=a(1+e)$

Here, r is the closest distance, a is the semi-major axis and e is the eccentricity.

Substituting the values in the above equation, we get-

$\begin{array}{c}\mathrm{R}=5.91\times {10}^{12}\text{\hspace{0.17em}m}\times (1+0.249)\\ =7.38\times {10}^{12}\text{\hspace{0.17em}m}\end{array}$

Thus, the closest distance of Pluto from the sun is $4.4\times {10}^{12}\text{\hspace{0.17em}m}$and the farthest distance of Pluto from the sun is $7.38\times {10}^{12}\text{\hspace{0.17em}m}$.