Question

The distance from the planet Pluto to the sun is\({\rm{4}}{\rm{.43 \times 1}}{{\rm{0}}^{\rm{9}}}\)km at perihelion and\({\rm{7}}{\rm{.37 \times 1}}{{\rm{0}}^{\rm{9}}}\) km at aphelion.

Use Exercise \({\rm{24}}\) to find the eccentricity of Pluto’s orbit.

Short answer

The eccentricity of Pluto’s orbit is\({\rm{e}} \approx {\rm{0}}{\rm{.249}}\).

Step-by-step solution

Expand the equation of perihelion and aphelion.

From 8 in the book we know that perihelion is

\(\begin{aligned}{c}{\rm{r = a(1 - e)}}\\{\rm{4}}{\rm{.43 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ = a(1 - e)}}\end{aligned}\)

Solve for\({\rm{a}}\)

\({\rm{a = }}\frac{{{\rm{4}}{\rm{.43 \times 1}}{{\rm{0}}^{\rm{9}}}}}{{{\rm{1 - e}}}}\)

Also, from \({\rm{8}}\) in the book we know that aphelion is

\(\begin{aligned}{c}{\rm{r = a(1 + e)}}\\{\rm{7}}{\rm{.37 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ = a(1 + e)}}\end{aligned}\)

Solve for\({\rm{a}}\)

\({\rm{a = }}\frac{{{\rm{7}}{\rm{.37 \times 1}}{{\rm{0}}^{\rm{9}}}}}{{{\rm{1 + e}}}}\)

Find the eccentricity of Pluto’s orbit.

Set the two equations equal to each other (since both equal\({\rm{a}}\) ), then solve for\({\rm{e}}\)

\(\frac{{{\rm{7}}{\rm{.37 \times 1}}{{\rm{0}}^{\rm{9}}}}}{{{\rm{1 + e}}}}{\rm{ = }}\frac{{{\rm{4}}{\rm{.43 \times 1}}{{\rm{0}}^{\rm{9}}}}}{{{\rm{1 - e}}}}\)

Cancel out \({\rm{1}}{{\rm{0}}^{\rm{9}}}\) then cross multiply.

\(\begin{aligned}{c}{\rm{7}}{\rm{.37(1 - e) = 4}}{\rm{.43(1 + e) }}\\{\rm{7}}{\rm{.37 - 7}}{\rm{.37 e = 4}}{\rm{.43 + 4}}{\rm{.43 e }}\\{\rm{ - 7}}{\rm{.37 e - 4}}{\rm{.43 e = 4}}{\rm{.43 - 7}}{\rm{.37}}\\{\rm{ - 11}}{\rm{.8e = - 2}}{\rm{.94}}\\{\rm{e = }}\frac{{{\rm{ - 2}}{\rm{.94}}}}{{{\rm{ - 11}}{\rm{.8}}}}\\ \approx {\rm{0}}{\rm{.249}}\\{\rm{e}} \approx {\rm{0}}{\rm{.249}}\end{aligned}\)

Therefore, the eccentricity of the Pluto’s orbit is\({\rm{e}} \approx {\rm{0}}{\rm{.249}}\).