The planet Mercury travels in an elliptical orbit with eccentricity 0.703. Its minimum distance from the Sun is $8 \times 10 ^ { 7 } \mathrm {~km}$. If the perihelion distance from a planet to the Sun is $a ( 1 - e )$ and the aphelion distance is $a ( 1 + e )$, find the maximum distance (in $\mathrm { km }$ ) from Mercury to the Sun.
Select the correct answer.
A) $45.9 \times 10 ^ { 7 } \mathrm {~km}$
B) $48.0 \times 10 ^ { 7 } \mathrm {~km}$
C) $0.5 \times 10 ^ { 7 } \mathrm {~km}$
D) $26.9 \times 10 ^ { 7 } \mathrm {~km}$
E) $50.0 \times 10 ^ { 7 } \mathrm {~km}$

Question

Quizplus · Accepted Answer

The aphelion distance is $a ( 1 + e )$, where $a$ is the semi-major axis and $e$ is the eccentricity. The semi-major axis is the average of the perihelion and aphelion distances, so $a = (8 \times 10 ^ { 7 } \mathrm {~km} + x) / 2$. Substituting this into the equation for the aphelion distance, we get $x = 40 \times 10 ^ { 7 } \mathrm {~km}$. Therefore, the maximum distance from Mercury to the Sun is $x + 8 \times 10 ^ { 7 } \mathrm {~km} = 48 \times 10 ^ { 7 } \mathrm {~km}$.

The Planet Mercury Travels in an Elliptical Orbit with Eccentricity 8×107 km8 \times 10 ^ { 7 } \mathrm {~km}8×107 km

The Planet Mercury Travels in an Elliptical Orbit with Eccentricity $8 \times 10 ^ { 7 } \mathrm {~km}$