Suppose a planet is discovered that revolves around its sun in an elliptical orbit with the sun at one focus. Its perihelion distance (minimum distance from the planet to the sun) is approximately $2.3 \times 10 ^ { 7 } \mathrm {~km}$, and its aphelion distance (maximum distance from the planet to the sun) is approximately $2.7 \times 10 ^ { 7 } \mathrm {~km}$. Approximate the eccentricity of the planet's orbit. Round to three decimal places.
A) $1.174$
B) $12.5$
C) $0.08$
D) $0.852$

Question

Quizplus · Accepted Answer

The eccentricity $e$ of an elliptical orbit is calculated using the formula $e = \frac{d}{l}$, where $d$ is the distance between the foci, and $l$ is the major axis length. The major axis length is the sum of the perihelion and aphelion distances, $l = 2.3 \times 10^7 + 2.7 \times 10^7 = 5.0 \times 10^7 \, \text{km}$. The distance between the foci $d$ is the difference between the aphelion and perihelion distances, $d = 2.7 \times 10^7 - 2.3 \times 10^7 = 0.4 \times 10^7 \, \text{km}$. Thus, $e = \frac{0.4 \times 10^7}{5.0 \times 10^7} = 0.08$.

Suppose a Planet Is Discovered That Revolves Around Its Sun 2.3×107 km2.3 \times 10 ^ { 7 } \mathrm {~km}2.3×107 km

Suppose a Planet Is Discovered That Revolves Around Its Sun $2.3 \times 10 ^ { 7 } \mathrm {~km}$