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 given by the formula $e = \frac{r_a - r_p}{r_a + r_p}$, where $r_a$ is the aphelion distance and $r_p$ is the perihelion distance. Substituting $r_a = 2.7 \times 10^7 \mathrm{km}$ and $r_p = 2.3 \times 10^7 \mathrm{km}$, we get $e = \frac{2.7 \times 10^7 - 2.3 \times 10^7}{2.7 \times 10^7 + 2.3 \times 10^7} = \frac{0.4 \times 10^7}{5.0 \times 10^7} = 0.08$.

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 $2.3 \times 10 ^ { 7 } \mathrm {~km}$