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 1.3 $\times 10 ^ { 7 }$ km, and its aphelion distance (maximum distance from the planet to the sun) is approximately 6.9 $\times 10 ^ { 7 }$ km. Approximate the eccentricity of the planet's orbit. Round to three decimal places.
A) 0.683
B) 5.308
C) 1.464
D) 0.188

Question

Quizplus · Accepted Answer

The eccentricity of an ellipse is given by the formula e = c/a, where c is the distance from the center of the ellipse to a focus and a is the semi-major axis of the ellipse. In this case, the semi-major axis is (1.3 + 6.9)/2 = 4.1 $\times 10 ^ { 7 }$ km and the distance from the center of the ellipse to a focus is 6.9 - 4.1 = 2.8 $\times 10 ^ { 7 }$ km. Therefore, the eccentricity is 2.8/4.1 = 0.683 (rounded to three decimal places).

Suppose a Planet Is Discovered That Revolves Around Its Sun ×107\times 10 ^ { 7 }×107

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