Question

The planet Mercury travels in an elliptical orbit with eccentricity 0.203. Its minimum distance from the Sun is 4.5 x 10^7 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 km) from Mercury to the Sun.Pick from the following:1. 7.7 x 10^7 km.2. 6.6 x 10^7 km.3. 6.8 x 10^7 km.

Answer 1

Answer:

Option C

Step-by-step explanation:

From the question we are told that:

Eccentricity $e=0.203$

Minimum distance from the Sun $d_s= 4.5 x 10^7 km$

Perihelion distance from a planet to the Sun is $r= a(1 - e)$

Aphelion distance $r'=a(1 + e)$

Generally the equation for Perihelion distance is mathematically given by

 $4.5 * 10^7= a(1 - 0.203)$

 $4.5 * 10^7 = 0.797a$

 $a = 56.46 * 10^6 km$

Generally the equation for Aperihelion distance is mathematically given by

 $r' = a(1 + e)$

 $r' = 56.4617 * 10^6 (1 + 0.203)$

 $r'=6.8 * 10^7 km$

Option C