With an eccentricity of 023 a certain planets orbit is the m

With an eccentricity of 0.23, a certain planet\'s orbit is the most eccentric in the solar system. The length of the minor axis of its orbit is approximately 14,000,000,000 km. Find the distance between this planet and the sun at perihelion and at aphelion. (Round your answers to the nearest hundredth.)

Solution

a = length of semi-major axis
b = length of semi-minor axis = 1/2 length of minor axis = 7,000,000,000
c = distance from centre of ellipse to foci = a² b² = ae
e = eccentricity

a² b² = ae
a² 7,000,000,000² = a.23

a^2 -0.23a - 7,000,000,000² =0

solve the quadratic:

neglect the -ve root

a = 7000000000.115 km

c = ae = 1610000000.02645 km

Distance between planet and sun at perihelion = a c

= 7000000000.115 - 1610000000.02645

= 5390000000.08855 km

Distance between planet and sun at aphelion = a + c

= 7000000000.115 +1610000000.02645

= 8610000000.02645 km