Question

Neptune moves in an elliptical orbit with the sun at one of the foci. The length of half of the major axis is kilometers, and the eccentricity is 0.0086. Find the perihelion distance of Neptune from the sun. Round your answer to the nearest kilometers.

Answer 1

Given that,

Eccentricity = 0.0086

Suppose, The length of half of the major axis is $1\times10^{9}\ km$

We need to calculate the distance from the center to the foci

Using formula of eccentricity

$e=\dfrac{c}{a}$

$c=e\times a$

Where, e = eccentricity

a = major axis

c = the distance from the center to the foci

Put the value in to the formula

$c=0.0086\times1\times10^{9}$

$c=8600000$

We need to calculate the perihelion distance of Neptune from the sun

Using formula of distance

$d=a-c$

Put the value into the formula

$d=1\times10^{9}-0.0086\times10^{9}$

$d=991400000\ km$

Hence, The perihelion distance of Neptune from the sun is 991400000 km.