Suppose halley's comet orbits the sun every 79 years with an eccentricity of 0.87. what is it's aphelion distance in
au's

it's aphelion distance  is 18.41 AU

Explanation:

Given data:

eccentricity 0.87

P is period of orbiting   = 79

from Kepler's third  Law.

P^2 = d^3

where P is the period of orbiting in years, and

d  it is orbit  semi-major axis in Astronomical Units

1 AU is the average distance between  earth and sun

so if P =79, we have

79^2 = d^3

6241 = d^3

d = (6241)^(1/3) = 18.41 AU

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