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28 June, 00:37

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

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  1. 28 June, 04:22
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    it's aphelion distance is 18.41 AU

    Explanation:

    Given dа ta:

    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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