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3 September, 18:54

E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a what is the asteroid's orbital radius?

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  1. 3 September, 20:39
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    To solve the problem, use Kepler's 3rd law:

    T² = 4π²r³ / GM

    Solved for r:

    r = [GMT² / 4π²]⅓

    but first covert 6.00 years to seconds:

    6.00years = 6.00years (365days/year) (24.0hours/day) (6 ...

    = 1.89 x 10^8s

    The radius of the orbit then is:

    r = [ (6.67 x 10^-11N∙m²/kg²) (1.99 x 10^30kg) (1.89 x 10^8s) ² / 4π²]⅓

    = 6.23 x 10^11m
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