Kepler\'s Law applications (a) Given that the Earth is 1.5 × 1011 m away from th
ID: 2305062 • Letter: K
Question
Kepler's Law applications (a) Given that the Earth is 1.5 × 1011 m away from the sun, and a certain comet orbits the sun in 923 Earth days, find the average distance from the sun to the comet. Hint: For the problem, treat the comet as a planet revolving around the sun (b) The period of the earth's orbit is approximately 365.25 days. Use this fact and Kepler's Third Law to find the length of the major axis of the earth's orbit. You will need the mass of the sun, M = 1.99 × 1030 kg, and the gravitational constant. G = 6.67 × 10-11 ?m2/kg"Explanation / Answer
(A) For average distance, assume path as circular then
Fg = m a_c
G M m / r = m w^2 r
r^3 = G M / w^2 and w = 2 pi / T
r^3 = G M T^2 / 4 pi^2
r^3 = (6.67 x 10^-11)(1.99 x 10^30)(923 x 24 x 3600)^2 / 4 pi^2
r = 2.78 x 10^11 m .....Ans
(B) From Kepler's third law,
T = 2 pi sqrt[ a^3 / G M ]
(365.25 x 24 x 3600 ) = 2 pi sqrt[ a^3 / (6.67 x 10^-11 x 1.99 x 10^30)]
a = 1.496 x 10^11 m . .semimajor axis
so length of major axis = 2 a = 2.992 x 10^11 m
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