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Problem: A result of Kepler’s harmonic law is that the mass of a planet (M) with a satellite is directly proportional to the cube of t
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Question 148179: Hope that you can help me with this problem.
Problem: A result of Kepler’s harmonic law is that the mass of a planet (M) with a satellite is directly proportional to the cube of the mean distance (d) from the satellite to the planet and inversely proportional to the square of the period of revolution (p). Early astronomers estimated the mass of the earth to be The mass of the earth is 5.976 x 10 to the 24th power. kg and observed that the moon orbited the earth with a period of 27.322 days at a mean distance of km.
(1) Write a formula for the mass of a planet according to Kepler.
(2) Find the proportionality constant using the observations and estimates for the earth and moon.
(3) Extend your knowledge across the solar system to find the mass of Mars based on observations that Phobos orbits Mars in 7.65 hours at a mean distance of 9330 km.
(1) What is the approximate ratio of the mass of the earth to the mass of Mars?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Problem: A result of Kepler’s harmonic law is that the mass of a planet (M) with a satellite is directly proportional to the cube of the mean distance (d) from the satellite to the planet and inversely proportional to the square of the period of revolution (p).
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Early astronomers estimated the mass of the earth to be 5.976 x 10 to the 24th power. kg and observed that the moon orbited the earth with a period of 27.322 days at a mean distance of 384,403 km.
(1) Write a formula for the mass of a planet according to Kepler.
Equation: M = k[d^3/p^2]
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(2) Find the proportionality constant (k) using the observations and estimates for the earth and moon.
5.978x10^24 = k[384,403^3/27.322^2]
5.978x10^24/7.609135732x1013 = k
k = 7.85635..x10^10
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(3) Extend your knowledge across the solar system to find the mass of Mars based on observations that Phobos orbits Mars in 7.65 hours at a mean distance of 9330 km.
M = k[d^3/p^2]
M = (7.85635x10^10)[9330^3/7.65^2]
M = 1.09029...x10^21 kg
(1) What is the approximate ratio of the mass of the earth to the mass of Mars?
r = (5.976 x 10^24th power. kg)/(1.09029...x10^21 kg) = 5481.1028:1
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Cheers,
Stan H.
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