Question

Newton's version of Kepler's third law of planetary motion can be written as P^2 = a^3 / (M_1 + M_2), where P is the orbital period of a planet in years, a is the semimajor axis of the orbit in AU, M_1 is the mass of the parent star, and M_2 is the mass of the planet (both masses given in solar masses). Suppose a star with a mass of 0.80 times the mass of the Sun had an Earth-like planet orbiting at a distance of 1.66 AU. How long (in years) would it take this planet to orbit the star? (Express your answer to 3 decimal places, with no units, for example, 0.123, 1.234, 12.345, 123.456, 1234.567, etc.)

          Newton's version of Kepler's third law of planetary motion can be written as P^2 = a^3 / (M_1 + M_2), where P is the orbital period of a planet in years, a is the semimajor axis of the orbit in AU, M_1 is the mass of the parent star, and M_2 is the mass of the planet (both masses given in solar masses). Suppose a star with a mass of 0.80 times the mass of the Sun had an Earth-like planet orbiting at a distance of 1.66 AU. How long (in years) would it take this planet to orbit the star? (Express your answer to 3 decimal places, with no units, for example, 0.123, 1.234, 12.345, 123.456, 1234.567, etc.)
        
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Newton's version of Kepler's third law of planetary motion can be written as P^2 = a^3 / (M1 + M2), where P is the orbital period of a planet in years, a is the semimajor axis of the orbit in AU, M1 is the mass of the parent star, and M2 is the mass of the planet (both masses given in solar masses). Suppose a star with a mass of 0.80 times the mass of the Sun had an Earth-like planet orbiting at a distance of 1.66 AU. How long (in years) would it take this planet to orbit the star? (Express your answer to 3 decimal places, with no units, for example, 0.123, 1.234, 12.345, 123.456, 1234.567, etc.)

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Elementary and Intermediate Algebra
Elementary and Intermediate Algebra
Alan S. Tussy, R. David Gustafson 5th Edition
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00:02 Okay, we want to know how long would it take this planet to orbit its star.
00:09 We're given newton's version of kepler's third law of planetary motion, the formula, and we're told the star is 0 .8 times the mass of the sun, and this is an earth -like planet, a distance of 1 .66 astronomical units away.
00:33 Way.
00:35 So first thing i did was just make sure i was understanding the formula correctly by using the earth and the sun.
00:46 So a is measured in au, astronomical units, which is defined to be the distance from the earth to the sun.
00:58 So that would just be one for the earth -sun system.
01:06 And then then the masses are measured in solar masses, which one solar mass is defined to be the mass of our sun.
01:20 So m1 here would just be one.
01:23 M2 would be the mass of the earth in solar masses.
01:29 But the earth is pretty massive, but compared to the sun, it's practically zero in mass.
01:41 Like we could well let me look it up it's um three times ten to the negative six so that would be point one two three four five six so yeah if we're rounding to to three decimal places it's not going to matter.
02:15 It's essentially zero...
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