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2 days ago · The sidereal period is the time it takes for a planet to orbit the Sun; for Earth it is one year, for Mars it is 1.881 years. When the periods of a planetary orbit are measured in years, and the semimajor axes that are the same as the average distance for the Sun are measured in astronomical units (AU), then Kepler’s third law becomes P 2 = a 3 .
May 2, 2024 · The movement of Mars was problematic – it didn’t quite fit the models as described by Greek philosopher and scientist Aristotle (384 to 322 B.C.E.) and Egyptian astronomer Claudius Ptolemy (about 100 C.E to 170 C.E.). Aristotle thought Earth was the center of the universe, and that the Sun, Moon, planets, and stars revolved around it.
A major problem with Copernicus’s theory was that he described the motion of the planet Mars as having a circular orbit. In actuality, Mars has one of the most eccentric orbits of any planet, with an eccentricity of 0.0935. (Earth’s orbit is quite circular, with an eccentricity of only 0.0167.)
Jul 29, 2023 · For instance, suppose you time how long Mars takes to go around the Sun (in Earth years). Kepler’s third law can then be used to calculate Mars’ average distance from the Sun. Mars’ orbital period (1.88 Earth years) squared, or P 2 P 2 , is 1.88 2 = 3.53 1.88 2 = 3.53 , and according to the equation for Kepler’s third law, this equals the cube of its semimajor axis, or a 3 a 3 .
For instance, suppose you time how long Mars takes to go around the Sun (in Earth years). Kepler’s third law can then be used to calculate Mars’ average distance from the Sun. Mars’ orbital period (1.88 Earth years) squared, or P 2 P 2 , is 1.88 2 = 3.53 1.88 2 = 3.53 , and according to the equation for Kepler’s third law, this equals the cube of its semimajor axis, or a 3 a 3 .
For instance, suppose you time how long Mars takes to go around the Sun (in Earth years). Kepler’s third law can then be used to calculate Mars’ average distance from the Sun. Mars’ orbital period (1.88 Earth years) squared, or P 2 , is 1.88 2 = 3.53, and according to the equation for Kepler’s third law, this equals the cube of its semimajor axis, or a 3 .
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Given that the moon orbits Earth each 27.3 days and that it is an average distance of 3.84 × 10 8 m 3.84 × 10 8 m from the center of Earth, calculate the period of an artificial satellite orbiting at an average altitude of 1,500 km above Earth’s surface.
