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What does the semimajor axis of a planet with a period of 12 earth years measure?
That can be calculated from Kepler's 3rd law which says if the period is T years the semimajor axis must be T2/3 astronomical units. So for a period of 12 years the s/m axis is 5.421 AU or 784 million km.
Suppose astronomers discover a new planet orbiting your sunthe orbit has a semimajor axis of 2.52 AU what is the planets period of revolution?
The period of revolution can be calculated using Kepler's Third Law: P^2 = a^3, where P is the period in years and a is the semimajor axis in astronomical units (AU). In this case, the period of revolution of the planet would be approximately 4.00 years.
How many earth days equal to 1 day in Ceres?
A day on Ceres is about 9.1 hours long.9 and a half hours. After one sunrise, the Sun sets about five hours later.The dwarf planet 1 Ceres rotates around its axis in about 9.1 hours.
How do you find the orbital period in earth years of a periodic comet if the furtherest from sun is 31.5 astronomical units while the closest is 0.5 astronomical units?
You can find the major axis, 0.5+31.5 or 32 AU. The semimajor axis is half that, 16 AU. Then you can use Keplers 3rd law to calculate the period, which is 161.5 or 64 years.
What is the revolution for Ceres dwarf planet?
It takes 4.6 years for Ceres to revolve around the sun.
How long are the days and years of the planet Ceres?
Ceres has a rotation period of 0.3781 days, and an orbital period of 4.6 years.
What does the semimajor axis of a planet with a period of 12 earth years measure?
That can be calculated from Kepler's 3rd law which says if the period is T years the semimajor axis must be T2/3 astronomical units. So for a period of 12 years the s/m axis is 5.421 AU or 784 million km.
How long are Ceres's days and years?
Ceres orbits the Sun once in 4.6 years and its rotation period is 9.1 hours.
Suppose astronomers discover a new planet orbiting your sunthe orbit has a semimajor axis of 2.52 AU what is the planets period of revolution?
The period of revolution can be calculated using Kepler's Third Law: P^2 = a^3, where P is the period in years and a is the semimajor axis in astronomical units (AU). In this case, the period of revolution of the planet would be approximately 4.00 years.
What is Ceres's orbital period?
4 years 220 days
How do Keplers 3 laws of motion combined with the parts of an ellipse explain the amount of time it takes for a planet to orbit the sun based on its length?
One of the parts of an ellipse is the length of its major axis. Half that is called the semimajor axis. Kepler's 3rd law says that the time to do one orbit is proportional to the 3/2 power of the semimajor axis. IF the semimajor axis is one astronomical unit the period is one year (the Earth). For a planet with a semimajor axis of 4 AUs the period would have to be 8 years, by Kepler-3.
How do Kepler's 3 laws of motion combined with the parts of an ellipse explain the amount of time it takes for a planet to orbit the sun based on its length?
One of the parts of an ellipse is the length of its major axis. Half that is called the semimajor axis. Kepler's 3rd law says that the time to do one orbit is proportional to the 3/2 power of the semimajor axis. IF the semimajor axis is one astronomical unit the period is one year (the Earth). For a planet with a semimajor axis of 4 AUs the period would have to be 8 years, by Kepler-3.
1.) Halley’s comet has an observed period of 76.3 years. Use Kepler’s 3rd law to calculate the semimajor axis of the comet’s orbit2.) The dwarf planet Makemake is located a distance 45.8 AU from the sun. What is its orbital period in years?
Semimajor Axis of Halley’s Comet Kepler’s 3rd law states: P^2 = a^3 where P is the orbital period in years, and a is the semimajor axis in astronomical units (AU). Given: P = 76.3 \text{ years} We need to solve for a : a^3 = P^2 a^3 = (76.3)^2 a^3 = 5821.69 a = \sqrt[3]{5821.69} a \approx 17.94 \text{ AU} So, the semimajor axis of Halley’s comet’s orbit is approximately 17.94 AU. Orbital Period of Makemake Given: a = 45.8 \text{ AU} Using Kepler’s 3rd law: P^2 = a^3 P^2 = (45.8)^3 P^2 = 96158.552 P = \sqrt{96158.552} P \approx 310 \text{ years} So, the orbital period of Makemake is approximately 310 years.
How many earth days does it take Ceres to orbit the sun?
Ceres follows an orbit between Mars and Jupiter, within the asteroid belt, with a period of 4.6 Earth years.
How long does Ceres get to orbit the sun?
The sun can't orbit itself. Stars in binary or trinary systems, however, do orbit each other. Depending on the distance and a few other factors, this can take hundreds if not thousands of years for one orbit.
What is the planet Ceres' rotation and revolution?
Ceres has a rotation period of about 9 hours, meaning it completes one full rotation on its axis in that amount of time. Its revolution period around the sun is approximately 4.6 Earth years, as it orbits the sun at an average distance of about 2.8 astronomical units.
How many earth days equal to 1 day in Ceres?
A day on Ceres is about 9.1 hours long.9 and a half hours. After one sunrise, the Sun sets about five hours later.The dwarf planet 1 Ceres rotates around its axis in about 9.1 hours.
