The period revolution of an orbiting body is directly related to its semimajor axis through Kepler's Third Law of Planetary Motion. This law states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semimajor axis (a) of its orbit, expressed mathematically as T² ∝ a³. This means that as the semimajor axis increases, the orbital period increases as well, indicating that objects further from a central body take longer to complete an orbit.
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.
Using Kepler's third law, the period (P) of an object in orbit can be calculated using the formula P^2 = a^3, where a is the semimajor axis in astronomical units (au). For Ceres with a semimajor axis of 2.77 au, the period of its orbit around the Sun is approximately 4.61 years.
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.
The major axis is the diameter across the widest part. The semimajor axis is half that, and for a planet it's the average of the maximum and minimum distances from the Sun .
The period of revolution of a planet is most closely related to its distance from the sun. The further a planet is from the sun, the longer it takes to complete one revolution.
the period of revolution is related to the semimajor axis.... :)
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.
Using Kepler's third law, the period (P) of an object in orbit can be calculated using the formula P^2 = a^3, where a is the semimajor axis in astronomical units (au). For Ceres with a semimajor axis of 2.77 au, the period of its orbit around the Sun is approximately 4.61 years.
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.
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.
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.
The major axis is the diameter across the widest part. The semimajor axis is half that, and for a planet it's the average of the maximum and minimum distances from the Sun .
The semimajor axis of a planet's orbit is important because it determines the size and shape of the orbit, as well as the distance of the planet from the sun. It helps us understand the planet's position in relation to the sun and other planets, and provides valuable information about the planet's orbital characteristics.
The period of revolution of a planet is most closely related to its distance from the sun. The further a planet is from the sun, the longer it takes to complete one revolution.
(I'm going to assume that when you said "first" you meant "fastest," because otherwise the question is nonsense.) Because of Kepler's Third Law. The orbital period for a body is related to the semimajor axis of its orbit. Mercury's orbit has the shortest semimajor axis of all the Solar planets, and therefore it has the shortest orbital period.
Not at all. The only thing that sets the orbital period is the semimajor axis, which is the average of the maximum and minimum distances from the Sun.
The major and minor axes of a circle are the same - either is any diameter. So a semimajor axis is half the diameter which is 12 cm.