To determine the semi-major axis of an orbit, you can measure the distance between the center of the orbit and one of its furthest points. This distance is half of the longest diameter of the elliptical orbit and is known as the semi-major axis.
To determine the period of orbit for a celestial body, one can use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. By measuring the semi-major axis of the orbit and the gravitational force acting on the celestial body, one can calculate the period of its orbit.
To determine the moments of inertia for an object, one can use mathematical formulas or physical experiments. The moment of inertia depends on the shape and mass distribution of the object. Common methods for deriving moments of inertia include integration, parallel-axis theorem, and the perpendicular-axis theorem. These methods involve calculating the distribution of mass around an axis to determine how the object resists rotational motion.
To determine the units of the y-intercept in a linear equation, you need to look at the units of the y-axis. The y-intercept represents the value of y when x is zero, so the units of the y-intercept will be the same as the units on the y-axis.
To calculate the orbital period using the semi-major axis, you can use Kepler's third law of planetary motion. The formula is T2 (42 / G(M1 M2)) a3, where T is the orbital period in seconds, G is the gravitational constant, M1 and M2 are the masses of the two objects in the orbit, and a is the semi-major axis of the orbit. Simply plug in the values for G, M1, M2, and a to find the orbital period.
Position is measured along one axis, and time is measured along the other one.
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.
Oh, an interesting question! Mercury's semimajor axis - which is the distance from the center of the Sun to the farthest point of Mercury's orbit - is about 0.39 astronomical units, or around 57.9 million kilometers. That's a nice and cozy space for our little Mercury to dance gracefully around the warm Sun. Nature has a way of creating beauty in all the details like this, doesn't it?
To determine the period of orbit for a celestial body, one can use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. By measuring the semi-major axis of the orbit and the gravitational force acting on the celestial body, one can calculate the period of its orbit.
You can look through a list of the planets' orbits to see which one has the smallest eccentricity, and the answer is Venus, 0.006787. But if you are an astronomer wanting to find it out by observing the planets in the sky, you need to have five accurate positions for each planet, spaced out across a good fraction of the orbit, and then you can calculate the 'elements' of the orbit. One of the elements is the eccentricity. The eccentricity and the semimajor axis determine the shape and size of the orbit. Then there is the longitude of perihelion, which sets the orientation of the major axis of the ellipse. Two more are the inclination of the plane of the orbit to the plane of the Earth's orbit, and then the longitude of the ascending node, which defines the line of intersection of the two planes. Finally the longitude at the Epoch and the date of the Epoch are required because they set the starting conditions for the model.
It does NOT orbit on its axis, but rotates on its axis. It takes 24 hours, one day, to make one complete rotation. However, It does ORBIT the Sun. It takes the Earth 365.25 days, one year, for make one complete orbit of the Sun. Whilst it is making this orbit it is also rotating on its axis, as above.
Keppler's third law of planetary motion, his Law of Periods states that the square of the period of any planet is proportional to the cube of the semimajor axis of its orbit. This means that the nearer a planet is to the sun, the shorter its orbit. SInce Mercury is at a far smaller distance from the sun compared to the earth, its orbital period is smaller.
"Rotate" on an axis. "Revolve" in an orbit.
Ah, the average distance from the sun to the planet Mars is about 225 million kilometers, my friend. Just imagine Mars floating peacefully in space, soaking up the sun's warm glow on its journey around the solar system. Isn't that just a happy little fact to ponder? Remember, in our universe, every planet has its own special place.
One year
rotation is a day or one turn on an axis and a revolution is a year or one orbit around the sun
Earth takes approximately 24 hours to complete one full rotation on its own axis. This rotation is what causes day and night on Earth.