For the purpose of the equation, ♫ is pi.
T^2=((4♫^2)(R^3))/(G)(Planetary Mass)
T^2 stands for the period, R is the radius of the orbit in metres.
G is the force of gravity, (6.67 X 10^-11), and the Planetary Mass is the mass of the object that is being orbited in kilograms.
Wiki User
∙ 13y agoThe orbital period of an object is calculated using Kepler's third law of planetary motion. The formula for calculating the orbital period is T^2 = (4π^2 * a^3) / (G * M), where T is the orbital period, a is the semi-major axis of the orbit, G is the gravitational constant, and M is the total mass of the object being orbited.
Wiki User
∙ 7y agoBriefly:
The length of Earth's orbit can be measured in time or space.
That is 365.25 days or approx 940,000,000 kilometers or 584,000,000 miles.
Time: Length of a Year
There are two lengths of the year in terms of time:
The time it takes the Earth to orbit around the Sun changes by a few minutes each year because the Earth's path is slightly affected by the other planets, mostly Saturn and Jupiter. In 2007 the year was about 11 minutes longer and in 2010 it was 12 minutes shorter, but in 2013, it will be within a minute.
Space: Circumference of Ellipse
Two ways to calculate the length.
1. Use the average speed times the length of a year.
For Earth, the orbital period (year) is 365.25 days which takes place at an average of distance of 149,600,000 km from the sun at an average speed of 29.8 km/s.
(365.25 days) * 29.8 * (km / s) ≈ 940 000 000 kilometers
2. Use geometry.
If the Earth's orbit were a circle, the length of the orbit would be the circumference,
C=2 pi r.
2 * pi * (149 600 000 kilometers) ≈ 940,000,000 kilometers
In miles the mean radius is 93,000,000 miles
→ length is approx 2 × π × 93,000,000 ≈ 584,000,000 miles.
Since the orbit is more nearly an ellipse, we could approximate the circumference as C= pi (a+b), but that does not actually improve the accuracy. The exact calculation of the circumference of an ellipse turns out to be difficult but the answer above is within one percent.
Wiki User
∙ 11y agoBriefly:
The length of Earth's orbit can be measured in time or space.
That is 365.25 days or 940,000,000 kilometers or 584,000,000 miles.
See related links from NASA.
Explanation:
Time: Length of Year
The length of time it takes the Earth to orbit the Sun is
365.24219 standard days, where a standard day is defined as 24 hours, not the time it takes the Earth to rotate about is axis.
The time it takes the Earth to orbit around the Sun changes by a few minutes each year because the Earth's path is slightly affected by the other planets, mostly Saturn and Jupiter. In 2007 the year was about 11 minutes longer and in 2010 it was 12 minutes shorter, but in 2013, it will be within a minute.
Space: Circumference of Ellipse
Two ways to calculate the length.
1. Use the average speed times the length of a year.
For Earth, the orbital period (year) is 365.25 days which takes place at an average of distance of 149,600,000 km from the sun at an average speed of 29.8 km/s.
(365.25 days) * 29.8 * (km / s) = 940 416 480 kilometers
2. Use geometry.
If the Earth's orbit were a circle, the length of the orbit would be the circumference,
C=2 pi r.
2 * pi * (149 600 000 kilometers) = 939 964 522 kilometers
Since the orbit is more nearly an ellipses, we could approximate the circumference as C= pi (a+b), but that does not actually improve the accuracy. The exact calculation of the circumference of an ellipse turns out to be difficult and has been a topic that has occupied the world's best mathematicians. But, for accuracies we need, our answer above is within one percent.
Wiki User
∙ 13y agoIn astrodynamics the orbital period T (in seconds) of a small body orbiting a central body in a circular or elliptic orbit is:
T = 2 x pi (a3/GM)1/2
a is length of orbit's semi-major axis
G is the gravitational constant
M is the mass of the central body
Wiki User
∙ 12y agoI think you're asking how to find the time period of one orbit, using the distance of the object (i expect planet) from its star.
Kepler showed through observation that the distance between a planet and star (a) is related to the period of orbit (P) by:
which Newton then proved mathematically and created an equation:
Where M is the mass of the object an G is the Universal Gravitational constant.
using this equation in various different forms you can find either the mass, period of orbit or the distance of an object from the star. You'll need to know the other two variables in SI units (kg for mass, seconds for P, m for a) and hey presto you've done it!
Hope this helps!
Wiki User
∙ 12y agoPlease help me calculate (step by step) the Average orbital speed for the object that has the following an Orbital Elements:
1) Semi-major axis: 541.429505752064 AU;
2) Eccentricity: 0.8590486217687489;
3) Perihelion distance: 76.31523505081853 AU;
4) Inclination: 11.92794476981208 (deg);
5) Longitude of the ascending node: 144.3772377555154 (deg);
6) Argument of perihelion: 310.9209931336452 (deg);
7) Mean anomaly: 358.1909214292181 (deg);
8) Orbital period: 4601625.87025175 Earth days or 12598.57 Earth years;
9) Mean motion: 7.823321803002312E-5 (deg/d);
10) Aphelion distance: 1006.543776453309 AU.
Wiki User
∙ 13y agoThe orbital period is the time taken for a given object to make one complete orbit about another object.
Wiki User
∙ 13y agoT= 2.Pi * sqrt(a3/GM)
The period of a planet's revolution can be used to calculate its orbital radius or distance from the sun using Kepler's third law of planetary motion. It can also be used to determine the planet's orbital speed or velocity if its mass is known. Additionally, the period of revolution helps in predicting future positions of the planet along its orbit.
Mars has an orbital period of approximately 687 Earth days.
Quaoar has an orbital period of about 285 years. It takes approximately 285 Earth years for Quaoar to complete one orbit around the Sun.
The distance of a planet from the sun affects its orbital period. Generally, the farther a planet is from the sun, the longer its orbital period will be. This relationship is described by Kepler's third law of planetary motion, which states that the square of a planet's orbital period is directly proportional to the cube of its average distance from the sun.
A planet's orbital period is related to its distance from the Sun by Kepler's third law, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit. For an orbital period of 3 million years, the planet would need to be located at a distance of approximately 367 AU from the Sun.
Orbital information. You need to know the size of the "semi-major axis". Then you can calculate the orbital period, using Kepler's Third Law.
The period of a planet's revolution can be used to calculate its orbital radius or distance from the sun using Kepler's third law of planetary motion. It can also be used to determine the planet's orbital speed or velocity if its mass is known. Additionally, the period of revolution helps in predicting future positions of the planet along its orbit.
The orbital period of Jupiter is 4332.71 days.
2007or10's orbital period is 552.52 years
Haumea's orbital period is 283 or 103,468 days
Quaoar has an orbital period of about 285 years. It takes approximately 285 Earth years for Quaoar to complete one orbit around the Sun.
Uranus has an orbital period of about 84 Earth years, or approximately 30,687 Earth days. It takes Uranus this long to complete one orbit around the sun.
Makemake's orbital period is about 309 Earth years. It takes Makemake nearly 310 years to orbit the sun once due to its distance from the sun and slower speed compared to inner planets.
A planet's orbital radius directly affects its orbital period through Kepler's third law of planetary motion. The farther a planet is from the star it orbits, the longer its orbital period will be, assuming all other factors remain constant. This relationship is expressed mathematically as T^2 ∝ r^3, where T is the orbital period and r is the orbital radius.
Orbital period is the time it takes a planet to go around its star once.
Ganymede's orbital period is approximately 7.2 Earth days.
the orbital period of Saturn in earth years are 89years