How do you calculate orbital period?

Answer 1

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.

Answer 2

Briefly:

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 Tropical year is the time taken between one vernal equinox and the next
• The Sidereal year is the time taken for the earth to make one orbit of the sun and return to its starting point.

In 2000 the mean tropical year was 365.24219 days and the mean sidereal year was 365.2564 days

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.

Answer 3

Briefly:

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.

Answer 4

In 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

Answer 5

I 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!

Answer 6

Please 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.

Answer 7

The 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.

Answer 8

The orbital period is the time taken for a given object to make one complete orbit about another object.

Answer 9

T= 2.Pi * sqrt(a3/GM)