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# Length of Year for Planets in Gravitational Orbit

by Ron Kurtus

You can derive the equation for the time it takes an object in a gravitational orbit to make one revolution, provided you know its orbital velocity, the separation between it and the center of the other object and the mass of each object.

Since the time it takes the Earth to make one revolution around the Sun is called a *year*, it is convenient to state the orbital period in terms of Earth years or even Earth days.

To simplify the calculations, the equation for the orbital period assumes that the orbit is circular. The equation can be verified by considering the period of the Moon around the Earth and how long it takes the Earth, as well as the planet Jupiter, to go around the Sun.

Questions you may have include:

- What velocity is necessary to be in orbit?
- What is the equation for the time for one revolution?
- What are some examples to verify the equations?

This lesson will answer those questions. Useful tool: Units Conversion

## Velocity to be in a circular orbit

The velocity of an object in circular orbit around another object is a function of the mass of each object and the separation between their centers. Astronomical objects actually orbit the center of mass (CM) between them.

(

See Derivation of Circular Orbits Around Center of Mass for more information.)

The orbital velocity of one object with respect to the other is a sum of their velocities around their CM.

(

See Orbital Motion Relative to Other Object for more information.)

The equation for the velocity of a circular orbit of one object around another is:

v_{T}= √[G(M + m)/R]

where

**v**is the tangential velocity of the object in orbit in kilometers/second (km/s)_{T}**G**is the Universal Gravitational Constant = 6.674*10^{−20 }km^{3}/kg-s^{2}**M**and**m**are the masses of the objects in kg**R**is the separation in km between the objects, as measured from their centers

Note: Because we are stating velocity in km/s, we converted the units ofGfrom N-m^{2}/kg^{2}to^{ }km^{3}/kg-s^{2}. Also, we considerRin km instead of meters.

In the case where one object has a much greater mass than the other (**M >> m**), the contribution of **m** is negligible. The velocity equation reduces to:

v_{T}= √(GM/R)

## Deriving equation for one revolution

Knowing the required velocity to be in a circular orbit allows you to determine the time it takes to make one revolution around the other object. The distance traveled in one revolution is the circumference of the circle of radius **R**:

C = 2πR

where

**C**is the circumference in km**π**stands for pi and equals 3.142...

But also distance equals velocity times time:

C = v_{T}T

where **T** is the time in seconds (s) it takes the object to make one revolution around the larger object; it is also called the orbital period

Combine the two equations for **C** and then solve for **T**:

v_{T}T = 2πR

T = 2πR/vseconds_{T}

Substitute **v _{T} = √[G(M + m)/R]** in the equation for

**T**:

T = 2πR/√[G(M + m)/R]

Multiply by ** √(R)/√(R)**

T = 2πR√(R)/√[G(M + m)]

T = 2π√(R^{3})/√[G(M + m)]

Thus, the equation for the orbital period is:

T = 2π√[Rseconds^{3}/G(M + m)]

### Verify units

It is a good practice to verify the units to make sure the equation is correct:

Ts= 2π√[Rkm^{3}^{3}/(Gkm^{3}/kg-s^{2})(Mkg +mkg)]s =

√(km^{3}/(km^{3}/kg-s^{2})kg)s =

√(kg-s^{2}/kg)s = s

### Convert from seconds

You usually calculate the orbit in Earth days or years, so you need to convert seconds to a different unit of measurement.

- 1 minute = 60 seconds
- 1 hour = 60 minutes
- 1 day = 24 hours
- 1 year = 365 days

### Equation for number of days

Thus, 1 day = (24 hours)*(60 minutes)*(60 seconds) = 8.640*10^{4} seconds/day. Divide by the number of seconds per day to get the orbit equation for days:

D = 2π√[R^{3}/G(M + m)]/8.640*10^{4}days

Since 2**π** = 6.284, you get:

D= 7.273*10^{−5}√[REarth days^{3}/G(M + m)]

If **M** is much greater than **m**, (**M >> m**), the CM is almost at the geometric center of the larger object. In such a case, the equation reduces to the simple version:

D= 7.273*10_{s}^{−5}√(REarth days^{3}/GM)

### Equation for number of years

Also, 1 year = (365.25 days/year)*(86400 seconds/day) =

31,557,600 seconds/year. That can be simplified to 3.156*10^{7} s/yr.

Y = 2π√[R3.156*10^{3}/G(M + m)]/^{7}years

Y =1.991*10^{−7}√[REarth years^{3}/G(M + m)]

If** M >> m**, the equation reduces to:

Y1.991*10_{s}=^{−7}√(REarth years^{3}/GM)

## Examples

You can calculate the number of days it takes the Moon to rotate around the Earth and the number of years it takes the Earth and the planet Jupiter to go around the Sun.

Note: Since the separationRand massMare not exact, as well as the fact that the orbits are not exactly circular, but are slightly elliptical, the time to complete an orbit is not exact. However, the calculations do come out fairly close to what is experienced.

### Moon orbits the Earth

The number of days that it takes the Moon to complete one revolution around the Earth is:

D= 7.273*10^{−5}√[REarth days^{3}/G(M + m)]

where

**R**= 3.844*10^{5}km (separation between centers)**G**= 6.674*10^{−20 }km^{3}/kg-s^{2}**M**= 5.974*10^{24}kg (mass of Earth)**m =**7.348*10^{22}kg = 0.073*10^{24}kg (mass of Moon)

Thus:

R= 5.680*10^{3}^{16}km^{3}

and

M + m =5.974*10^{24}kg + 0.073*10^{24}kg

M + m =6.047*10^{24}kg

Entering in values:

D =7.273*10^{−5}√[5.680*10^{16}/(6.674*10^{−20}*6.047*10^{24})]

D =7.273*10^{−5}√(14.07*10^{10})

D =7.273*10^{−5}*3.752*10^{5}

D =27.28 days

That is close to the average of 27.322 days for the Moon to orbit the Earth, as determined by NASA.

### Earth orbits the Sun

You can verify the number of days it takes the Earth to orbit the Sun. The center of mass or barycenter between the Earth and the Sun is almost at the Sun's geometric center.

(

See Circular Planetary Orbits for more information.)

Thus, the simple equation for the number of days can be used:

D= 7.273*10_{s}^{−5}√(REarth days^{3}/GM)

where

**R**= 1.496*10^{8}km (separation between centers)**G**= 6.674*10^{−20 }km^{3}/kg-s^{2}**M**= 1.989*10^{30}kg (mass of Sun)

Thus:

R= 3.348*10^{3}^{24 }km^{3}

and:

D7.273*10_{s}=^{−5}√[3.348*10^{24}/(6.674*10^{−20}*1.989*10^{30})]

D7.273*10_{s}=^{−5}√[3.348*10^{24}/(6.674*10^{−20}*1.989*10^{30})]

D7.273*10_{s}=^{−5}√(25.221*10^{12})

D7.273*10_{s}=^{−5}*5.022*10^{6}

D365.25 days_{s}=

This value corresponds to most measurements.

The number of years for the Earth to orbit the Sun is:

Y =1.991*10^{−7}√(R^{3}/GM)

Y= (1.991*10^{−7})(5.022*10^{6}) years

Y= 0.999 years

It takes one year for the Earth to orbit the Sun.

Orbital periods of Earth and Jupiter

### Jupiter orbits the Sun

The average separation between the planet Jupiter and the Sun is **R =** 7.786*10^{11} m. Thus:

R472*10^{3}=^{33}m^{3}

The mass of the Sun is **M** = 1.989*10^{30} kg. Thus:

GM= (6.67*10^{−11})(1.9891*10^{30}) = 13.27*10^{19}m^{3}/s^{2}

R= 472*10^{3}/GM^{33}/13.27*10^{19}= 35.57*10^{14}s^{2}

√(R= 5.96*10^{3}/GM)^{7}s

Substitute into **Y = **2*10^{−7}**√(R ^{3}/GM)**:

Y= (2*10^{−7})(5.96*10^{7}) years

Y= 11.9 years

This corresponds to the measured time it takes Jupiter to orbit the Sun. 11.86 yrs

## Summary

If you know the velocity of an object in orbit and its separation from the center of the much larger object, as well as the mass of the larger object, you can calculate how long it takes to make one revolution in Earth days or Earth years.

The equation can be verified by considering the period of the Moon around the Earth and how long it takes the Earth and Jupiter to go around the Sun.

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## Resources and references

### Websites

**Acceleration due to Gravity Calculations** - from Western Washington University

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**Top-rated books on Gravitation**

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