Length of Year for Objects in Gravitational Orbit
About 7 min read
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 distance to 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. There is a mini-quiz near the end of the lesson.
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 distance between their centers. Astronomical objects actually orbit the center of mass (CM) between them. The orbital velocity of one object with respect to the other is a sum of their velocities around their CM.
The equation for the velocity of a circular orbit of one object around another is:
v = √[G(M + m)/R]
where
- v is the tangential velocity of the object in orbit in kilometers/second (km/s)
- G is the Universal Gravitational Constant (6.674*10−20 km3/kg-s2)
- M and m are the masses of the objects in kg
- R is the distance in km between the objects, as measured from their centers
In the case where one object has a much greater mass than the other, such that its effect is negligible, the velocity equation reduces to:
v = √(GM/R)
where M is the mass of the much larger object.
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 = vT
where
- v was previously defined in km/s
- 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:
vT = 2πR
T = 2πR/v s
Substitute v = √[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√(R3)/√[G(M + m)]
Thus, the equation for the orbital period is:
T = 2π√[R3/G(M + m)] s
Verify units
It is a good practice to verify the units to make sure the equation is correct:
T s = 2π√[R3 km3/(G km3/kg-s2)(M kg + m kg)]
s = √(km3/(km3/kg-s2)kg)
s = √(kg-s2/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*104 seconds/day. Divide by the number of seconds per day to get the orbit equation for days:
D = 2π√[R3/G(M + m)]/8.640*104 days
Since 2π = 6.284, you get:
D = 7.273*10−5√[R3/G(M + m)] Earth days
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:
Ds = 7.273*10−5√(R3/GM) Earth days
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*107 s/yr.
Y = 2π√[R3/G(M + m)]/3.156*107 years
Y = 1.991*10−7√[R3/G(M + m)] Earth years
If M >> m, the equation reduces to:
Ys = 1.991*10−7√(R3/GM) Earth years
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.
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√[R3/G(M + m)] Earth days
where
- R = 3.844*105 km (distance between centers)
- G = 6.674*10−20 km3/kg-s2
- M = 5.974*1024 kg (mass of Earth)
- m = 7.348*1022 kg = 0.073*1024 kg (mass of Moon)
(See Circular Gravitational Orbits for more information.)
Thus:
R3 = 5.680*1016 km3
and
M + m = 5.974*1024 kg + 0.073*1024 kg
M + m = 6.047*1024 kg
Entering in values:
D = 7.273*10−5√[5.680*1016/(6.674*10−20*6.047*1024)]
D = 7.273*10−5√(14.07*1010)
D = 7.273*10−5*3.752*105
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. As was seen in Circular Gravitational Orbit Examples, the center of mass or barycenter between the Earth and the Sun is almost at the Sun’s geometric center. Thus, the simple equation for the number of days can be used:
Ds = 7.273*10−5√(R3/GM) Earth days
where
- R = 1.496*108 km (distance between centers)
- G = 6.674*10−20 km3/kg-s2
- M = 1.989*1030 kg (mass of Sun)
Thus:
R3 = 3.348*1024 km3
and:
Ds = 7.273*10−5√[3.348*1024/(6.674*10−20*1.989*1030)]
Ds = 7.273*10−5√[3.348*1024/(6.674*10−20*1.989*1030)]
Ds = 7.273*10−5√(25.221*1012)
Ds = 7.273*10−5*5.022*106
Ds = 365.25 days
This value corresponds to most measurements.
The number of years for the Earth to orbit the Sun is:
Y = 1.991*10−7√(R3/GM)
Y = (1.991*10−7)(5.022*106) years
Y = 0.999 years
It takes one year for the Earth to orbit the Sun.
Jupiter orbits the Sun
The average distance of the planet Jupiter from the Sun is R = 7.786*1011 m. Thus:
R3 = 472*1033 m3
The mass of the Sun is M = 1.989*1030 kg. Thus:
GM = (6.67*10−11)(1.9891*1030) = 13.27*1019 m3/s2
R3/GM = 472*1033/13.27*1019 = 35.57*1014 s2
√(R3/GM) = 5.96*107 s
Substitute into Y = 2*10−7√(R3/GM):
Y = (2*10−7)(5.96*107) 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 distance to 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
The following resources provide information on this subject:
Websites
Acceleration due to Gravity Calculations – from Western Washington University
Gravity and Gravitation Resources
Mini-quiz to check your understanding
Self-check your understanding
If many orbits are elliptical, why is the circular orbit equation used?
Why is the orbital period converted into Earth years?
What would the Earth's orbital period be if it was 4 times as far from the Sun than it is now?
Pick an answer to see instant feedback. Re-read the lesson if you miss one.
If you got all three correct, you are on your way to becoming a Champion in Physics. If you had problems, you had better look over the material again.
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