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Length of Year for Objects in Gravitational Orbit
by Ron Kurtus (revised 8 July 2010)
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, its distance to the center of the much larger object about which it is in orbit, as well as the mass of the larger object.
The time it takes the Earth to make one revolution around the Sun is called a year, so 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.
Useful tools: Metric-English Conversion | Scientific Calculator.
Velocity to be in a circular orbit
The velocity of an object in circular orbit is a function of the mass of the larger object and the distance between their centers. The equation for the orbital velocity is derived by equating the Universal Gravitational Equation with the centrifugal inertial force of the object.
(See Circular Gravitational Orbits for more information.)
The equation for the velocity of a circular orbit is:
v = √(GM/R)
where
- v is the tangential velocity of the object in orbit in meters/second (m/s)
- G is the universal gravitational constant (6.67*10−11 m3/kg-s2)
- M is the mass of the larger object in kg
- R is the distance in meters (m) between the objects, as measured from their centers of mass
Note that the mass of the object in orbit is not a factor in this equation.
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 larger object. The distance traveled in one revolution is the circumference of the circle of radius R:
C = 2πR
where
- C is the circumference in meters (m)
- π stands for pi and equals 3.14...
Note: Circumference is in meters since G is in meters.
Since distance equals velocity times time:
C = vT
where
- v was previously defined in m/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
Substitute in C and solve for T:
vT = 2πR
T = 2πR/v
Substitute v = √(GM/R):
T = 2πR/√(GM/R)
T = 2π√(R2)/√(GM/R)
Multiply by √(R3/GM)/√(R3/GM) = 1 to facilitate calculations:
T = 2π√(R3/GM)
Verify units
Verify the equation is correct for the units used:
T s = 2π√[(R3 m3)/(G m3/kg-s2)(M kg)]
s = √(s2)
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) = 86400 seconds. Divide by the number of seconds per day to get the orbit equation for days:
D = 2π√(R3/GM)/86400
Since 2π = 6.28, you get:
D = (7.27*10−5)√(R3/GM) Earth days
Equation for number of years
Also, 1 year = (365 days)*(86400 seconds) = 31,536,000 seconds. That can be simplified to 3.15*107 s/yr.
Y = [2π√(R3/GM)]/3.15*107
Y = 2*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 planet Jupiter to go around the Sun.
Note: Since the distance R and mass M are 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 average distance from the center of the Earth to the center of the Moon is 384,403 kilometers or R = 3.84*108 m.
R3 = 56.62*1024 m3
The mass of the Earth is M = 5.9736*1024 kg.
GM = (6.67*10−11)(5.97*1024) = 39.8*1013 m3/s2
R3/GM = 56.62*1024/39.8*1013 = 1.423*1011 s2
Make the exponent even to facilitate taking square root:
R3/GM = 14.23*1010 s2
√(R3/GM) = 3.77*105 s
Substitute into D = (7.27*10−5)√(R3/GM)
D = (7.27*10−5)(3.77*105) days = 27.4 days
That is close to the average of 28 days for the Moon to orbit the Earth.
Earth orbits the Sun
Let's test these equations by checking the time it takes the Earth to rotate around the Sun.
The average distance from
the center of the Sun to the center of the Earth is
R = 1.496*1011 m.
R3 = 3.348*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 = 3.348*1033/13.27*1019 = 0.2522*1014 s2
Increase number to facilitate taking square root:
R3/GM = 25.22*1012 s2
√(R3/GM) = 5.02*106 s
Substitute into D = (7.27*10−5)√(R3/GM):
D = (7.27*10−5)(5.02*106) days = 364.95 days
Substitute into Y = 2*10−7√(R3/GM):
Y = (2*10−7)(5.02*106) years
Y = 1.004 years
It takes one year for the Earth to orbit the Sun.
Earth and Jupiter orbit 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.
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
Books
Top-rated
books on Simple Gravity Science
Top-rated
books on Advanced Gravity Physics
Mini-quiz to check your understanding
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