Computing The Orbital Speed of the Moon
Computing the mean orbital speed of the moon is not very difficult.
To make the computation we need certain data. Let:
APPLIED WORKING DATA: 1 statute mile = 1609.344 meters pi = 3.141592653589793 R = Mean radius of lunar orbit = 384399 km T = Mean orbital time period = 27.321582 days = 655.717968 hours r = Mean radius of the moon = 1737.10 km TO BE COMPUTED: C = Mean circumference of lunar orbit v = Mean orbital speed t = Time for the moon to move a distance of 1 diameter in its orbitThe concept is simple. We divide a distance by a time interval to obtain a mean rate of speed. The distance here is the mean circumference of the lunar orbit and the time interval is the mean orbital time.
The mean circumference of the lunar orbit is:
C = 2 * pi * R = 6.283185307179586 * 384399 km = 2415250.148894526 kmNow we divide the circumference (C) by the orbital time-period (T) to obtain:
v = 2 * pi * R / T v = C / T v = 2415250.148894526 km / 655.717968 hours = 3683.367342001105 km/h (2288.738356747287 mi/h)So, we have determined that the moon travels at a mean speed of about
Now, let's determine how this might appear to us in the sky by computing how long it would take the moon to move a distance in its orbit equal to its diamenter. To do this we simply divide the diameter of the moon (d) by the orbital speed (v).
d = Lunar diameter = 2 * 1737.10 km = 3474.2 km and t = d / v t = 3474.2 km / (3683.4 km/hr) = 0.943205 h and 0.943205 h = 56.5923 minutesSo, in the sky, on the average, the moon moves about 1 diameter eastward in its orbit in about
Variation in the Orbital Speed of the Moon
We have computed the mean orbital speed of the moon, but by how much does the speed of the moon vary during its orbit? To find out, we now need to determine the general range or the minimum and maximum speeds. This requires one more piece of information called the eccentricity (e). The eccentricity is a measure of how much the orbital ellipse deviates from a perfect circle, where a perfect circle has zero eccentricity while the eccentricity of an ellipse is between zero and 1 (limits exclusive).
In the case of the moon:
vMean = 3683.4 km/h (2288.7 mi/h) and e = 0.0549 we find (1 - e) = 0.9451 (for the minumum) (1 + e) = 1.0549 (for the maximum) vMin = vMean * (1 - e) = 3481.2 km/h (2163.1 mi/h) vMax = vMean * (1 + e) = 3885.6 km/h (2414.4 mi/h)Rounding the results to the nearest unit we have:
vMin = 3481 km/h (2163 mi/h) vMean = 3683 km/h (2289 mi/h) vMax = 3887 km/h (2414 mi/h) diff = (3887 - 3481) = 406 km = 252 miFrom this, we might reasonably say that at any given moment, the orbital speed (
v = vMean ± diff/2 = 3683 ± 203 km/h (2289 ± 126 mi/h)
© Jay Tanner - PHP Science Labs - 2010