Here is a demonstration of calculating the mean orbital speed of the moon based on Kepler's second law and applied lunar data.

The data we will apply is:

APPLIED DATA:

363301 km (Min = Perigee) 225745 mi
384401 km (Mean) 238856 mi
405502 km (Max = Apogee) 251967 mi

Mean lunar orbital period = 27.32166 days = 655.71984 hours

pi = 3.1415926535897932

Kepler's second law states that an orbiting body sweeps out equal areas on the orbital ellipse plane in equal time intervals.  This value is a constant for the given orbit.  The time interval we will use is simply one hour.

So, the first step is to compute the total area (A) of the lunar orbital ellipse plane using the data given above.

Given an ellipse with a minimum radius (r) and a maximum radius (R), the area of the ellipse is found by

Area = pi * r * R

Respectively substituting the perigee distance (d) and apogee distance (D) of the moon for the minimum radius (r) and maximum radius (R) of the orbital ellipse plane, we obtain the result

d = 363301 km = Perigee distance = Minimum
D = 405502 km = Apogee distance  = Maximum

A = Total area of lunar orbital ellipse plane
  = pi * d * D
  = pi * (363301 km) * (405502 km)
  = 462817174383.765513 km²

Now we divide the total area (A) of the lunar orbital ellipse plane by the mean time (T) taken for one orbit to obtain the total area (a) swept out in one hour.  This value remains constant regardless of the continually varying distance.

T = 27.32166 days = 655.71984 hours

A = 462817174383.765513 km²

a = Total area swept out in one orbital hour
  = A / T
  = 462817174383.76551 km² / 655.71984 hours
  = 705815420.17055 km² per hour

This (a) value is constant for the orbital ellipse.

At the distance of the moon, we can treat the short distance it moves in one hour as a line segment for our purposes here.  This line will be the base of a tall, slender triangle with the Earth at the apex.

The area of the triangle is computed by

a = 1/2 * base * distance

2*a = base * distance

base = 2*a / distance

Substituting the computed value of (a), we obtain:

At perigee, d = 363301

Substituting the lunar perigee distance gives:

a = 705815420.17055 = 1/2 * (base * d)

2 * a = 1411630840.3411 = base * d

base = 2 * a / d
     = 1411630840.3411 km²/h / 363301 km
     = 3885.6 km/h = 2414.4 mi/h

Doing the same for apogee gives

At apogee, D = 405502

Substituting the lunar apogee distance gives:

a = 705815420.17055 = 1/2 * (base * D)

2 * a = 1411630840.3411 = base * D

base = 2 * a / d
     = 1411630840.3411 km²/h / 405502 km
     = 3481.2 km/h = 2163.1 mi/h

So, applying Kepler's second law with the given data, we have determined that at perigee, the speed of the moon in its orbit is about 3885.6 km/h = 2414.4 mi/h and at apogee, about 3481.2 km/h = 2163.1 mi/h.

Taking the difference between the speeds shows that when the moon is at perigee, it is moving about 404.4 km/h = 251.3 mi/h faster than when it is at apogee.

For the mean orbital speed:

Mean orbital speed = (3481.2 km/h + 3885.6 km/h) / 2
                   = 3683.4 km/h = 2288.8 mi/h