Because the moon's orbit is not perfectly circular, its angular size can vary from 0.49 degrees at the moon's apogee to 0.558 degrees at its perigee. On January 28, 2013, the angular diameter of the full moon for an observer in Wellington, New Zealand, was 0.504 degrees.
The moon's angular velocity (west-to-east) = 2*pi radians/(27.3*24 hr)
The earth's angular velocity (west-to-east) = 2*pi radians/(24 hr)
The net moon angular velocity = - 0.252 rad/hr = - 14.5 deg/hr = - 0.241 deg/min
Thus it takes 2.09 min (126 sec.) for the earth observer (or a Wellington, New Zealand, video) to see the moon move (east-to-west) across the sky by one moon diameter, which is what the video shows, given the camera timing and video/atmospheric resolution.
Rick Strickert
Austin, TX
-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@phys-l.org] On Behalf Of Bob Sciamanda
Sent: Monday, March 03, 2014 10:24 AM
To: Bob Sciamanda; Phys-L@Phys-L.org
Subject: Re: [Phys-L] Trick photography?
I'm still learning (a good sign)! ==>
I am just realizing that the apparent moon speed is NOT calculable as the moon's linear orbital speed minus the linear spin speed of an earth surface point.
IE., Vap is NOT= {(6.28*250000)/(30*24) - (6.28*4000)/24} miles/hr. This is grossly wrong - even in direction!
Instead, Vap = {6.28/(30*24)-6.28/24}*250000 miles/hr. => the relative ANGULAR velocity times the moon's orbital radius.
It is incorrect to treat the earth observer as viewing a linear motion from a linear moving frame. We must acknowledge that the earth observer is viewing a rotational motion from a rotational frame. Note that this turns the moon's actual EASTWARD motion into the observed WESTWARD motion. The linear mistreatment fails to even do this.
Again, thanks to John M for leading me to this realization.
Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
treborsci@verizon.net http://sciamanda.com