- **Astronomy Misc**
(*http://www.spacebanter.com/forumdisplay.php?f=12*)

- - **What is the cycle for when the maximum amount of the Moon's apparent area is sunlit?**
(*http://www.spacebanter.com/showthread.php?t=226650*)

What is the cycle for when the maximum amount of the Moon's apparent area is sunlit?The Moon is defined as full when its ecliptic longitude is 180° distant
from the Sun's. But what is the cycle for when the maximum proportion of the Moon's apparent area is sunlit, and for observers at what latitudes of the Earth's surface is this maximum realised? Note that the plane of the Moon's orbit around the Earth is inclined at an angle of 5° to the plane of the Earth's around the Sun. To get a handle on why this affects the Moon's apparent fullness, imagine if it were 90°. Given the inclination I am sceptical as to whether the proportion ever reaches 50%. |

What is the cycle for when the maximum amount of the Moon's apparent area is sunlit?On Sat, 11 Jan 2020 20:43:08 -0000 (UTC), Harold Davis
wrote: The Moon is defined as full when its ecliptic longitude is 180° distant from the Sun's. But what is the cycle for when the maximum proportion of the Moon's apparent area is sunlit, and for observers at what latitudes of the Earth's surface is this maximum realised? Note that the plane of the Moon's orbit around the Earth is inclined at an angle of 5° to the plane of the Earth's around the Sun. To get a handle on why this affects the Moon's apparent fullness, imagine if it were 90°. Given the inclination I am sceptical as to whether the proportion ever reaches 50%. Since the Sun's diameter exceeds the Moon's, the Sun always illuminates slightly more than 50% of the Moon's surface. (To visualize, draw the external common tangents from a larger circle to a smaller one.) Using 98M miles as the distance, 432K miles as the Sun's radius, and 1K miles for the Moon's, the Sun illuminates almost 7 miles into the opposite hemisphere. Any observer a finite distance from the Moon can only ever see less than 50% of the Moon's surface. The further away the observer is, the closer the observable percentage is to 50. (To visualize, draw the two tangents from an external point to a circle.) However, when the Moon is in the plane of the ecliptic, an observer at the north pole can see approximately 12 miles over the Moon's north pole into the opposite hemisphere. So he can see some 5 miles beyond the terminator. Similar geometry causes his view of the Moon's south pose to fall short of the terminator. However, for a person standing in line with the ecliptic, his view in any direction falls 4 miles short of the terminator and that is the maximum you seek. When the Moon is within 4K miles of the ecliptic, an observer on Earth at he same offset from the ecliptic will have the view described above. When the Moon is more than 4K miles away from the ecliptic, all Earth-bound observers see some non-illuminated portion and therefore do not see the maximum you want. Not only is the Moon's orbit tilted with respect to the ecliptic, but the direction of the tilt wobbles like a child's top. I have no idea how to factor that into an equation that would compute the time between one visible maximum and another.. -- Remove del for email |

What is the cycle for when the maximum amount of the Moon's apparent area is sunlit?Barry Schwarz wrote in
: On Sat, 11 Jan 2020 20:43:08 -0000 (UTC), Harold Davis wrote: The Moon is defined as full when its ecliptic longitude is 180° distant from the Sun's. But what is the cycle for when the maximum proportion of the Moon's apparent area is sunlit, and for observers at what latitudes of the Earth's surface is this maximum realised? Note that the plane of the Moon's orbit around the Earth is inclined at an angle of 5° to the plane of the Earth's around the Sun. To get a handle on why this affects the Moon's apparent fullness, imagine if it were 90°. Given the inclination I am sceptical as to whether the proportion ever reaches 50%. Since the Sun's diameter exceeds the Moon's, the Sun always illuminates slightly more than 50% of the Moon's surface. (To visualize, draw the external common tangents from a larger circle to a smaller one.) Using 98M miles as the distance, 432K miles as the Sun's radius, and 1K miles for the Moon's, the Sun illuminates almost 7 miles into the opposite hemisphere. Any observer a finite distance from the Moon can only ever see less than 50% of the Moon's surface. The further away the observer is, the closer the observable percentage is to 50. (To visualize, draw the two tangents from an external point to a circle.) However, when the Moon is in the plane of the ecliptic, an observer at the north pole can see approximately 12 miles over the Moon's north pole into the opposite hemisphere. So he can see some 5 miles beyond the terminator. Similar geometry causes his view of the Moon's south pose to fall short of the terminator. However, for a person standing in line with the ecliptic, his view in any direction falls 4 miles short of the terminator and that is the maximum you seek. When the Moon is within 4K miles of the ecliptic, an observer on Earth at he same offset from the ecliptic will have the view described above. When the Moon is more than 4K miles away from the ecliptic, all Earth-bound observers see some non-illuminated portion and therefore do not see the maximum you want. Not only is the Moon's orbit tilted with respect to the ecliptic, but the direction of the tilt wobbles like a child's top. I have no idea how to factor that into an equation that would compute the time between one visible maximum and another.. Hi Barry, I am very grateful to you for this detailed reply. Can you possibly tell me what proportion of the Moon's apparent disc is sunlit for an observer at Earth latitude theta at the moment of "full" moon if the Earth-Moon line is inclined at say 2.5 degrees to the plane of the Earth's orbit around the Sun? (I'm taking 2.5 degrees as an "average" of 0 degrees and the maximum of 5 degrees.) That would give a handle on how "full" an "average" full moon is, viewed from Earth latitude theta. (I'm a bit embarrassed about asking this, because I could work it out myself, but my geometry is so rusty that it would probably take me about six hours!) Thanks again, Harry |

What is the cycle for when the maximum amount of the Moon's apparent area is sunlit?Barry Schwarz wrote in
: On Sat, 11 Jan 2020 20:43:08 -0000 (UTC), Harold Davis wrote: The Moon is defined as full when its ecliptic longitude is 180° distant from the Sun's. But what is the cycle for when the maximum proportion of the Moon's apparent area is sunlit, and for observers at what latitudes of the Earth's surface is this maximum realised? Note that the plane of the Moon's orbit around the Earth is inclined at an angle of 5° to the plane of the Earth's around the Sun. To get a handle on why this affects the Moon's apparent fullness, imagine if it were 90°. Given the inclination I am sceptical as to whether the proportion ever reaches 50%. Since the Sun's diameter exceeds the Moon's, the Sun always illuminates slightly more than 50% of the Moon's surface. (To visualize, draw the external common tangents from a larger circle to a smaller one.) Using 98M miles as the distance, 432K miles as the Sun's radius, and 1K miles for the Moon's, the Sun illuminates almost 7 miles into the opposite hemisphere. Any observer a finite distance from the Moon can only ever see less than 50% of the Moon's surface. The further away the observer is, the closer the observable percentage is to 50. (To visualize, draw the two tangents from an external point to a circle.) However, when the Moon is in the plane of the ecliptic, an observer at the north pole can see approximately 12 miles over the Moon's north pole into the opposite hemisphere. So he can see some 5 miles beyond the terminator. Similar geometry causes his view of the Moon's south pose to fall short of the terminator. However, for a person standing in line with the ecliptic, his view in any direction falls 4 miles short of the terminator and that is the maximum you seek. When the Moon is within 4K miles of the ecliptic, an observer on Earth at he same offset from the ecliptic will have the view described above. When the Moon is more than 4K miles away from the ecliptic, all Earth-bound observers see some non-illuminated portion and therefore do not see the maximum you want. Not only is the Moon's orbit tilted with respect to the ecliptic, but the direction of the tilt wobbles like a child's top. I have no idea how to factor that into an equation that would compute the time between one visible maximum and another.. Barry, Will it be OK if I copy your post into an answer to the same question posted at the Stack Exchange astronomy website at https://tinyurl.com/tze2pwm ? Harry |

What is the cycle for when the maximum amount of the Moon's apparent area is sunlit?On Sun, 12 Jan 2020 14:43:50 -0000 (UTC), Harold Davis
wrote: Barry Schwarz wrote in : Barry, Will it be OK if I copy your post into an answer to the same question posted at the Stack Exchange astronomy website at https://tinyurl.com/tze2pwm ? Harry Sure. Maybe somebody there will be able to confirm my calculations. -- Remove del for email |

What is the cycle for when the maximum amount of the Moon's apparent area is sunlit?On Sun, 12 Jan 2020 14:35:00 -0000 (UTC), Harold Davis
wrote: Barry Schwarz wrote in : On Sat, 11 Jan 2020 20:43:08 -0000 (UTC), Harold Davis wrote: The Moon is defined as full when its ecliptic longitude is 180° distant from the Sun's. But what is the cycle for when the maximum proportion of the Moon's apparent area is sunlit, and for observers at what latitudes of the Earth's surface is this maximum realised? Note that the plane of the Moon's orbit around the Earth is inclined at an angle of 5° to the plane of the Earth's around the Sun. To get a handle on why this affects the Moon's apparent fullness, imagine if it were 90°. Given the inclination I am sceptical as to whether the proportion ever reaches 50%. Since the Sun's diameter exceeds the Moon's, the Sun always illuminates slightly more than 50% of the Moon's surface. (To visualize, draw the external common tangents from a larger circle to a smaller one.) Using 98M miles as the distance, 432K miles as the Sun's radius, and 1K miles for the Moon's, the Sun illuminates almost 7 miles into the opposite hemisphere. Any observer a finite distance from the Moon can only ever see less than 50% of the Moon's surface. The further away the observer is, the closer the observable percentage is to 50. (To visualize, draw the two tangents from an external point to a circle.) However, when the Moon is in the plane of the ecliptic, an observer at the north pole can see approximately 12 miles over the Moon's north pole into the opposite hemisphere. So he can see some 5 miles beyond the terminator. Similar geometry causes his view of the Moon's south pose to fall short of the terminator. However, for a person standing in line with the ecliptic, his view in any direction falls 4 miles short of the terminator and that is the maximum you seek. When the Moon is within 4K miles of the ecliptic, an observer on Earth at he same offset from the ecliptic will have the view described above. When the Moon is more than 4K miles away from the ecliptic, all Earth-bound observers see some non-illuminated portion and therefore do not see the maximum you want. Not only is the Moon's orbit tilted with respect to the ecliptic, but the direction of the tilt wobbles like a child's top. I have no idea how to factor that into an equation that would compute the time between one visible maximum and another.. Hi Barry, I am very grateful to you for this detailed reply. Can you possibly tell me what proportion of the Moon's apparent disc is sunlit for an observer at Earth latitude theta at the moment of "full" moon if the Earth-Moon line is inclined at say 2.5 degrees to the plane of the Earth's orbit around the Sun? (I'm taking 2.5 degrees as an "average" of 0 degrees and the maximum of 5 degrees.) That would give a handle on how "full" an "average" full moon is, viewed from Earth latitude theta. (I'm a bit embarrassed about asking this, because I could work it out myself, but my geometry is so rusty that it would probably take me about six hours!) Consider a plane that passes through the centers of the Earth, Moon, and Sun perpendicular to the ecliptic. In that plane, the Sun is located at (0,0) and is a circle with radius 432e3. The Earth is located at (98e6,0) and is a circle with radius 4e3. The Moon is located at (98.25e6,10.9e3) and is a circle with radius 1e3. The angle from the (center of the) Earth to the (center of the) Moon is 2.5 degrees. So the angle from the Sun to the Moon is almost .0064 degrees. Let us define the term boundary to be a line passing through the center of the Moon that is perpendicular to a particular line of site. Each of these will divide the Moon into two semicircles. The first one is the N-S boundary that is perpendicular to the ecliptic (basically the vertical line x=98.25e6). The next is the S-M boundary perpendicular to the line from the Sun to the Moon. Since this line makes an angle of .0064 degrees with the ecliptic, then the S-M boundary makes the same angle with the N-S boundary. The third one is the O-M boundary perpendicular to the line from the Observer to the Moon. The angle it makes with the N-S boundary depends on the observer's latitude. Our observer will also be in this plane at latitude theta. His distance from the Moon (D) is given by sqrt(62516e6 - 2e9 * cos(theta)) His elevation (Z) above the ecliptic is 4e3 * sin(theta) The angle the O-M boundary makes with the N-S boundary is atan( (10.9e3 - Z) / (250e3 - 4e3 * cos(theta) From this and the data in my previous message, you should be able to determine if the observer can view any non-illuminated area beyond the northern end of the S-M boundary. But the full answer to your question is much more complicated. For example, as the observer shifts is gaze to the eastern and western edges to the Moon's disk, his latitude becomes less and less relevant and completely irrelevant at due east and west.. -- Remove del for email |

All times are GMT +1. The time now is 04:26 AM. |

Powered by vBulletin® Version 3.6.4

Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

Copyright ©2004 - 2006 SpaceBanter.com. Space photos on this page are credited to NASA and ESA