madmac Posted May 8, 2017 Share Posted May 8, 2017 (edited) 1. What is the kg of the gravitationally effective mass of the Sun's photons blocked by the Moon at eclipse ??. Wikileaks says that an average of 1426 watts of the Sun's radiation hits the Moon per square m (ie 1426 joules/sm/sec). And that 89,875,517,873,681,764 joules (of radiation energy??) are equivalent to 1 kg of mass (i guess that this is based on E=mcc). The Moon's radius is 1,737,100 m. Distance tween Moon & Earth is 380,000,000 m. Speed of light is 300,000,000 m/sec. I calculate that this equates to 6.842 kg of the Sun's photons (blocked by the Moon) in each metre tween Moon & Earth. And the total mass of photons along the 380,000,000 m is 1.3 billion kg. Is this calculation & logic ok ?? 2. What is the loss of g on Earth in the zone of total shadow, ie due to the "loss" of mass ?? I found an equation for the g at the end of a long cylinder due to the mass of the cylinder. g=2(PI)G(d)L where d is the density & L the length. If the area of the ray of shadow were 1 sq m the density is 6.842 kg/m^{3} & the change in g would be 0.178 m/s^{2 }(eg 9.8 goes to 9.978) . For an area of 100 sq km the change in g would be 1.78/10^7 m/s^{2 }(eg 9.8 goes to 9.800 000 0178). Anyhow this is so small that it aint worth calculating the actual shape & size of the Moon's shadow. The loss of photonic mass & the associated gain in g wouldn't affect the azimuth of the plane of Allais' pendulum much at all (but it would affect a good g-meter). Or is my calc & logic wrong ?? Edited May 8, 2017 by madmac Link to comment Share on other sites More sharing options...

Strange Posted May 8, 2017 Share Posted May 8, 2017 Is this calculation & logic ok ?? Hard to say, as you don't show how you calculate anything! But the numbers look way to big to me. Let's see: Surface area of moon = 38 million m^{2} Total energy falling on moon per second = 1/2 x 38x10^{6} m^{2} x 1426 J/s (watts) = 27x10^{9} joules/s Total mass falling on moon per second (from e=mc^{2}) = 27x10^{9} / c^{2} = 3x10^{-7} kg/s Not sure how to equate that to the total mass between the Earth and Moon, but maybe multiplying by light travel time (1.3s) would be about right. So we are talking about a total mass of about half a milligram. As you say, not really significant. More significant could be the mass of the Moon directly above. Link to comment Share on other sites More sharing options...

swansont Posted May 8, 2017 Share Posted May 8, 2017 Or is my calc & logic wrong ?? Bingo! Link to comment Share on other sites More sharing options...

madmac Posted May 8, 2017 Author Share Posted May 8, 2017 (edited) Visible (2D) area of moon = PI r^{2} (where r = 1,737,100 m) = 9.4798 * 10^{12} m^{2}. Multiply that area by 1426 w/m^{2} = 1.3518*10^{16} joule/sec. (this is where i made my mistake earlier). If I kg = 9*10^{16} joule (if E=mcc) then moon blocks 0.15020 kg/sec. 380,000,000 m divided by speed of light 300,000,000 m/sec = 1.27 sec. So total loss of light tween moon & earth at eclipse is 1.27 by 0.1502 = 0.191 kg (at any instant). Logic was ok i think, but i messed up the number crunching. Anyhow, the moon's shadow at eclipse has very little mass, & thusly almost zero direct effect re Earth's g & the Allais Effect (or Effects). If desperate i could add the loss of the mass of photons usually emitted by Earth, but i seem to recall that this loss is less than 10% ie less than 142 w/m^{2}, ie less than 0.0191 kg (at any instant)(probably not even half that actually). Edited May 8, 2017 by madmac Link to comment Share on other sites More sharing options...

imatfaal Posted May 8, 2017 Share Posted May 8, 2017 Hard to say, as you don't show how you calculate anything! But the numbers look way to big to me. Let's see: Surface area of moon = 38 million m^{2} Total energy falling on moon per second = 1/2 x 38x10^{6} m^{2} x 1426 J/s (watts) = 27x10^{9} joules/s Total mass falling on moon per second (from e=mc^{2}) = 27x10^{9} / c^{2} = 3x10^{-7} kg/s Not sure how to equate that to the total mass between the Earth and Moon, but maybe multiplying by light travel time (1.3s) would be about right. So we are talking about a total mass of about half a milligram. As you say, not really significant. More significant could be the mass of the Moon directly above. "Surface area of moon = 38 million m^{2}" that should be km^2 - so that final figure needs to be lifted by x10^6 Link to comment Share on other sites More sharing options...

Strange Posted May 8, 2017 Share Posted May 8, 2017 (edited) "Surface area of moon = 38 million m^{2}" that should be km^2 - so that final figure needs to be lifted by x10^6 Ah yes. Well spotted. (Are you proof-reading and fact-checking all my posts today ...) Visible (2D) area of moon = PI r^{2} (where r = 1,737,100 m) = 9.4798 * 10^{12} m^{2}. And that is a much better way of calculating it. Edited May 8, 2017 by Strange Link to comment Share on other sites More sharing options...

imatfaal Posted May 8, 2017 Share Posted May 8, 2017 Ah yes. Well spotted. (Or you proof-reading and fact-checking all my posts today ...) today? I realised it must look like that - but I couldn't see where the discrepancy was so had to dig till it came to light and then I couldn't stay schtum Link to comment Share on other sites More sharing options...

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