Recent content by Jimmy B

How low would the Suns intensity have to go before it started global cooling? Using Boltsman's radiation law, and putting in 5855 Kelvin for the Suns temperature, I get an average intensity at the top of the Earths atmosphere of 1369.39 Watts/m^2. This is obviously incorrect according to the...

Could this reduction in the Suns energy have any effect on the Earths climate? On the TV the other day, a weather man for the BBC said that we might be heading for another medieval ice age.

Is the radiated energy that reaches the Earth from the Sun diminishing? Some meteorologists think it is due to the reduction of Sun spots. Is this the normal 11 year cycle, or is there more to it?

I also got M = 6.2832173611570 radians (360.001836557620 degrees) using this equation from wiki. M = n t = sqrt ( G M / a^3) * t My values were, G M = 1.33E+11, a^3 = 149598261.0 km^3, t = 31558432.541760 seconds in a anomalistic year. There does seem to be a relationship. Getting...

Wikipedia has the mean anomaly of 357.51716 degrees, is this correct? If it is, the only way I can satisfy this equation “M = Mo + nt” that is on Wikipedia is, Mo = 357.5176 degrees, n = 0.98560025850 degrees, and “t” = 365.259644606150 days. “t” has a very near value of a anomalistic year...

Mean Anomaly Hi, all. I’m back with another question. If you want to find the Earths mean anomaly using this equation “M = n t”. What value should “n” be, the daily motion, if “t = 0”? Should it be 360/Sidereal year, 360/Tropical year, 360/Anomalistic year, or is it none of these?

Same here, if you ever find out please let me know.

Thanks for the reply, I’ll discount that one. Here’s another. Average speed of the Earth around the Sun = (2*pi)*(a/p) = 29784.8132 m/s. a = 1.49598E+11 Earths semi-major axis, and p = 31558149.76 seconds in a sidereal year. Earths orbital perimeter = 365.256363007 days in a Sidereal year *...

This is one of many. vp=\sqrt((1+e)\mu/(1-e)a) va=\sqrt((1-e)\mu/(1+e)a) vp = velocity at perihelion and va = velocity at aphelion. vp = 30.25375649 km and va =29.32291236 km mean velocity = vp+va/2 I got a mean velocity of 29.7883344272490 km. have I done it right...

What is the Earths average velocity around the Sun? I just plucked that value of the internet. Each website that I went on had a slightly different average velocity. I was spoilt for choice. They also had their own method of calculating it.

Now you’ve mentioned ellipses. How far does the Earth travel around the Sun each year? 365.25636004 days * Earths average velocity 29784.813 m/s = 939.9536 million kilometres. Or? (360 degrees / 365.25636004 days) * 939.9536 million kilometres = 939.89623 million kilometres

I use this ephemeris. Ephemeris Type ELEMENTS. Target Body Earth Moon Barycenter [EMB] [3]. Center Sun [Bodycenter]. Time Span Start=2013-03-04 Stop=1 d. Table settings default. Display Output default. Formatted HTML. http://ssd.jpl.nasa.gov/horizons.cgi#results This ephemeris gives...

Ask Gordon Ramsey, he’ll know that one. Do you know how much the Sidereal year varies D H? According to NASA it now is 365.246007919206 days. See post 32. What fixed stars are used as a reference for the Earth to complete one orbit? They did use the star Sirius until they found it had proper...

Thanks Jim for replying. It is very confusing how we keep time. The Gregorian calendar is a prime example. The time it takes the Earth to orbit once around the sun with respect to the fixed stars is called a Sidereal year. Now as far as I know the precise length of this year is unknown...

While checking a generated ephemeris on the JPL horizons system, I notice the PR Sidereal orbital period seemed low at 365.2460079192060 days. The sidereal year was equal to 365.256360417 days at noon on the 1 January 2000; this value is nearly 15 minutes less. Why does NASA use this value for...