![\[\begin{array}{|l|l|l|l|l|l|l|l|l|}\hline Name& a&\epsilon&i&\omega&\Omega&d_0&t_0$&$T_{days}\\ \hline Mars&1.52371&0.09339&1.850&286.5&49.6&2018:9:16& 07:20&686.92971\\ \hline Merc&0.3870993&0.20564&7.005&29.13&48.3&2017:3:23& 10:33& 87.969257\\ \hline Venus& 0.723336&0.00678&3.3947&54.9& 76.7&2017:2:20& 05:19&224.700799\\ \hline Earth& 1.000003& 0.01671& 0.0& 102.9& 0.0&2017:01:04& 00:30& 365.25636\\ \hline Jup&5.2029&0.0484&1.304&274.3&100.4&2011:3:13& 05:19&4330.67043\\ \hline Sat&9.537&0.0539&2.486&338.9&113.7&2003:7:11& 13:30&10747.12745\\ \hline Uran&19.189&0.04726&0.773&96.9&74.02&1966:9:9&00:00&30589.273040\\ \hline Nept&30.0699&0.00859&1.7700&273.2&131.784&2046:11:13&00:00&59800.943732\\ \hline Pluto&39.4821&0.24883&17.14001&113.76&110.30&1990:1:13&00:00&90591.174612 \\ \hline\end{array}\]](http://abadonna.physics.wisc.edu/wp-content/ql-cache/quicklatex.com-7d26dedfe98cf5823b6753e49112c0fd_l3.png "Rendered by QuickLaTeX.com")
(date of perihelion passage) is 
, 
(time of perihelion passage) is 
.
is in 
, all anomalies 
are in degrees. is date of perihelion passage, is time of perihelion passage UTC. 
is orbital period in days.
![\[1.0\, AU=1.495978707\times 10^{11}m=92.956\times 10^6 \, mi, \qquad 1.0\, day=86400\, s\]](http://abadonna.physics.wisc.edu/wp-content/ql-cache/quicklatex.com-0c3d3237c12a5552555b4c846d789656_l3.png "Rendered by QuickLaTeX.com")
Julian time
![\[1.0 \, JD=86400 \, s, \qquad 1.0 \, JY=365.25 \, JD\]](http://abadonna.physics.wisc.edu/wp-content/ql-cache/quicklatex.com-31f7861bb554dd3fb2f91b9d44e57a87_l3.png "Rendered by QuickLaTeX.com")
and a zero-point, taken to be 
for January 
, 
. The Montenbruck algorithm lets us transform a calendar day to Julian.
Let 
be year, month, day and UT time according to your calendar and watch. Compute
![\[\begin{array}{lllll} y=Y-1 & \mbox{and} & m=M+12 & \mbox{if} & M\le 2\\ y=Y & \mbox{and} & m=M & \mbox{if} & M> 2\end{array}\]](http://abadonna.physics.wisc.edu/wp-content/ql-cache/quicklatex.com-a377215f5c9bbfa94a0d16c11b1d00bc_l3.png "Rendered by QuickLaTeX.com")
and
![\[\begin{array}{ll} B=-2 & \mbox{up to/including 4 Oct 1582}\\ B=floor(y/400)-floor(y/100) & \mbox{from and including 15 Oct 1582}\end{array}\]](http://abadonna.physics.wisc.edu/wp-content/ql-cache/quicklatex.com-f71a51c87b7fe7e8ccecf93d6014a0e7_l3.png "Rendered by QuickLaTeX.com")
since those are days/years in which Europe played fiddles with the calendar system. Then
![\[JD=floor(365.25 \, y)+floor(30.6001 (m+1))+B\]](http://abadonna.physics.wisc.edu/wp-content/ql-cache/quicklatex.com-e958df4a91d4b649219485dea7e2fe5b_l3.png "Rendered by QuickLaTeX.com")
![\[+1720996.5 +D +UT/24\]](http://abadonna.physics.wisc.edu/wp-content/ql-cache/quicklatex.com-403c90a23624a254a327ee24af333bb4_l3.png "Rendered by QuickLaTeX.com")
For example, today’s date and current time of loading this page is the Julian time
Note that you also need the universal time (UT). For us that’s local time plus 5 hrs.
Convert this to UT in the formula above via (Hr+min/60+sec/3600). In the formula for Julian date/time the result is expressed in Julia days, hence the penultimate division by 24.0.
You can run this application to compute Julian date/time for (semi) arbitrary dates:
Julian date=