Is there a way to find out the current Earth-sun distance?

[Johnny5]

Does anyone know if there a way to find out the current distance between the earth, and the sun in real time, using an online resource?

 

 

I need to know the distance between the earth and sun now.

 

Regards

[swansont]

Between this data and the aphelion/perihelion table I gave in the "If any of these definitions are wrong, please let me know" thread, you should be able to calculate it.

[Johnny5]

Between this data[/url'] and the aphelion/perihelion table I gave in the "If any of these definitions are wrong, please let me know" thread, you should be able to calculate it.

 

Here is the data from the table in the other thread:

 

2005

 

Phenomenon----- Date--hour:minute

Perihelion-------- Jan 2, 01:00

Vernal Equinox--- Mar 20, 12:33

Autmn Equinox--- Sept 22, 22:23

Aphelion--------- July 5, 05:00

Summer Solstice- June 21, 06:46

Winter Solstice-- Dec 21, 18:35

 

 

 

Vernal equinox

 

Ok, let me sort of fumble towards the answer.

 

Nice picture

 

Analemma

 

Newton/Kepler mathematical connection

 

 

PS: I don't mind doing the calculation at all Dr swanson, it's good practice, plus it forces me to improve the simple solar system model I started off with, but what I really was looking for, was a site which has the current values continuously computed. Something professional which I can rely on, because i am going to need it again, and I cannot stop and do the calculation every time I need it.

 

Regards

[Janus]

Go to

 

http://ssd.jpl.nasa.gov/cgi-bin/eph

 

Choose the target as the Sun. Set the date/time range you want. Generate the ephemerides.

The distance to the sun will be under the heading "delta" and in AU. One AU is 149,598,550,000 meters.

[Johnny5]

Go to

 

http://ssd.jpl.nasa.gov/cgi-bin/eph'>http://ssd.jpl.nasa.gov/cgi-bin/eph

 

Choose the target as the Sun. Set the date/time range you want. Generate the ephemerides.

The distance to the sun will be under the heading "delta" and in AU. One AU is 149' date='598,550,000 meters.[/quote']

 

 

Thank you

 

ssd=solar system dynamics jpl =jet propulsion laboratories

Here are the results:

 

Ephemeris Settings

Target Body: Sun [sol]

Observer Location: New York, NY

Coordinates: 73°59'39.1''W, 40°45'06.1''N

 

From: A.D. 2005-04-26 00:00 UT

To: A.D. 2005-04-27 00:00

Step: 1 day

Format: Calendar Date and Time

 

Output Quantities: 1-2,4,9-10,13-14,19-21,24,29,32

Ref. Frame, RA/Dec Format: J2000, HMS

Apparent Coordinates Model: Airless

 

 

--------------------------------------------------------------------------------

 

HORIZONS Generated Ephemeris

*******************************************************************************

Ephemeris / WWW_USER Tue Apr 26 06:47:56 2005 Pasadena, USA / Horizons

*******************************************************************************

Target body name: Sun (10) {source: DE405}

Center body name: Earth (399) {source: DE405}

Center-site name: (User Defined Site)

*******************************************************************************

Start time : A.D. 2005-Apr-26 00:00:00.0000 UT

Stop time : A.D. 2005-Apr-27 00:00:00.0000 UT

Step-size : 1440 minutes

*******************************************************************************

Center geodetic : 286.005800, 40.7517, -0.00{E-lon(deg),Lat(deg),Alt(km)}

Center cylindric: 286.005800, 4838.6361, 4141.57{E-lon(deg),Dxy(km),Dz(km)}

Center pole/equ : High-precision EOP model {East-longitude +}

Center radii : 6378.1 x 6378.1 x 6356.8 km {Equator, meridian, pole}

Target pole/equ : IAU_SUN {East-longitude +}

Target radii : 696000.0 x 696000.0 x 696000.0 k{Equator, meridian, pole}

Target primary : Sun {source: DE405}

Interfering body: MOON (Req= 1737.400) km {source: DE405}

Deflecting body : Sun, EARTH {source: DE405}

Deflecting GMs : 1.3271E+11, 3.9860E+05 km^3/s^2

Atmos refraction: NO (AIRLESS)

RA format : HMS

Time format : CAL

EOP file : eop.050422.p050714

EOP coverage : DATA-BASED 1962-JAN-20 TO 2005-APR-22. PREDICTS-> 2005-JUL-13

Units conversion: 1 AU= 149597870.691 km, c= 299792.458 km/s, 1 day= 86400.0 s

Table cut-offs 1: Elevation (-90.0deg=NO ),Airmass (>38.000=NO), Daylight (NO )

Table cut-offs 2: Solar Elongation ( 0.0,180.0=NO )

*********************************************************************************************************************************************************************************************************************************

Date__(UT)__HR:MN R.A._(ICRF/J2000.0)_DEC R.A.__(a-apparent)__DEC Azi_(a-appr)_Elev APmag S-brt Illu% Ang-diam Ob-lon Ob-lat r rdot delta deldot 1-way_LT S-T-O Cnst N.Pole-RA N.Pole-DC

*********************************************************************************************************************************************************************************************************************************

2005-Apr-26 00:00 C 02 13 56.94 +13 27 05.5 02 14 12.29 +13 28 29.8 290.9710 -3.3065 -26.73 -10.59 100.0 1907.386 298.31 -4.65 0.0000000000 0.00000 1.0062406299 0.78301 8.368648 0.0000 Ari 286.13000 63.87000

2005-Apr-27 00:00 C 02 17 43.68 +13 46 20.6 02 17 59.08 +13 47 44.1 291.2317 -3.1151 -26.73 -10.59 100.0 1906.880 285.10 -4.55 0.0000000000 0.00000 1.0065076293 0.77970 8.370869 0.0000 Ari 286.13000 63.87000

*********************************************************************************************************************************************************************************************************************************

Column meaning:

 

TIME

 

Prior to 1962, times are UT1. Dates thereafter are UTC. Any 'b' symbol in

the 1st-column denotes a B.C. date. First-column blank (" ") denotes an A.D.

date. Calendar dates prior to 1582-Oct-15 are in the Julian calendar system.

Later calendar dates are in the Gregorian system.

 

The uniform Coordinate Time scale is used internally. Conversion between

CT and the selected non-uniform UT output scale has not been determined for

UTC times after the next July or January 1st. The last known leap-second

is used over any future interval.

 

NOTE: "n.a." in output means quantity "not available" at the print-time.

 

SOLAR PRESENCE

Time tag is followed by a blank, then a solar-presence symbol:

 

'*' Daylight (refracted solar upper-limb on or above apparent horizon)

'C' Civil twilight/dawn

'N' Nautical twilight/dawn

'A' Astronomical twilight/dawn

' ' Night OR geocentric ephemeris

 

LUNAR PRESENCE

The solar-presence symbol is immediately followed by a lunar-presence symbol:

 

'm' Refracted upper-limb of Moon on or above apparent horizon

' ' Refracted upper-limb of Moon below apparent horizon OR geocentric

ephemeris

 

R.A._(ICRF/J2000.0)_DEC =

J2000.0 astrometric right ascension and declination of target. Corrected

for light-time. Units: HMS (HH MM SS.ff) and DMS (DD MM SS.f)

 

R.A._(a-apparent)__DEC. =

Airless apparent right ascension and declination of the target with respect

to the Earth true-equator and the meridian containing the Earth true equinox of

date. Corrected for light-time, gravitational deflection of light, stellar

aberration, precession & nutation. Units:HMS (HH MM SS.ff) and DMS (DD MM SS.f)

 

Azi_(a-appr)_Elev =

Airless apparent azimuth and elevation of target. Corrected for light-time,

the gravitational deflection of light, stellar aberration, precession and

nutation. Azimuth measured North(0) -> East(90) -> South(180) -> West(270),

elevation with respect to plane perpendicular to local zenith direction.

TOPOCENTRIC ONLY. Units: DEGREES

 

APmag S-brt =

Sun's approximate apparent visual magnitude & surface brightness.

APmag= M - 5 + 5*log10(d), where M= 4.83 and d= distance from Sun in parsecs.

Units: none & VISUAL MAGNITUDES PER SQUARE ARCSECOND

 

Illu% =

Fraction of target circular disk illuminated by Sun (phase), as seen by

observer. Units: PERCENT

 

Ang-diam =

The equatorial angular width of the target body full disk, if it were

fully visible to the observer. Units: ARCSECONDS

 

Ob-lon Ob-lat =

Apparent planetographic ("geodetic") longitude and latitude (IAU2000 model)

of the center of the target disk seen by the observer at print-time. Light

travel-time from target to observer is taken into account. For the gas giants

Jupiter, Saturn, Uranus and Neptune, IAU2000 longitude is based on the

"Set III" prime meridian rotation angle of the magnetic field. By contrast,

pole direction (thus latitude) is relative to the body dynamical equator.

There can be an offset between the magnetic pole and the dynamical pole of

rotation. Units: DEGREES

 

r rdot =

Target apparent heliocentric range ("r") and range-rate ("rdot") as seen

by observer. Units: AU and KM/S

 

delta deldot =

Target apparent range ("delta") and range-rate ("delta-dot") relative to

observer. Units: AU and KM/S

 

1-way_LT =

Target 1-way light-time, as seen by observer. The elapsed time since light

(observed at print-time) left or reflected off the target. Units: MINUTES

 

S-T-O =

Sun-Target-Observer angle; target's apparent PHASE ANGLE as seen at

observer's location at print time. Units: DEGREES

 

Cnst =

Constellation ID; the 3-letter abbreviation for the constellation name of

target's astrometric position, as defined by IAU (1930) boundary delineation.

See documentation for list of abbreviations.

 

N.Pole-RA N.Pole-DC

ICRF/J2000.0 Right Ascension and Declination (IAU2000 rotation model)

of target body's North Pole direction at the time light left the body to

be observed at print time. Units: DEGREES

 

 

Computations by ...

Solar System Dynamics Group, Horizons On-Line Ephemeris System

4800 Oak Grove Drive, Jet Propulsion Laboratory

Pasadena, CA 91109 USA

Information: http://ssd.jpl.nasa.gov/

Connect : http://telnet://ssd.jpl.nasa.gov:6775 (via browser)

telnet ssd.jpl.nasa.gov 6775 (via command-line)

Author : Jon.Giorgini@jpl.nasa.gov

 

 

 

Definition: Ephemeris is a set of parameters used by a global navigation satellite receiver to predict the location of a satellite and its clock behavior. Each satellite contains and transmits ephemeris data about its own orbit and clock. Ephemeris data is more accurate than the almanac data but is applicable over a short time frame from four to six hours. Ephemeris data is transmitted by the satellite every 30 seconds.

 

 

 

I need to make sure I understand this properly.

 

I went to the column, and selected two numbers, the first labeled delta, and the second labeled delta dot. Here they are:

 

Delta = 1.0062406299

 

Delta dot = 0.78301

 

So I am thinking that 'dot' is a derivative with respect to time, so that if delta represents distance, then delta-dot is a speed.

 

Now, the two locations I entered were:

 

1. New York

2. Sun

 

So I presume that delta gives me the distance from New York to the center of mass of the sun.

 

So now, I want to understand this distance in meters, but the explanation of the table says that delta is in astronomical units AU.

 

Here is what link to nasa has:

 

1 AU = 149,597,870.691 kilometers

 

and 1 kilometer equals 1000 meters.

 

Now, the ephermeris generator gave the following number for the distance between New York and the Sun, at time 0:00:00 today...

 

Delta = 1.0062406299

 

And here is what the explanation as to the meaning of "Delta" is:

 

delta deldot =

Target apparent range ("delta") and range-rate ("delta-dot") relative to

observer. Units: AU and KM/S

 

So the units of delta are astronomical units, which are units of distance, and the units of delta-dot are kilometers per second, which are units of speed.

 

Very good. Ok so...

 

I see that I can compute the distance from New York to the center of mass of the sun, in meters, at 0:00:00 this morning. Let me try and figure it out...

 

[math] 1.0062406299 AU = (1.0062406299 AU) \frac{149,597,870.691 Km}{AU} [/math]

 

[math] (1.0062406299 AU) \frac{149,597,870.691 Km}{AU} = 150531455.635811 Km [/math]

 

But I wanted the distance in meters...

 

[math] 150531455.636 Km \frac{1000 meters}{Km} = 150,531,455,636 \ \ \text{meters} [/math]

 

so 150 billion meters

 

or 150 million kilometers

 

I wonder what it is in parsecs.

 

Here is the conversion formula, to go from astronomical units AU, to parsecs:

 

206265 AU = 3.26 light year

 

Here is the definition of PARSEC

 

Let me see if I can make sense out of this.

 

What the hell is the definition of AU again, average distance from earth to sun i think, but best to check...

 

Here is the exact definition of an Astronomical Unit (AU) straight from NASA:

 

Its(astronomical unit) formal definition is the radius of an unperturbed circular orbit a massless body would revolve about the sun in 2*(pi)/k days (i.e.' date=' 365.2568983.... days), where k is defined as the Gaussian constant exactly equal to 0.01720209895.

[/quote']

 

Hmm... that's clever.

 

Here is the formula for the circumference of a circle...

 

[math] C = 2 \pi R [/math]

 

C= circumference(distance around circle's perimeter)

 

R = radius (distance from center of circle to any point on its perimeter)

 

pi = 3.14159... (a pure number)

 

We all know that there are 365 days per year, but this isn't super-precise.

 

NASA has the orbital period of earth at 365.2568983 days.

 

Let me check NASA vs something else.

 

Here is something interesting...

The Tropical and the Anomalistic Year

 

Most importantly, it says that the "ellipse is rotating with respect to the fixed stars"

 

Anyways, back to the goal, figure out the distance from New York to the center of mass of the sun, in PARSECS, at 12:00 AM this morning.

 

Right now, I am trying to figure out NASA's definition of astronomical unit.

 

Oh yeah, I need to figure out how they got that value for number of days per year, all the decimals.

 

Oh great, this sourceHermetic systems on the Julian & Gregorian Calendars says there are competing values for the year:

 

The main competing values seem to be the "mean tropical year" of 365.2422 days ("mean solar days") and the "vernal equinox year" of 365.2424 days.

 

And NASA has is at: 365.2569 days

 

So we have three competing values.

 

I think I'll go with NASA' date=' they had more decimal places.

 

Here is Dr Alan Cooper talking about NASA's value:

What does the AU mean?

 

he makes the following important comments on the Gaussian constant:

 

Ignoring the details and big numbers' date=' we have reached the crucial point. The Gaussian constant, and therefore the AU, is a number decreed by the International Astronomical Unit, it is not a measurement. (It is not the only constant-by-decree; the same is true of the speed of light in a vacuum, it is now a defined number. It will not change even if someone finds a way of measuring it to 20 places of decimals.)

 

It is worth going one detail further. The defined value of the Gaussian constant is 0.01720209895 (AU)1.5. Ignore the number and look at the power of 1.5. This comes straight from Kepler's third Law T2 is proportional to R3 i.e. T is proportional to R1.5, so the definition is based on Newtonian dynamics, which are well known to differ from the more accurate General Relativistic treatment. The definition still needs further refinement![/quote']

 

Let me have a look at Kepler's third law:

 

Square of any planet's orbital period (sidereal) is proportional to cube of its mean distance (semi-major axis) from Sun

 

Mathematical statement: T = kR3/2 , where T = sideral period, and R = semi-major axis

 

Here is a very good article on ellipses at wikipedia

 

Here is a derivation of the cartesian formula for an ellipse:

 

Derviation of formula for ellipse in a rectangular coordinate system

 

I'm sure there's a simpler proof.

 

Here is something about the International Astronomical Union

 

 

 

here is something about a new program by the NAVY. They mention that the program performs calculations in the center of mass frame of the solar system, at least if I understood the article correctly, and they mention something about VLBI (very long baseline interferometry).

 

NOVAS: Naval observatory vector astronomy subroutines

 

If anyone reads down, you will see this remark:

 

The International Celestial Reference System (ICRS)

 

Reference data for positional astronomy, such as the data in astrometric star catalogs (e.g., Hipparcos) or barycentric planetary ephemerides (e.g., JPL’s DE405) are now specified within the International Celestial Reference System (ICRS). The ICRS is a coordinate system whose origin is at the solar system barycenter and whose axis directions are effectively defined by the adopted coordinates of about 600 extragalactic radio sources observed by VLBI (see Section H of The Astronomical Almanac). These radio sources (quasars and active galactic nuclei) are assumed to have no observable intrinsic angular motions. Thus, the ICRS is a “space-fixed” system (more precisely, a kinematically non-rotating system) without an associated epoch. However, the ICRS closely matches the conventional dynamical system defined by the Earth’s mean equator and equinox of J2000.0; the alignment difference is at the 0.02 arcsecond level, negligible for many applications.

 

[math] \beta \alpha \rho \iota [/math]

 

Pronounced "var-ee" in Greek, "barry" in English. Greek word for "heavy." So i guess i was right, "barycenter of solar system" means center of mass of the solar system.

 

International Celestial Reference System

 

Professionally speaking, this is the best reference frame to understand the motion of the planets, and sun in, relative to the "fixed stars"

 

But still the rest frame of the sun is a good approximation to the ICRS.

 

Right now I am trying to understand this "Gaussian constant."

 

---------------------------------------------------------------

 

Ok, looks like I'm finally making sense out of things here.

 

This site really helped a lot:

 

Definition of sidereal year

 

Also useful at that site are:

1) Gaussian year

2) Astronomical unit

 

So this is really important:

 

Sidereal year

 

Screw it, I'm just going to let NASA teach me.

 

NASA on our solar system

[swansont]

Oh great' date=' this sourceHermetic systems on the Julian & Gregorian Calendars says there are competing values for the year:

 

"The main competing values seem to be the "mean tropical year" of 365.2422 days ("mean solar days") and the "vernal equinox year" of 365.2424 days."

 

And NASA has is at: 365.2569 days

 

So we have three competing values.

 

 

Not competing, because they are defined differently. The NASA definition says it's for an idealized circular orbit of a particular radius - the links says Since an AU is based on radius of a circular orbit, one AU is actually slightly less than the average distance between the Earth and the Sun so there is no reason to think that the year will end up being exactly the same as for the real orbit.

 

The mean tropical year and the vernal equinox year aren't the same, either.

[Johnny5]

Not competing' date=' because they are defined differently. The NASA definition says it's for an idealized circular orbit of a particular radius - the links says [i']Since an AU is based on radius of a circular orbit, one AU is actually slightly less than the average distance between the Earth and the Sun[/i] so there is no reason to think that the year will end up being exactly the same as for the real orbit.

 

The mean tropical year and the vernal equinox year aren't the same, either.

 

What should I do here?

 

I want to end up understanding things the best way possible. Right now, I just want to understand what NASA is doing, and that will probably take 2 hours.

 

I liked what Dr Cooper said.

 

Also, I see 'sidereal' time coming in now, and I never did understand that.

[swansont]

Also, I see 'sidereal' time coming in now, and I never did understand that.

 

A sidereal day is how long it takes for the earth to rotate 360 degrees.

 

Approximately how far does the earth rotate in a solar day (from solar noon to solar noon, i.e. the sun is overhead, or on a N-S line projecting to that point)? Why is that different than 360 degrees?

 

That should spell out the difference. A sidereal day is important to astronomers because the stars essentially don't move - they are a good coordinate system for the 360 degree motion upon which the sidereal day is based.

[Johnny5]

A sidereal day is how long it takes for the earth to rotate 360 degrees.

 

Approximately how far does the earth rotate in a solar day (from solar noon to solar noon' date=' i.e. the sun is overhead, or on a N-S line projecting to that point)? Why is that different than 360 degrees?

 

That should spell out the difference. A sidereal day is important to astronomers because the stars essentially don't move - they are a good coordinate system for the 360 degree motion upon which the sidereal day is based.[/quote']

 

I get it, now I just have to do the math.

 

To a good approximation, the stars are fixed in the celestial sphere.

But the stars appear to orbit the earth, in the rest frame of the earth.

 

But actually, the following statement is true:

 

"The earth is spinning."

 

The earth buldges at the equator.

 

The axis of spin, runs through the poles of earth.

 

Called "poles" because the north star is called "Polaris" from the greek word for colt, because the little dipper looks like a colt, if you look at it right, with the north star for its head.

 

[math] \pi \omega \lambda \alpha \rho \iota o \nu [/math]

 

Pronunciation: (paw-lah-rion)

 

So the earth is spinning, around this axis which pretty much stays fixated on the north star, ignoring precession. The time it takes for one quarter of precession, is on the order of 6000 years (i think).

 

But the point is, that in the rest frame of the earth, the stars appear to orbit the earth, and since they seem fixed to an extremely good approximation in the celestial sphere, the time of orbit is fairly well defined. By sensory perception alone, its about 24 hours.

 

So, to measure a sidereal day, one would pick a star in say the constellation orion, just falling below the horizon, at sunrise, and measure the time for the same phenomenon to occur next. Which would be close to 24 hours.

 

But if you do this every day, you wont always get the exact same reading, the key point being, that the amount of time in a "day" varies subtly, over the course of a year.

 

So now to answer Dr Swanson's question.

 

Approximately how far does the earth rotate in a solar day (from solar noon to solar noon, i.e. the sun is overhead, or on a N-S line projecting to that point)? Why is that different than 360 degrees?

 

I'm not sure what you are asking.

 

solar day... hmmm

sidereal day...

 

Apparently not the same thing.

 

The amount of time it takes the earth to spin 360 degress in the rest frame of the stars, is called a sidereal day.

 

 

i got it.

 

 

Suppose you use the sun, to measure the value of a day, instead of a star in the constellation Orion, as I first suggested.

 

Then you are not using a star in the celestial sphere, to define "day" you are using our star(the sun) which is much much closer to us than any star in the celestial sphere.

 

But the sun is not really "fixed." It too, is orbiting the center of mass of our solar system. Kepler approximated things, by treating planetary motion in a frame in which the sun was at rest.

 

So naturally, a solar day will not be equivalent to a sidereal day.

 

Thank you Dr.

 

There is still an unanswered question, "Approximately how far does the earth rotate in a solar day?"

 

I am reading this article here.

 

That animation was exactly the answer.

 

In order to understand the answer given in the animation, it is necessary to understand "celestial meridian," so I am now reading this (Celestial sphere).

[swansont]

In order to understand the answer given in the animation, it is necessary to understand "celestial meridian,"

 

The N-S line that passes through a point directly overhead.

 

 

Here's a followup, when you're ready: How many days are there in a year, rounded to the nearest whole number? Now, how many times does the earth rotate in a year, again rounded to the nearest whole number?

 

This also points out the difference.

[Johnny5]

The N-S line that passes through a point directly overhead.

 

Dr Swanson, I think that's incorrect.

 

I think the north south line intersects the umm celestial meridian, which is a great circle on the celestial sphere.

 

But would different earth observers necessarily have different celestial meridians?

 

Thank you

[swansont]

Dr Swanson' date=' I think that's incorrect.

 

I think the north south line intersects the umm celestial meridian, which is a great circle on the celestial sphere.

 

But would different earth observers necessarily have different celestial meridians?

 

Thank you[/quote']

 

A meridian is a line of longitude, i.e. a N-S line.

 

Different earth observers at different longitudes would have their own celestial meridian. If they looked straight up, they would be looking at different points (which ties in with the "longitude problem" of navigation, and its solution with Harrison's chronometer)