Background
Several years ago I was was watching the Science channel, and I heard the commentator It could have been Sagan or Kaku say to the affect; “When I finish saying this sentence it will take 2.25x10^8 years to return to this same point in the Galaxy.
For each orbit, the Sun traveled 150,000 light years of distance. Speed of the Sun (http://foxyurl.com/6Fz)

My question on this is, how far have I traveled from my Origin Point, My Birth in Light Years in that arc?
Or should it be considered as a straight line?

Known Factors and Bibliography
Sun travels at c.150mi/sec c.240km/sec
Speed of the Sun (http://foxyurl.com/6Fz)

The duration from my Birthday to, lets say 12:00 midnight June 20, 2009. 1 hr and 45min before the start of the Summer Solstice, is 21,130days
Date Duration Calculator: Days between two dates (http://foxyurl.com/6Fq)

wiki:
the speed of light
Speed of light - Wikipedia, the free encyclopedia (http://foxyurl.com/6N7)

Methodology
Please bear with me, I did make it to Geometry in high school, but that was a while back.

d = s x t

d = 150mi/sec x 21,130days, at this point days need to be converted to seconds

d = 150mi/sec x (1,825,632,000sec. = 21,130days)

d = 273,844,800,000mi

To convert to LY divide by 5.879×10^12 (58,790,000,000)
Light-year - Wikipedia, the free encyclopedia (http://foxyurl.com/6Na)

LY = 273,844,800,000 / 58,790,000,000 = 4.6580166695016159210750127572716 Light Years from Origin Point.

Correct? If it is I have to change my Profile Location.


Regards... KUXJ

Your answer is roughly correct with respect to the center of the milky way (you might get a slightly better answer considering the sun's path as an arc). However the milky way itself will be moving much faster, swamping the local speed of the sun, i.e. your absolute speed with reference to the center of the universe (if such a place exists) is likely to be much faster.

your answer is roughly correct with respect to the center of the milky way (you might get a slightly better answer considering the sun's path as an arc). However the milky way itself will be moving much faster, swamping the local speed of the sun, i.e. your absolute speed with reference to the center of the universe (if such a place exists) is likely to be much faster.I knew this was going to make my head hurt. :rolleyes:

It may not be the correct representation, but using the attached image...

* The Solar system is the small circle.
* The Milky Way is the larger circle.
* The X,Y,Z co-ordinates represent the center of the Universe "if such a place
* exists" or could it stand for "my origin point".

If I understand correctly:

I have the relationship between the Solar system and Milky Way "roughly correct". :cool: Thank You.

What I need to do now is apply the Hubble constant to my result, if I can determine, within the Universe, the distance of "my origin point" to 21,130 days later, and counting.

Edited at 2:00pm edt, 19:00z

This last statement is incorrect as to "if I can determine, within the Universe, the distance of "my origin point" to 21,130 days later, and counting." all I would need to do is apply the H-constant, I think...

Background

d = 273,844,800,000mi


So far so good...


[
To convert to LY divide by 5.879×10^12 (58,790,000,000)
Light-year - Wikipedia, the free encyclopedia (http://foxyurl.com/6Na)

LY = 273,844,800,000 / 58,790,000,000 = 4.6580166695016159210750127572716 Light Years from Origin Point.

Correct?

Not quite - you divided the distance by 5.879 x 10^10 instead of 5.879 x 10^12. Consequently your answer is too big by a factor of 100. You should have an answer of 0.0465 LY. Remember that a light year is a very, very long distance - it just didn't seem possible that you could travel over 4 light years in only 57 years.

By the way, given that the accuracy of the speed of light is given to only 4 digits, there is no point in writing out the results of the division to anything more than 4 decimal places.

it just didn't seem possible that you could travel over 4 light years in only 57 years.Oh... I don't know, I'll be 58 in Aug. and the older I get, the faster time seems to go by ;)

Thank You for the insight,

Regards

KUXJ

Just an update:

I sent an email to NDTyson at the Hayden 'bout my question,
And we are in dialog whether the H0 or Newton's laws of orbital Mechanics should be used.

That's how it was signed. Might just be a member of his staff.


Home | Neil deGrasse Tyson (http://www.haydenplanetarium.org/tyson/)

K

Have received answer:

The Hubble constant remains irrelevant at these scales of distance and time. Simply Earth's rotation, Earth's revolution around the Sun, the Sun's revolution around the MIlky Way galaxy's enter, and the falling of the Milky way towards the Andromeda galaxy. There is no other relevant motion to your needs.

-NDTyson


So for now, it remains .0465 LY