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How to measure the Distance from Earth to Sun?

  1. May 20, 2004 #1
    how can I measure?
    I have'nt got any advanced tools (I have got a telescope only)!
    I think it is possible, because it was done in past.

  2. jcsd
  3. May 22, 2004 #2


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    Hi Deepsky,

    In the past I'm pretty sure it was done via triangulation. If you know what angle the sun is at a certain day & time in two different locations, you can use trigonometry to solve for the distance. I don't know of any ways to find it sitting in one place, unfortunately, but that's not to say there aren't any.
  4. May 22, 2004 #3
    Hi enigma
    Could you give me a document for that?
  5. May 22, 2004 #4


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    Here are two methods, the first is ancient and works in theory, but not in practice, the second was the source of the first accurate value in the 18th century.

    1. Aristarchus' method. Wait till the moon is EXACTLY half full. Then the Earth, the Moon, and the Sun form a right triangle with the right angle at the moon. Measure the angle between the Moon and the Sun, and with the Moon's distance (previously found by triangulation), derive the Sun's distance as Moon's distance divided by the cosine of the angle. The uncertainties in that "exactly' make this a shaky method. Aristarchus himself got the Sun 30 times as far away as the Moon, when the correct value is nearly 400 times.

    2. Transit of Venus method. You know the relative sizes of the orbits of Venus and the Earth from their revolution times (observed) and Kepler's third law (also derived from observations - Tycho Brahe's observations). So when Venus crosses the face of the Sun (one of those transits is coming up) measure the time it takes and be careful to note the position of the transit path on the disc of the Sun. From this you can work similar triangles to solve for the distance from the Earth to the sun.

    (Edited to correct the trig formula in the Aristarchus method.)
    Last edited: May 22, 2004
  6. May 22, 2004 #5
    thanks, but I want to measure it at different times (not exactly when venuse transit)
    any other idea?
    Last edited: May 23, 2004
  7. May 22, 2004 #6


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    Actually measuring the distance from the Earth to the Sun is one of the harder jobs in observational astronomy. Until modern times with satellites and radar there weren't any good triangulation methods available. Trying for parallax from two well separated points on Earth would be best, but it requires super accurate pointing. The angle subtended by the Earth from the Sun is arc tangent 8000/93,000,000, or .005o.
  8. May 22, 2004 #7
    The sun spins on its axis. You can see how fast it spins by watching sunspots move across the disc. You find (because the sun is not solid) that different latitudes have different rotation periods.

    So concentrate on the area near the Sun's equator and measure the length of its day. Then measure the red shift between the approaching side and the retreating side of the equator, and this will allow you to work out how big the Sun actually is. Once you know how big it is, and the angle the Sun's disc subtends as viewed from Earth, you can calculate its distance.
  9. May 22, 2004 #8
    I think you can get radar echoes from planets (e.g. venus) to measure distance any time. Distance to sun is then only a matter of math.

    BTW, there will be a venus transit on june 8, 2004. Happens only every 100 years or so.
  10. May 23, 2004 #9
    Another way is to measure the times of the eclipses of Jupiter's moons. Because of the finite speed of light, the eclipses seem to happen earlier than expected when the Earth is on the same side of the Sun as Jupiter, and later than expected when we are on the opposite side. Say you find the maximum variation is about 16 minutes, Then the Earth's orbit is 16 light minutes in diameter, and therefore the Earth is eight light minutes from the Sun. Knowing the speed of light, that allows you to calculate the distance.
  11. May 23, 2004 #10
    But we know that the earth orbit around sun is not a circle, in fact I desided to measure the parameters of the earth orbit, so I need a more acurate method.
    I remember you again that it was done in past.

    how about the planets? Is it easier to measure the distances for planets or moon :confused:
  12. May 24, 2004 #11
    Once you have measured the distance to any one of the planets, or the Sun, you can work out all the other distances, using Kepler's laws.

    The best way to work out the shape of the orbits - the eccentricitity and inclination etc., is just to plot the movement of all the planets against the background stars, then do the math. This allows you to calculate the shape and relative size of the orbits very exactly, but you need just one distance measurement to calibrate the scale.

    Measuring the distance to the moon doesn't help much in this scale calibration. The planet's orbit sizes are related to their orbital periods, as they all go around the sun, but the moon goes around the Earth - unless the relative masses of the Earth and Sun are known, the moon distance wouldn't allow you to calculate the sun distance.

    Edited to fix spelling.
    Last edited: May 24, 2004
  13. May 24, 2004 #12


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    By comparing the times it takes Mars and the Earth to complete one orbit, you can determine the relative distance each must be from the Sun (Kepler's third law).

    You can measure the distance between the Earth and Mars using just two measurements twelve hours apart (this would work better in the winter). You'll have the angle - side - angle of a triangle (the Earth's diameter being the side).

    Once you have the distance between Earth and Mars, you can measure the angle between the Sun and Mars. Now you have the side - angle of a triangle, which, by itself, normally wouldn't do you a lot of good. However, you do know the proportion between two sides of the triangle. The side opposite the known angle is about 1.5 something times as big as the Earth-Sun distance. If you use the law of sines to find the angle between the Earth and Sun as viewed from Mars, the Earth-Sun distance will be the side opposite.

    [tex]\frac{a}{sin A} = \frac{1.5a}{sin B}[/tex]
    [tex]sin B = \frac{1.5a sin A}{a}[/tex]

    The a is unknown, but it divides out, so you can find the Earth-Sun angle as viewed from Mars. Now you have a known angle - side - angle and can determine the missing lengths.

    Simplified explanation at:

    http://www.astrosociety.net/pubs/mercury/32_04/sidebar.html [Broken]
    Last edited by a moderator: May 1, 2017
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