I am trying to use the transit of Mercury to estimate the distance between the Earth and the Sun, or 1 AU.

I know that I need to observe mercury from two antipodes, but I do not understand any of the underlying math behind this, or how to derive the astronomical unit from this.

Any help is appreciated!


Thanks to the Mercury transit, you can measure the parallax from the Earth. That happens due to TRACE , which tracks the transit of Mercury along the polar diameter of the Earth.


During that tracking, the transit of Mercury goes like that:

[Lockheed Martin/TRACE[2]

Now notice that, if TRACE remained stationary, the transit would be a straight line. So, if you calculate the maximum separation vertically of the center of Mercury and divide by two (how it's another different story), you got the parallax angle $\theta$:

Lockheed Martin/TRACE

You know $\theta$, you know Earth radius, so you know D, the distance from Earth to Mercury (in km) because

$$\tan\theta = R/D$$

Now, if you wanna know how many km a AU is, just realise that Mercury is 0.56 AU from us, so divide D/0.56 and you got it!

All images credited to NASA/IMAGE and/or Lockheed Martin/TRACE

  • 1
    $\begingroup$ Please credit the image source and explain what TRACE was. $\endgroup$ – Mike G Feb 4 '20 at 8:47
  • $\begingroup$ This is an interesting answer, but links break and rot over time, so it would be great if you included a discussion of what TRACE is rather than make each reader go off site. Thanks! btw I can't figure out why the heck that transit trajectory is wiggly in that way. It really does need to be explained. $\endgroup$ – uhoh Feb 4 '20 at 10:54
  • 1
    $\begingroup$ The transit trajectory (seen from Earth) is not wiggly, but it is the sight from TRACE. It's wiggly because of the position of the spacecraft, the norther it is the further down the disk goes, and vice versa. We see an ondulation because I assume that TRACE makes several revolutions to the Earth, not just one, so it goes up and down and up and down, etc. $\endgroup$ – Carlos Vázquez Monzón Feb 4 '20 at 11:29
  • $\begingroup$ Oh! I understand. Because of the its sun synchronous orbit the Sun (and Mercury) are continuously visible as the spacecraft circled the Earth about three times! Okay got it. So was that an accident, or did they plan it that way? $\endgroup$ – uhoh Feb 4 '20 at 13:08
  • 1
    $\begingroup$ Because $\theta$ is measured like in the picture, and you calculate the maximum separation that spans from the south Pole to the north Pole, so, you need to divide by two $\endgroup$ – Carlos Vázquez Monzón Feb 4 '20 at 20:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.