by **jjb2011** » Mon May 09, 2016 6:23 pm

Since we have a transit of Mercury today, it is interesting to calculate what the maximum transit time would be. The maximum transit time would occur when Mercury passes along a diameter of the sun rather than a chord.

1. Mercury orbits the Sun in 88 days so in a day the angular motion of Mercury as seen from the Sun would be 360/88 = 4.09degrees per day, which is 4.09/24 = 0.170 degrees/hr

2. The Earth orbits the Sun in 365 days so the angular motion of the Earth as seen from the Sun is 360/365 = 0.986degrees per day or 0.986/24 = 0.041degree/hr

3. The difference is 0.170 – 0.041 = 0.129 degrees per hour

4. Now we observe the Mercury transit from the Earth not the Sun, so the apparent angular speed will be reduced by the ratio of the distance of Mercury from the Sun to the distance of Mercury from the Earth. Mercury is 57.9 Mkm from the Sun and the Earth is 150Mkm from the Sun, so the Earth to Mercury distance is 92.1 and the ratio is 57.9/92.1 = 0.629.

5. The apparent angular speed of Mercury is therefore 0.129x0.629 = 0.081degrees per hour.

6. The Sun has an apparent diameter of 0.53 degrees as seen from the Earth so the maximum transit time will be, 0.53/0.081 = 6.54 hours

Of course this calculation uses averaged values for the distances to the Sun for Mercury and the Earth. The largest variation comes from the eccentricity of the orbit of Mercury, which is 0.21 resulting in a possible variation of +/- 0.65hrs.

JJB

**Hi, you are viewing as a guest, You can only see this first post in this topic. if you sign up you get access to other goodies you can't even see as a guest, including video tutorials on imaging and processing, scope modifications and even member discounts on gear! **

**So, give us a try, what have you got to lose! Oh, and if you stay, when you reach 50 posts you get access to MORE goodies! What are you waiting for!**