<p>Yes, 27.8 is the maximum that can be achieved.  <a href="http://www.jgiesen.de/skymap/MercuryVenus" rel="nofollow noreferrer">jgiesen.de</a> calculates this from the eccentricity of the orbit of Mercury:</p>
<p><span class="math-container">$$\sin β = a(1+e)/(1\, \text{AU})= 0.467,\ β = 27.8°$$</span></p>
<p>For Mercury a=0.387 and e=0.206.</p>
<p>You could make a small correction for the Earth's eccentricity, but this is very much less than the eccentricity of Mercury, so has less effect</p>

CC BY-SA 4.0
2022-09-27T20:42:02.807Z
2022-09-27T20:42:02.807Z
James K
<p>I have been messing around with the elongation for Mercury, and I have been getting 27.8 degrees at maximum in <a href="https://www.heavens-above.com/PlanetSummary.aspx" rel="nofollow noreferrer">https://www.heavens-above.com/PlanetSummary.aspx</a>. Is there any history record that Mercury could go higher than 27.8 degrees or is 27.8 degrees the absolute maximum elongation that could be ever achieved?</p>

CC BY-SA 4.0
2022-09-27T21:23:05.730Z
The title asks about altitude, the text asks about elongation.  You should clarify which you're interested in.
Greg Miller
<the-sun><mercury>
At what exact date and location does Mercury appear at the highest altitude from Earth in history?
InfinitySwordofDiamond
# At what exact date and location does Mercury appear at the highest altitude from Earth in history?

(Asked by: username_0 on 2022-09-27)

I have been messing around with the elongation for Mercury, and I have been getting 27.8 degrees at maximum in <https://www.heavens-above.com/PlanetSummary.aspx>. Is there any history record that Mercury could go higher than 27.8 degrees or is 27.8 degrees the absolute maximum elongation that could be ever achieved?

**Tags:** the-sun, mercury
> **Comments:**
> **username_1**: The title asks about altitude, the text asks about elongation.  You should clarify which you're interested in.  

---

## 1 answers

### Answer by username_2 on 2022-09-27. Score: 3

Yes, 27.8 is the maximum that can be achieved. [jgiesen.de](http://www.jgiesen.de/skymap/MercuryVenus) calculates this from the eccentricity of the orbit of Mercury:

$$\sin β = a(1+e)/(1\, \text{AU})= 0.467,\ β = 27.8°$$

For Mercury a=0.387 and e=0.206.

You could make a small correction for the Earth's eccentricity, but this is very much less than the eccentricity of Mercury, so has less effect


---

