<p>This is a cool question; I'll give it a go.</p>

<blockquote>
  <p>Or is it that obvious that Venus transit will be longer because its orbital speed is less than Mercury?</p>
</blockquote>

<p>No, nothing here is obvious!</p>

<p><strong>tl;dr:</strong> Times for the transit of Mercury can range from 5.37 to 8.30 hours, and for Venus they range from 7.94 to 8.12 hours under assumptions of coplanarity but allowing for eccentricity.</p>

<p>Let's assume all orbits are in one plane. We can get the orbital speed of body <span class="math-container">$i$</span> with semimajor axis <span class="math-container">$a$</span> and current distanc <span class="math-container">$r$</span> from the <a href="https://en.wikipedia.org/wiki/Vis-viva_equation" rel="nofollow noreferrer">vis-viva equation</a>;</p>

<p><span class="math-container">$$v_i = \sqrt{GM\left( \frac{2}{r_i} - \frac{1}{_i} \right)}$$</span></p>

<p>where <span class="math-container">$GM$</span> is the <a href="https://en.wikipedia.org/wiki/Standard_gravitational_parameter" rel="nofollow noreferrer">standard gravitational parameter</a> of the Sun which is about 1.327124E+20 m<sup>3</sup>/s<sup>2</sup>.</p>

<pre><code>               a          ecc        pei           apo       v_peri   v_apo
Earth    1.49598E+11   0.016709  1.470984E+11  1.520976E+11  30286.6  29291.1
Venus    1.08208E+11   0.006772  1.074752E+11  1.089408E+11  35258.8  34784.4
Mercury  5.79091E+10   0.205630  4.600125E+10  6.981695E+10  58976.4  38858.6
</code></pre>

<p>Imagine Venus and Mercury crossing a line fixed at one end to the Sun and the other end to the center of the Earth. The linear speed of a planet <span class="math-container">$i$</span> relative to that line will be <span class="math-container">$v_i$</span> minus the speed of the line at the distance of planet <span class="math-container">$i$</span> or <span class="math-container">$v_E r_i / r_E$</span>, and the angular velocity seen from Earth will be that divided by (<span class="math-container">$r_E - r_i)$</span> or</p>

<p><span class="math-container">$$\dot{\theta}_i = \frac{v_i - (r_i/r_E) v_E}{r_E - r_i} = \frac{r_e v_i - r_i v_E}{r_E^2 - r_i r_E}$$</span></p>

<pre><code>(seconds)     Venus          Mercury
Earth       peri   apo      peri   apo
peri       28570  29242    19335  29884
apo        28060  28626    19384  29646

(hours)       Venus          Mercury
Earth       peri   apo      peri   apo
peri       7.936  8.123    5.371  8.301
apo        7.794  7.952    5.384  8.235
</code></pre>

CC BY-SA 4.0
2020-03-31T16:21:10.870Z
Sorry, but I got lost where you converted $v_p-v_E$ to the term with the square root GM etc. Are you sure that conversion is correct?
JohnHoltz
CC BY-SA 4.0
2020-03-31T19:52:21.673Z
@JohnHoltz yikes! Okay I'll fix that asap!
uhoh
CC BY-SA 4.0
2020-04-11T09:43:38.710Z
@JohnHoltz I've just substantially revised my answer.
uhoh
CC BY-SA 4.0
2020-03-31T13:18:43.207Z
2020-04-11T09:43:04.347Z
2020-04-11T09:43:04.347Z
uhoh
<p>Is it even possible to give a definite answer to this question? There are a lot of factors involved in this like Earth's position and motion relative to the planets and of course their own motion and apparent sizes in earth's sky.</p>

<p>Or is it that obvious that Venus transit will be longer because its orbital speed is less than Mercury?</p>

CC BY-SA 4.0
2020-03-30T15:04:22.447Z
Question is ill-defined. What are you keeping the same? Same observation point on Earth, same transit latitude across the Sun?
ProfRob
CC BY-SA 4.0
2020-03-30T16:02:25.370Z
Yes, it is possible to give a definitive answer since all of the factors you mentioned (position, distance, motion) are all known or easily calculated. Depending on where the question is from, it may be sufficient to assume the mean orbit of each planet.
JohnHoltz
CC BY-SA 4.0
2020-03-30T18:04:31.630Z
same path on solar disk means same transit latitude across the Sun.
peakcipher
CC BY-SA 4.0
2020-04-05T03:17:37.663Z
[As seen from Earth, only transits of the inner planets Mercury and Venus are possible. A transit of Mercury can last for up to 8 hours, while a transit of Venus can slightly exceed 8 hours at its maximum.](http://xjubier.free.fr/en/site_pages/transits.html)
Keith McClary
<planetary-transits>
Between Mercury and Venus, which planet produces a longer transit? Assuming both planets describe same path on the solar disk
peakcipher
# Between Mercury and Venus, which planet produces a longer transit? Assuming both planets describe same path on the solar disk

(Asked by: username_0 on 2020-03-30)

Is it even possible to give a definite answer to this question? There are a lot of factors involved in this like Earth's position and motion relative to the planets and of course their own motion and apparent sizes in earth's sky.

Or is it that obvious that Venus transit will be longer because its orbital speed is less than Mercury?

**Tags:** planetary-transits

> **Comments:**
> **username_1**: Question is ill-defined. What are you keeping the same? Same observation point on Earth, same transit latitude across the Sun?  
> **username_2**: Yes, it is possible to give a definitive answer since all of the factors you mentioned (position, distance, motion) are all known or easily calculated. Depending on where the question is from, it may be sufficient to assume the mean orbit of each planet.  
> **username_0**: same path on solar disk means same transit latitude across the Sun.  
> **username_3**: [As seen from Earth, only transits of the inner planets Mercury and Venus are possible. A transit of Mercury can last for up to 8 hours, while a transit of Venus can slightly exceed 8 hours at its maximum.](http://xjubier.free.fr/en/site_pages/transits.html)  

---

## 1 answers

### Answer by username_4 on 2020-03-31. Score: 2

This is a cool question; I'll give it a go.

> Or is it that obvious that Venus transit will be longer because its orbital speed is less than Mercury?

No, nothing here is obvious!

**tl;dr:** Times for the transit of Mercury can range from 5.37 to 8.30 hours, and for Venus they range from 7.94 to 8.12 hours under assumptions of coplanarity but allowing for eccentricity.

Let's assume all orbits are in one plane. We can get the orbital speed of body $i$ with semimajor axis $a$ and current distanc $r$ from the [vis-viva equation](https://en.wikipedia.org/wiki/Vis-viva_equation);

$$v\_i = \sqrt{GM\left( \frac{2}{r\_i} - \frac{1}{\_i} \right)}$$

where $GM$ is the [standard gravitational parameter](https://en.wikipedia.org/wiki/Standard_gravitational_parameter) of the Sun which is about 1.327124E+20 m3/s2.

```
               a          ecc        pei           apo       v_peri   v_apo
Earth    1.49598E+11   0.016709  1.470984E+11  1.520976E+11  30286.6  29291.1
Venus    1.08208E+11   0.006772  1.074752E+11  1.089408E+11  35258.8  34784.4
Mercury  5.79091E+10   0.205630  4.600125E+10  6.981695E+10  58976.4  38858.6

```

Imagine Venus and Mercury crossing a line fixed at one end to the Sun and the other end to the center of the Earth. The linear speed of a planet $i$ relative to that line will be $v\_i$ minus the speed of the line at the distance of planet $i$ or $v\_E r\_i / r\_E$, and the angular velocity seen from Earth will be that divided by ($r\_E - r\_i)$ or

$$\dot{\theta}\_i = \frac{v\_i - (r\_i/r\_E) v\_E}{r\_E - r\_i} = \frac{r\_e v\_i - r\_i v\_E}{r\_E^2 - r\_i r\_E}$$

```
(seconds)     Venus          Mercury
Earth       peri   apo      peri   apo
peri       28570  29242    19335  29884
apo        28060  28626    19384  29646

(hours)       Venus          Mercury
Earth       peri   apo      peri   apo
peri       7.936  8.123    5.371  8.301
apo        7.794  7.952    5.384  8.235

```

> **Comments:**
> **username_2**: Sorry, but I got lost where you converted $v_p-v_E$ to the term with the square root GM etc. Are you sure that conversion is correct?  
> **username_4**: @username_2 yikes! Okay I'll fix that asap!  
> **username_4**: @username_2 I've just substantially revised my answer.  

---

