<p>Let us first consider the Earth a fixed point and Mars moving around the Sun on a circular orbit with angular velocity equal to the relative angular velocity (<span class="math-container">$\omega_\bigoplus - \omega_♂)$</span>
The distance between Mars and the Earth can be described as the square root of <span class="math-container">$R_E^2+R_M^2-2R_ER_Mcos(\theta)$</span>, and when this is equal to <span class="math-container">$R_E^2$</span>: <br> <span class="math-container">$cos(\theta) = \frac{R_E^2-R_E^2+R_M^2}{2R_ER_M} = 0.75$</span>,<br> meaning this happens when the elongation of Mars is around 0.72 and -0.72 radians (40 and -40 degrees). For smaller angles than 40, the cosine is bigger making the distance smaller (same goes for angles between <span class="math-container">$-40$</span> and <span class="math-container">$0$</span>). Therefore, Mars is closer for <span class="math-container">$80/360 = 22$</span> percent of the synodic period of Mars, and for the remaining 78% or around 600 days.</p>

CC BY-SA 4.0
2019-12-06T20:54:17.000Z
2019-12-06T20:54:17.000Z
Tosic
<p>For a diagramatic answer, draw a circle around the Earth of 1AU radius, place the Sun on that circle then draw Mars' orbit. The part of Mars' orbit that is outside the 1AU circle is the part where Mars is farther away than the Sun.
I expect that the two current answers are equivalent.</p>

CC BY-SA 4.0
2019-12-07T08:19:03.910Z
okay, as a diagramatic answer it may be an equivalent way to express the fraction, but without discussing the synodic period it doesn't technically answer "How long..."
uhoh
CC BY-SA 4.0
2019-12-08T01:26:17.613Z
Of course it doesn't, but it's a good visualization tool, useful for sanity checking the analytical approach.
stolenmoment
CC BY-SA 4.0
2019-12-07T01:02:52.693Z
2019-12-07T01:02:52.693Z
stolenmoment
<p>A <a href="https://space.stackexchange.com/q/40252/109">recent question at Space.se</a> has generated a few comments that are interesting.  </p>

<p>But the answer to one particular comment to might address a lot of misconceptions. </p>

<blockquote>
  <p>I imagine the actual halfway point is close to the sun <a href="https://space.stackexchange.com/questions/40252/is-a-midspace-space-station-between-earth-and-mars-practical#comment130615_40252">Source</a></p>
</blockquote>

<p>As the two planets orbit the sun at different speeds, I assume it must be a significant amount.</p>

<p>How long is the Sun closer to Earth then Mars is?</p>

<orbit><solar-system><earth><mars>
How long is the Sun closer to Earth then Mars is?
James Jenkins
# How long is the Sun closer to Earth then Mars is?

(Asked by: username_0 on 2019-12-06)

A [recent question at Space.se](https://space.stackexchange.com/q/40252/109) has generated a few comments that are interesting.

But the answer to one particular comment to might address a lot of misconceptions.

> I imagine the actual halfway point is close to the sun [Source](https://space.stackexchange.com/questions/40252/is-a-midspace-space-station-between-earth-and-mars-practical#comment130615_40252)

As the two planets orbit the sun at different speeds, I assume it must be a significant amount.

How long is the Sun closer to Earth then Mars is?

**Tags:** orbit, solar-system, earth, mars


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## 2 answers

### Answer by username_1 on 2019-12-06. Score: 5

Let us first consider the Earth a fixed point and Mars moving around the Sun on a circular orbit with angular velocity equal to the relative angular velocity ($\omega\_\bigoplus - \omega\_♂)$
The distance between Mars and the Earth can be described as the square root of $R\_E^2+R\_M^2-2R\_ER\_Mcos(\theta)$, and when this is equal to $R\_E^2$:   
 $cos(\theta) = \frac{R\_E^2-R\_E^2+R\_M^2}{2R\_ER\_M} = 0.75$,  
 meaning this happens when the elongation of Mars is around 0.72 and -0.72 radians (40 and -40 degrees). For smaller angles than 40, the cosine is bigger making the distance smaller (same goes for angles between $-40$ and $0$). Therefore, Mars is closer for $80/360 = 22$ percent of the synodic period of Mars, and for the remaining 78% or around 600 days.


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### Answer by username_2 on 2019-12-07. Score: 2

For a diagramatic answer, draw a circle around the Earth of 1AU radius, place the Sun on that circle then draw Mars' orbit. The part of Mars' orbit that is outside the 1AU circle is the part where Mars is farther away than the Sun.
I expect that the two current answers are equivalent.

> **Comments:**
> **username_3**: okay, as a diagramatic answer it may be an equivalent way to express the fraction, but without discussing the synodic period it doesn't technically answer "How long..."  
> **username_2**: Of course it doesn't, but it's a good visualization tool, useful for sanity checking the analytical approach.  

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