<p>Let's assume that Earth and Venus have circular, coplanar orbits.
The spacecraft advances 180&deg; in heliocentric ecliptic longitude in 143 days; meanwhile, Venus advances 241&deg; and Earth advances 141&deg;.
When the spacecraft departs from Earth (left), Venus should be 61&deg; behind Earth as seen from the Sun.
When the spacecraft arrives at Venus (right), Venus should be 39&deg; ahead of Earth.</p>

<p><img src="https://i.sstatic.net/qjO5h.png" alt="Earth-Venus departure and arrival"></p>

<p>The Sun-Earth-Venus angle marked in green in your second figure is Venus's elongation, 43&deg; east of the Sun as seen from Earth at departure time.</p>

<p>In another context, "phase angle" could refer to the Sun-Venus-Earth angle.
At departure time, an Earth-based observer would see Venus in a 76&deg; phase angle or 62% illuminated.</p>

CC BY-SA 4.0
2020-03-24T21:35:47.117Z
2020-03-24T21:35:47.117Z
Mike G
<p>This is Question 23 from USAAAO First Round 2015. The answer is D. Can someone show how to get the answer?</p>
<blockquote>
<p>Calculate the planetary phase angle (counterclockwise from Earth, a =
1.0 AU) that a probe may correctly complete a Hohmann transfer orbit to Venus (a = 0.7 AU)</p>
<ul>
<li>a. 141 degrees</li>
<li>b. 17.5 degrees</li>
<li>c. 121 degrees</li>
<li>d. 241 degrees</li>
<li>e. 343 degrees</li>
</ul>
</blockquote>
<p>What I tried:</p>
<ul>
<li>Semimajor axis of Hohmann orbit: 0.85 AU</li>
<li>Period of Hohmann orbit: 0.7836 yr</li>
<li>Time of flight: 0.3918 yr</li>
<li>Period of Venus: 0.5856 yr</li>
<li>Angular distance travelled by Venus during flight: 241°</li>
</ul>
<p>The last one looks suspiciously similar to the answer, but using that angular distance the phase angle should be 61°. Feel free to correct me anywhere that I’m wrong</p>
<p><img src="https://i.sstatic.net/EO7s7m.jpg" alt="Phase angle=61°" /><img src="https://i.sstatic.net/4A1tlm.jpg" alt="Phase angle=48.99°" /></p>

CC BY-SA 4.0
2020-03-23T20:26:18.003Z
I'm not sure what you are asking.  It seems to be "Here is a test question. The answer is D. I did some working and I got D.  Please tell me what I did right." That is sort of an odd question to ask.  Could you clarify.
James K
CC BY-SA 4.0
2020-03-23T20:29:15.967Z
I think the question here is why is the angular distance Venus travelled in that time the answer
Tosic
CC BY-SA 4.0
2020-03-24T00:25:36.877Z
James K,  thanks for replying. I have added a diagram to show what I thought of the question.
Chooi Je Qin
CC BY-SA 4.0
2020-03-24T00:42:38.570Z
In the first diagram, the phase angle is 61°. In the second, it is 49°
Chooi Je Qin
CC BY-SA 4.0
2020-03-24T06:16:29.067Z
Both of your diagrams appear to be incorrect. There's no need for the fourth unlabeled object, and the Earth-Sun-Venus path should not be a straight line.
David Hammen
CC BY-SA 4.0
2020-03-24T08:01:28.017Z
@DavidHammen Thanks for replying. It would be great if you can show us the correct diagram.
Chooi Je Qin
CC BY-SA 4.0
2020-03-24T17:47:07.603Z
I edited the images just for readability (I couldn't read them easily as they were so huge) - if you think it causes issues please revert.
Rory Alsop
<astrophysics><orbital-mechanics><artificial-satellite>
Planetary phase angle of Hohmann transfer
Chooi Je Qin
# Planetary phase angle of Hohmann transfer

(Asked by: username_0 on 2020-03-23)

This is Question 23 from USAAAO First Round 2015. The answer is D. Can someone show how to get the answer?

> Calculate the planetary phase angle (counterclockwise from Earth, a =
> 1.0 AU) that a probe may correctly complete a Hohmann transfer orbit to Venus (a = 0.7 AU)
>
> * a. 141 degrees
> * b. 17.5 degrees
> * c. 121 degrees
> * d. 241 degrees
> * e. 343 degrees

What I tried:

* Semimajor axis of Hohmann orbit: 0.85 AU
* Period of Hohmann orbit: 0.7836 yr
* Time of flight: 0.3918 yr
* Period of Venus: 0.5856 yr
* Angular distance travelled by Venus during flight: 241°

The last one looks suspiciously similar to the answer, but using that angular distance the phase angle should be 61°. Feel free to correct me anywhere that I’m wrong

![Phase angle=61°](https://i.sstatic.net/EO7s7m.jpg)![Phase angle=48.99°](https://i.sstatic.net/4A1tlm.jpg)

**Tags:** astrophysics, orbital-mechanics, artificial-satellite
> **Comments:**
> **username_1**: I'm not sure what you are asking.  It seems to be "Here is a test question. The answer is D. I did some working and I got D.  Please tell me what I did right." That is sort of an odd question to ask.  Could you clarify.  
> **username_2**: I think the question here is why is the angular distance Venus travelled in that time the answer  
> **username_0**: James K,  thanks for replying. I have added a diagram to show what I thought of the question.  
> **username_0**: In the first diagram, the phase angle is 61°. In the second, it is 49°  
> **username_3**: Both of your diagrams appear to be incorrect. There's no need for the fourth unlabeled object, and the Earth-Sun-Venus path should not be a straight line.  
> **username_0**: @DavidHammen Thanks for replying. It would be great if you can show us the correct diagram.  
> **username_4**: I edited the images just for readability (I couldn't read them easily as they were so huge) - if you think it causes issues please revert.  

---

## 1 answers

### Answer by username_5 on 2020-03-24. Score: 3

Let's assume that Earth and Venus have circular, coplanar orbits.
The spacecraft advances 180° in heliocentric ecliptic longitude in 143 days; meanwhile, Venus advances 241° and Earth advances 141°.
When the spacecraft departs from Earth (left), Venus should be 61° behind Earth as seen from the Sun.
When the spacecraft arrives at Venus (right), Venus should be 39° ahead of Earth.

![Earth-Venus departure and arrival](https://i.sstatic.net/qjO5h.png)

The Sun-Earth-Venus angle marked in green in your second figure is Venus's elongation, 43° east of the Sun as seen from Earth at departure time.

In another context, "phase angle" could refer to the Sun-Venus-Earth angle.
At departure time, an Earth-based observer would see Venus in a 76° phase angle or 62% illuminated.


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