<p>Your picture isn't quite right.  The spacecraft trailing Earth by 20 degrees a year is in a wider orbit.  The spacecraft leading Earth is in a smaller orbit.  Here is a <strong>Not To Scale</strong> diagram: <a href="https://i.sstatic.net/rtY5t.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/rtY5t.png" alt="enter image description here" /></a></p>
<p>Kepler's third law states that <span class="math-container">$a^3/T^2$</span> is constant, where <span class="math-container">$a$</span> is the semi-major axis, and T is the orbital period.  For Earth, <span class="math-container">$a=1$</span> AU and <span class="math-container">$T=1$</span> year.  For the leading spacecraft, the orbital period <span class="math-container">$T = 360/380 = 18/19$</span> years.  Use this orbital period along with Kepler's third law to calculate the Semi-major axis of the leading spacecraft's orbit.  Since the orbit is assumed circular, this is the distance to the Sun.</p>
<p>Calculate the semi-major axis of the trailing spacecraft with a similar calculation.  Their different distances to the Sun will then allow you to calculate the time delay necessary to apply to composing &quot;a picture&quot;.</p>

CC BY-SA 4.0
2021-10-29T19:36:05.903Z
2021-10-29T19:36:05.903Z
Connor Garcia
<p>I'm trying to solve this task:</p>
<blockquote>
<p>Astronomers recently managed to get an image of the entire surface of the Sun for the first time. This became possible thanks to the STEREO project.</p>
</blockquote>
<blockquote>
<p>In 2006, two identical spacecraft were launched from the Earth - probes STEREO A and STEREO B. Both probes made a gravitational maneuver in the gravitational field of the Moon, after which one of them began to move in an orbit around the Sun, located a little closer to Earth, the other in orbit, located a little further. One spacecraft began to overtake the Earth in its movement around the Sun at a speed of 20 ° per year, while the other began to lag behind it by 20 ° per year. And on February 6, 2011, the spacecraft found themselves in diametrically opposite points of their orbits, which made it possible to obtain an image of the entire surface of the Sun in real time.</p>
</blockquote>
<blockquote>
<p>Calculate what time delay astronomers on Earth should have taken into account at this moment when combining data from two spacecraft into one &quot;picture&quot;? Consider the orbit of the Earth circular, the orbits of the probes - circular and lying in the plane of the ecliptic.</p>
</blockquote>
<p>I assumed that the picture should be something like that:</p>
<p><a href="https://i.sstatic.net/Ocdoz.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Ocdoz.jpg" alt="enter image description here" /></a></p>
<p>Here S - Sun, A - stereo A, B - stereo B, E - Earth. If they move the same distance from the Earth then triangle ABE is isosceles (AE=AB). Also SA=SB=SE=1au. That means AE=BE=√2. In kilometers that would be √2 * 15 * 10^7. After that we can divide this value by the speed of light and get the time for the signal to reach Earth. When I calculated it I got 707 seconds or 11,8 minutes.</p>
<p>Are these idea and answer correct?</p>

CC BY-SA 4.0
2021-10-29T18:21:20.190Z
Well, but in the task it is said that one spacecraft was a little closer to the Sun than Earth and another - a little more far. Also, it is said that they were in the end located in two diametrically opposite points of their orbits, but if their speed differs from Earth's by 20 degrees, in 5 years they should have be 100 degrees away from Earth each. But this doesn't make them be diametrically opposite... I'm a little confused with this and couldn't find any other way to do some calculations. Maybe this is what causes the delay?
ALiCe P.
<the-sun><earth><light><artificial-satellite>
Calculation of time delay in NASA's STEREO project
ALiCe P.
# Calculation of time delay in NASA's STEREO project

(Asked by: username_0 on 2021-10-29)

I'm trying to solve this task:

> Astronomers recently managed to get an image of the entire surface of the Sun for the first time. This became possible thanks to the STEREO project.

> In 2006, two identical spacecraft were launched from the Earth - probes STEREO A and STEREO B. Both probes made a gravitational maneuver in the gravitational field of the Moon, after which one of them began to move in an orbit around the Sun, located a little closer to Earth, the other in orbit, located a little further. One spacecraft began to overtake the Earth in its movement around the Sun at a speed of 20 ° per year, while the other began to lag behind it by 20 ° per year. And on February 6, 2011, the spacecraft found themselves in diametrically opposite points of their orbits, which made it possible to obtain an image of the entire surface of the Sun in real time.

> Calculate what time delay astronomers on Earth should have taken into account at this moment when combining data from two spacecraft into one "picture"? Consider the orbit of the Earth circular, the orbits of the probes - circular and lying in the plane of the ecliptic.

I assumed that the picture should be something like that:

[![enter image description here](https://i.sstatic.net/Ocdoz.jpg)](https://i.sstatic.net/Ocdoz.jpg)

Here S - Sun, A - stereo A, B - stereo B, E - Earth. If they move the same distance from the Earth then triangle ABE is isosceles (AE=AB). Also SA=SB=SE=1au. That means AE=BE=√2. In kilometers that would be √2 \* 15 \* 10^7. After that we can divide this value by the speed of light and get the time for the signal to reach Earth. When I calculated it I got 707 seconds or 11,8 minutes.

Are these idea and answer correct?

**Tags:** the-sun, earth, light, artificial-satellite

> **Comments:**
> **username_0**: Well, but in the task it is said that one spacecraft was a little closer to the Sun than Earth and another - a little more far. Also, it is said that they were in the end located in two diametrically opposite points of their orbits, but if their speed differs from Earth's by 20 degrees, in 5 years they should have be 100 degrees away from Earth each. But this doesn't make them be diametrically opposite... I'm a little confused with this and couldn't find any other way to do some calculations. Maybe this is what causes the delay?  

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## 1 answers

### Answer by username_1 on 2021-10-29. Score: 1 | Accepted answer

Your picture isn't quite right. The spacecraft trailing Earth by 20 degrees a year is in a wider orbit. The spacecraft leading Earth is in a smaller orbit. Here is a **Not To Scale** diagram: [![enter image description here](https://i.sstatic.net/rtY5t.png)](https://i.sstatic.net/rtY5t.png)

Kepler's third law states that $a^3/T^2$ is constant, where $a$ is the semi-major axis, and T is the orbital period. For Earth, $a=1$ AU and $T=1$ year. For the leading spacecraft, the orbital period $T = 360/380 = 18/19$ years. Use this orbital period along with Kepler's third law to calculate the Semi-major axis of the leading spacecraft's orbit. Since the orbit is assumed circular, this is the distance to the Sun.

Calculate the semi-major axis of the trailing spacecraft with a similar calculation. Their different distances to the Sun will then allow you to calculate the time delay necessary to apply to composing "a picture".


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