<p>Given one body orbiting one another, the radial distance between the two and the true anomaly are the polar coordinates of the orbiting body. This wouldn't be all that useful if there was no way to predict where the orbiting body will be at some point in the future.</p>

<p>The mean anomaly along with Kepler's laws of motion do just that. Kepler's first law says that the planets orbits about the Sun are ellipses. His second law says that the rate at which a planet sweeps out area (with the Sun as the central point) is constant. His third law says that the period of a planet's orbit depends on the semi-major axis length but not eccentricity.</p>

<p>This means that a planet in an eccentric orbit and a planet in a circular orbit, both with the same semi-major axis length, will, on average, exhibit the same motion. In particular, their periods will be exactly the same. It's easy to predict where the planet in a circular orbit will be at some point in the future because that planet's true anomaly is a linear function of time. Calling this the mean anomaly, this linear relationship means that
<span class="math-container">$$M(t) = M(t_0) + (t-t_0)n$$</span>
where <span class="math-container">$n$</span> is the <a href="https://en.wikipedia.org/wiki/Mean_motion" rel="noreferrer">mean motion</a>:
<span class="math-container">$$n = \frac{2\pi}{T} = \sqrt{\frac{\mu}{a^3}}$$</span>
Here, <span class="math-container">$T$</span> is the period, <span class="math-container">$a$</span> is the semi-major axis length, and <span class="math-container">$\mu$</span> is the constant of proportionality in Kepler's third law.</p>

<p>What's needed is a bridge between the mean anomaly and the true anomaly that makes the true anomaly satisfy Kepler's second law. This bridge is the <a href="https://en.wikipedia.org/wiki/Eccentric_anomaly" rel="noreferrer">eccentric anomaly</a>. Mean anomaly and eccentric anomaly are related via <a href="https://en.wikipedia.org/wiki/Kepler%27s_equation" rel="noreferrer">Kepler's equation</a>,
<span class="math-container">$$M = E - e \sin E$$</span>
Here, <span class="math-container">$E$</span> is the eccentric anomaly and <span class="math-container">$e$</span> is the eccentricity. Note that by this equation, it's trivial to calculate the mean anomaly given the eccentric anomaly and the eccentricity. Simply plug the values in the right hand side and evaluate. What's wanted is a means to calculate the eccentric anomaly given the mean anomaly and the eccentricity. Unfortunately, there is no close-formed inverse of Kepler's equation in the elementary functions. Fortunately, there are many ways to solve Kepler's equation for the eccentric anomaly. One easy way is to use Newton-Raphson iteration.</p>

<p>What's still needed is a relationship between eccentric anomaly and true anomaly. This relationship can be represented in closed form:
<span class="math-container">$$\tan\frac\theta2 = \sqrt{\frac{1+e}{1-e}}\tan\frac E2$$</span></p>

CC BY-SA 4.0
2019-06-07T10:58:17.557Z
2019-06-07T10:58:17.557Z
David Hammen
<p>Why is a true and average anomaly of the planet important? What useful information do they give us?</p>

<orbit><planet>
Why is true and average anomaly of planet important?
Thom
# Why is true and average anomaly of planet important?

(Asked by: username_0 on 2019-06-07)

Why is a true and average anomaly of the planet important? What useful information do they give us?

**Tags:** orbit, planet


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## 1 answers

### Answer by username_1 on 2019-06-07. Score: 6 | Accepted answer

Given one body orbiting one another, the radial distance between the two and the true anomaly are the polar coordinates of the orbiting body. This wouldn't be all that useful if there was no way to predict where the orbiting body will be at some point in the future.

The mean anomaly along with Kepler's laws of motion do just that. Kepler's first law says that the planets orbits about the Sun are ellipses. His second law says that the rate at which a planet sweeps out area (with the Sun as the central point) is constant. His third law says that the period of a planet's orbit depends on the semi-major axis length but not eccentricity.

This means that a planet in an eccentric orbit and a planet in a circular orbit, both with the same semi-major axis length, will, on average, exhibit the same motion. In particular, their periods will be exactly the same. It's easy to predict where the planet in a circular orbit will be at some point in the future because that planet's true anomaly is a linear function of time. Calling this the mean anomaly, this linear relationship means that
$$M(t) = M(t\_0) + (t-t\_0)n$$
where $n$ is the [mean motion](https://en.wikipedia.org/wiki/Mean_motion):
$$n = \frac{2\pi}{T} = \sqrt{\frac{\mu}{a^3}}$$
Here, $T$ is the period, $a$ is the semi-major axis length, and $\mu$ is the constant of proportionality in Kepler's third law.

What's needed is a bridge between the mean anomaly and the true anomaly that makes the true anomaly satisfy Kepler's second law. This bridge is the [eccentric anomaly](https://en.wikipedia.org/wiki/Eccentric_anomaly). Mean anomaly and eccentric anomaly are related via [Kepler's equation](https://en.wikipedia.org/wiki/Kepler%27s_equation),
$$M = E - e \sin E$$
Here, $E$ is the eccentric anomaly and $e$ is the eccentricity. Note that by this equation, it's trivial to calculate the mean anomaly given the eccentric anomaly and the eccentricity. Simply plug the values in the right hand side and evaluate. What's wanted is a means to calculate the eccentric anomaly given the mean anomaly and the eccentricity. Unfortunately, there is no close-formed inverse of Kepler's equation in the elementary functions. Fortunately, there are many ways to solve Kepler's equation for the eccentric anomaly. One easy way is to use Newton-Raphson iteration.

What's still needed is a relationship between eccentric anomaly and true anomaly. This relationship can be represented in closed form:
$$\tan\frac\theta2 = \sqrt{\frac{1+e}{1-e}}\tan\frac E2$$


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