<p>If $r_p$ is the perihelion distance (the closest distance of the planet in its elliptical orbit from the Sun), and if $r_a$ is the aphelion distance (the farthest distance from the Sun), the major axis is<br>
$$
2a = r_a+r_p
$$
where $a$ is the <em>semi-major axis</em> (half of the major axis).
The <em>eccentricity</em> of the orbit can be calculated from the perihelion and aphelion distances using
$$
e = \frac{r_a-r_p}{r_a+r_p}
$$
Once the eccentricity is known, the <em>semi-minor axis</em> can be calculated from
$$
b = a\sqrt{1-e^2}
$$
The minor axis, of course, is twice the semi-minor axis, or $2b$.</p>

CC BY-SA 3.0
2017-05-24T13:47:06.963Z
Just to a source for these equations, check out [this section of the Wiki page on ellipses](https://en.wikipedia.org/wiki/Ellipse#Planetary_orbits).
zephyr
CC BY-SA 3.0
2017-05-24T11:10:54.640Z
2017-05-24T11:10:54.640Z
Christo
<p>I've studied that the major axis of a planets orbit = ( radius of perihelion + radius of aphelion) . But then  how can we calculate the value of the minor axis if we are provided with the values of radius of perihelion and radius of aphelion only ?</p>

CC BY-SA 3.0
2017-05-24T10:17:06.613Z
Look at the ratio of perihelion and aphelion radius. If they are equal, you have a circular orbit, if they are very different, a highly elliptical one -- for the same major axis size. Basic trigonometry should get you the rest of the way.
Alex
<planet><astrophysics>
Property of elliptical orbits
Munj Patel
# Property of elliptical orbits

(Asked by: username_0 on 2017-05-24)

I've studied that the major axis of a planets orbit = ( radius of perihelion + radius of aphelion) . But then how can we calculate the value of the minor axis if we are provided with the values of radius of perihelion and radius of aphelion only ?

**Tags:** planet, astrophysics

> **Comments:**
> **username_1**: Look at the ratio of perihelion and aphelion radius. If they are equal, you have a circular orbit, if they are very different, a highly elliptical one -- for the same major axis size. Basic trigonometry should get you the rest of the way.  

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## 1 answers

### Answer by username_2 on 2017-05-24. Score: 6

If $r\_p$ is the perihelion distance (the closest distance of the planet in its elliptical orbit from the Sun), and if $r\_a$ is the aphelion distance (the farthest distance from the Sun), the major axis is  
$$
2a = r\_a+r\_p
$$
where $a$ is the *semi-major axis* (half of the major axis).
The *eccentricity* of the orbit can be calculated from the perihelion and aphelion distances using
$$
e = \frac{r\_a-r\_p}{r\_a+r\_p}
$$
Once the eccentricity is known, the *semi-minor axis* can be calculated from
$$
b = a\sqrt{1-e^2}
$$
The minor axis, of course, is twice the semi-minor axis, or $2b$.

> **Comments:**
> **username_3**: Just to a source for these equations, check out [this section of the Wiki page on ellipses](https://en.wikipedia.org/wiki/Ellipse#Planetary_orbits).  

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