How Do You Calculate the Eccentricity of an Elliptical Orbit?

[Dustinsfl]

##\mu_{sun} = 132712000000##
##\mu_{earth} = 398600##
##\mu_{mars} = 42828##
##R_{earth} = 149.6\times 10^6##
##R_{mars} = 227.9\times 10^6##
##r_{earth} = 6378##
##r_{mars} = 3396##

The spacecraft will make 3 rev in 2 Earth years. I found the semi-major axis of the ellipse which is
$$
a = 1.14162979\times 10^8
$$
How can I determine the eccentricity of the ellipse with this information? I have been looking at every formula but can figure it out.

http://img20.imageshack.us/img20/182/orbit2.png

 

[gneill]

Please make use of the posting template when you start a thread. This is required.

--------

If you know the aphelion distance and the length of the major axis then you can determine the perihelion distance. Having both r_a and r_p you can determine eccentricity.

 

[Dustinsfl]

I wasn't thinking thanks.

 

[I like Serena]

From your drawing the aphelion of the spacecraft is the same as the radius of Earth's orbit.
And the sum of aphelion and perihelion is equal to the length of the major axis.

 

[rezeew]

I would first like to commend you on your efforts in finding the semi-major axis of the ellipse. This is an important step in determining the eccentricity of the orbit. To determine the eccentricity, we can use the following formula:

$$
e = \sqrt{1 - \frac{b^2}{a^2}}
$$

where a is the semi-major axis and b is the semi-minor axis of the ellipse. To find the semi-minor axis, we can use the following formula:

$$
b = a\sqrt{1-e^2}
$$

Now, substituting the value of the semi-major axis (a) that you have found, we get:

$$
b = 1.14162979\times 10^8 \times \sqrt{1-e^2}
$$

To solve for e, we need to have the value of b. To find b, we need the value of e. This may seem like a circular problem, but we can use an iterative approach to find the value of e. We can start with an initial guess for e (say 0.5) and use that to find b. Then, we can use the value of b to refine our guess for e and repeat the process until we get a more accurate value for e.

Alternatively, we can also use the vis-viva equation to find the eccentricity of the orbit. This equation relates the semi-major axis (a) and the specific orbital energy (ε) of the orbit:

$$
ε = -G\frac{\mu}{2a}
$$

where G is the gravitational constant and μ is the standard gravitational parameter of the central body.

Using the values given in the question, we can calculate the specific orbital energy for Earth and Mars:

$$
ε_{earth} = -G\frac{\mu_{earth}}{2a_{earth}} = -1.758\times 10^{11} \frac{m^2}{s^2}
$$

$$
ε_{mars} = -G\frac{\mu_{mars}}{2a_{mars}} = -1.870\times 10^{11} \frac{m^2}{s^2}
$$

Now, we can use the following equation to find the eccentricity:

$$
e = \sqrt{1 + \frac{2εh^2}{G^2m^3