<p>Supposing that the mass of the object is negligible compared with the mass of the Earth, you can derive the orbital period $T$ from the 3rd Keplero's law:</p>

<p>$\frac{T^2}{a^3} = \frac{4\pi^2}{G(m_E + m_b)} \approx \frac{4\pi^2}{Gm_E},$</p>

<p>where $a$ is the semi-major. With $T$, for each time istant you also know the mean anomaly $M$, given by (suppose $t = 0$ at perigee):</p>

<p>$M(t) = \frac{2\pi}{T}t$.</p>

<p>Solving numerically the Keplero's equation for the eccentric anomaly $E$ (where $e$ is the eccentricity)</p>

<p>$M = E - e\sin E$</p>

<p>and then use the following equation to derive the true anonaly $\nu$, which is the angle between the direction of periapsis and the current position of the body, as seen from the Earth:</p>

<p>$ \cos \nu = \frac{\cos E - e}{1 - e\cos E}$
and
$ \sin \nu = \frac{\sqrt{1-e^2}\sin E}{1 - e\cos E}$.</p>

<p>The distance from the Earth is just given by the orbit equation</p>

<p>$ r = \frac{a(1-e^2)}{1 + e\cos \nu}$.</p>

<p>If I'm not wrong with calculation, it should be:</p>

<p>$ T = 9364 s = 2.6 h.$</p>

<p>$ M(90min) = 207.60°$</p>

<p>$ E(90min) = 203.11°$</p>

<p>$ \nu(90min) = 198.95°$</p>

<p>$ r = 11'366 km$</p>

<p>To get the covered distance you should calculate the line integral of the orbit equation.</p>

CC BY-SA 3.0
2016-02-11T15:32:42.540Z
2016-02-11T15:32:42.540Z
Dario Panarello
<p>I am trying to obtain the distance covered by an object in orbit around the earth within a specified amount of time past its perigee passage. The object is in an elliptical orbit with eccentricity 0.2 and has a semi-major axis of 9600km. Where would I look to find the objects position 90 mins after it passes its perigee? Thanks for any advice</p>

<orbit><astrophysics><fundamental-astronomy><orbital-elements><stellar-astrophysics>
Determining distance from semi-major axis and eccentricity
Carl Jonson
# Determining distance from semi-major axis and eccentricity

(Asked by: username_0 on 2016-02-09)

I am trying to obtain the distance covered by an object in orbit around the earth within a specified amount of time past its perigee passage. The object is in an elliptical orbit with eccentricity 0.2 and has a semi-major axis of 9600km. Where would I look to find the objects position 90 mins after it passes its perigee? Thanks for any advice

**Tags:** orbit, astrophysics, fundamental-astronomy, orbital-elements, stellar-astrophysics


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## 1 answers

### Answer by username_1 on 2016-02-11. Score: 1

Supposing that the mass of the object is negligible compared with the mass of the Earth, you can derive the orbital period $T$ from the 3rd Keplero's law:

$\frac{T^2}{a^3} = \frac{4\pi^2}{G(m\_E + m\_b)} \approx \frac{4\pi^2}{Gm\_E},$

where $a$ is the semi-major. With $T$, for each time istant you also know the mean anomaly $M$, given by (suppose $t = 0$ at perigee):

$M(t) = \frac{2\pi}{T}t$.

Solving numerically the Keplero's equation for the eccentric anomaly $E$ (where $e$ is the eccentricity)

$M = E - e\sin E$

and then use the following equation to derive the true anonaly $\nu$, which is the angle between the direction of periapsis and the current position of the body, as seen from the Earth:

$ \cos \nu = \frac{\cos E - e}{1 - e\cos E}$
and
$ \sin \nu = \frac{\sqrt{1-e^2}\sin E}{1 - e\cos E}$.

The distance from the Earth is just given by the orbit equation

$ r = \frac{a(1-e^2)}{1 + e\cos \nu}$.

If I'm not wrong with calculation, it should be:

$ T = 9364 s = 2.6 h.$

$ M(90min) = 207.60°$

$ E(90min) = 203.11°$

$ \nu(90min) = 198.95°$

$ r = 11'366 km$

To get the covered distance you should calculate the line integral of the orbit equation.


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