<p>I think you're not too far from understanding the formula. However, there are a couple details you're missing: </p>

<ol>
<li>From what I understand from the document you linked to, $t_0 = t_{oe}$ is the time at the beginning of the week, or reference epoch.</li>
<li>When computing an angle from a angular velocity, you usually need to compute it with respect to an initial angle and an initial time: when the Earth and the satellite rotate between $t_0$ and $t$, the angle they rotated of is $\dot{\Omega}(t - t_0)$ or $\dot{\Omega}_E(t - t_0)$.<br>
The only exception to this rule seems to be the initial Greenwich longitude with respect to the vernal equinox, which seems to be computed from $t = 0$.</li>
<li>$w_0$, the initial Greenwich longitude wrt the vernal equinox, is $\dot{\Omega}_E t_{oe}$. From this formula, it seems like the document considers an origin of time $t = 0$ when the longitude of Greenwich wrt to the vernal equinox at this time is 0.</li>
</ol>

<p>To summarize: the formula would have been more explicit and comprehensible written as:
$$
\Omega(t) = \Omega_0 - \dot{\Omega}_E t_{oe} + (\dot{\Omega} - \dot{\Omega}_E)(t - t_{oe})
$$
with:  </p>

<ul>
<li>$\Omega_0$ = LAN wrt the vernal equinox at $t_{oe}$</li>
<li>$\dot{\Omega_E}$ = angular velocity of the Earth wrt the vernal
equinox</li>
<li>$-\dot{\Omega}_Et_{oe}$ = Longitude of Greenwich at $t_{oe}$</li>
<li>$\dot{\Omega}$ = angular velocity of the satellite wrt the vernal
equinox</li>
<li>$(\dot{\Omega} - \dot{\Omega}_E)(t - t_{oe})$ = relative angular
displacement between the satellite and Greenwich, between time $t$ and
$t_{oe}$</li>
</ul>

<p>That said, it seems like really bad practice to name $\Omega(t)$ the longitude with respect to Greenwich and $\dot{\Omega}$ the angular velocity with respect to the vernal equinox. This causes quite a bit of confusion, since mathematically speaking, for any linear function $f$, $f(t) = f(t_0) + \dot{f}\times(t - t_0)$</p>

CC BY-SA 3.0
2014-08-12T22:06:44.160Z
2014-08-12T22:06:44.160Z
Etienne Pellegrini
<p>I've been looking into the document <a href="http://www.gps.gov/technical/icwg/IS-GPS-200G.pdf" rel="nofollow">IS-GPS-200H</a> to understand how to calculate satellite location in the ECEF coordinate.</p>

<p>I am having problem understanding the formula to derive $\Omega$, the longitude of the ascending node (LAN) relative to Greenwich at given time $t$:</p>

<p>$$
\Omega = \Omega_0 + \left( \dot{\Omega - w} \right)\times t_k - w \times t_{oe}
$$</p>

<p>where:
$$
\Omega_0: \text{LAN relative to vernal equinox, at the beginning of the week}\\
\dot{\Omega}: \text{angular velocity for LAN, relative to vernal equinox.}\\
w: \text{angular velocity of earth, relative to vernal equinox.}\\
t_k: t - t_{oe}\\
t_{oe}: \text{ephemeris reference epoch}\\
$$
(and let us denote the beginning of the week as $t_0$ for brevity).</p>

<p>But if I try to work out this from scratch:</p>

<ol>
<li>At $t = t_0$, LAN was $\Omega_0$. But since what we really need is the difference of LAN and longitude of Greenwich, we also need to know $w_0$, the initial longitude of Greenwich at $t = t_0$.
$$
\Omega(t = t_0) = \Omega_0 - w_0
$$</li>
<li>At the ephemeris reference epoch time $t = t_{oe}$, LAN and the earth both rotate with their respective angular momentum and hence:
$$
\Omega(t = toe) = \Omega_0 + w_0 + (\dot{\Omega} - w) \times t_{oe}
$$</li>
<li>As time varies from $toe$ to $t$, again LAN and the earth both rotate with their own respective angular momentum and hence
$$
\Omega(t) = \Omega_0 + w_0 + (\dot{\Omega} - w) \times t_{oe} + (\dot{\Omega} - w) \times t_k
$$
which obviously differs from the right formula by $w_0 + \dot{\Omega} \times t_{oe}$.</li>
</ol>

<p>My question is where am I making mistakes/misunderstanding the eqution? 
Explain also why we don't need to know $w_0$ or equivalent input, that would be greatly appreciated.</p>

CC BY-SA 3.0
2014-07-13T04:47:22.183Z
I just edited your questions check out [this guide](http://astronomy.stackexchange.com/editing-help) which explains how you can enter equations for future reference.
harogaston
<orbit><orbital-elements><satellite>
Derivation of the formula for longitude of ascending node for a satellite
GM3D
# Derivation of the formula for longitude of ascending node for a satellite

(Asked by: username_0 on 2014-07-13)

I've been looking into the document [IS-GPS-200H](http://www.gps.gov/technical/icwg/IS-GPS-200G.pdf) to understand how to calculate satellite location in the ECEF coordinate.

I am having problem understanding the formula to derive $\Omega$, the longitude of the ascending node (LAN) relative to Greenwich at given time $t$:

$$
\Omega = \Omega\_0 + \left( \dot{\Omega - w} \right)\times t\_k - w \times t\_{oe}
$$

where:
$$
\Omega\_0: \text{LAN relative to vernal equinox, at the beginning of the week}\\
\dot{\Omega}: \text{angular velocity for LAN, relative to vernal equinox.}\\
w: \text{angular velocity of earth, relative to vernal equinox.}\\
t\_k: t - t\_{oe}\\
t\_{oe}: \text{ephemeris reference epoch}\\
$$
(and let us denote the beginning of the week as $t\_0$ for brevity).

But if I try to work out this from scratch:

1. At $t = t\_0$, LAN was $\Omega\_0$. But since what we really need is the difference of LAN and longitude of Greenwich, we also need to know $w\_0$, the initial longitude of Greenwich at $t = t\_0$.
   $$
   \Omega(t = t\_0) = \Omega\_0 - w\_0
   $$
2. At the ephemeris reference epoch time $t = t\_{oe}$, LAN and the earth both rotate with their respective angular momentum and hence:
   $$
   \Omega(t = toe) = \Omega\_0 + w\_0 + (\dot{\Omega} - w) \times t\_{oe}
   $$
3. As time varies from $toe$ to $t$, again LAN and the earth both rotate with their own respective angular momentum and hence
   $$
   \Omega(t) = \Omega\_0 + w\_0 + (\dot{\Omega} - w) \times t\_{oe} + (\dot{\Omega} - w) \times t\_k
   $$
   which obviously differs from the right formula by $w\_0 + \dot{\Omega} \times t\_{oe}$.

My question is where am I making mistakes/misunderstanding the eqution?
Explain also why we don't need to know $w\_0$ or equivalent input, that would be greatly appreciated.

**Tags:** orbit, orbital-elements, satellite

> **Comments:**
> **username_1**: I just edited your questions check out [this guide](http://astronomy.stackexchange.com/editing-help) which explains how you can enter equations for future reference.  

---

## 1 answers

### Answer by username_2 on 2014-08-12. Score: 1

I think you're not too far from understanding the formula. However, there are a couple details you're missing:

1. From what I understand from the document you linked to, $t\_0 = t\_{oe}$ is the time at the beginning of the week, or reference epoch.
2. When computing an angle from a angular velocity, you usually need to compute it with respect to an initial angle and an initial time: when the Earth and the satellite rotate between $t\_0$ and $t$, the angle they rotated of is $\dot{\Omega}(t - t\_0)$ or $\dot{\Omega}\_E(t - t\_0)$.  
   The only exception to this rule seems to be the initial Greenwich longitude with respect to the vernal equinox, which seems to be computed from $t = 0$.
3. $w\_0$, the initial Greenwich longitude wrt the vernal equinox, is $\dot{\Omega}\_E t\_{oe}$. From this formula, it seems like the document considers an origin of time $t = 0$ when the longitude of Greenwich wrt to the vernal equinox at this time is 0.

To summarize: the formula would have been more explicit and comprehensible written as:
$$
\Omega(t) = \Omega\_0 - \dot{\Omega}\_E t\_{oe} + (\dot{\Omega} - \dot{\Omega}\_E)(t - t\_{oe})
$$
with:

* $\Omega\_0$ = LAN wrt the vernal equinox at $t\_{oe}$
* $\dot{\Omega\_E}$ = angular velocity of the Earth wrt the vernal
  equinox
* $-\dot{\Omega}\_Et\_{oe}$ = Longitude of Greenwich at $t\_{oe}$
* $\dot{\Omega}$ = angular velocity of the satellite wrt the vernal
  equinox
* $(\dot{\Omega} - \dot{\Omega}\_E)(t - t\_{oe})$ = relative angular
  displacement between the satellite and Greenwich, between time $t$ and
  $t\_{oe}$

That said, it seems like really bad practice to name $\Omega(t)$ the longitude with respect to Greenwich and $\dot{\Omega}$ the angular velocity with respect to the vernal equinox. This causes quite a bit of confusion, since mathematically speaking, for any linear function $f$, $f(t) = f(t\_0) + \dot{f}\times(t - t\_0)$


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