Goal
We want a clear, intuitive answer to: given the true anomaly \(\theta\), how can we tell from the geometry whether a satellite in an elliptical orbit is moving toward perigee or away from it?
We will first give a short, intuitive argument (no heavy algebra), then derive an explicit formula for the angle \(\psi\) between the position vector \(\mathbf{r}\) and velocity vector \(\mathbf{v}\).
Quick, intuitive argument (sign of radial velocity)
Decompose the velocity into radial and transverse components: \[ \mathbf{v} = v_r\,\hat{\mathbf r} + v_\theta\,\hat{\boldsymbol\theta}. \tag{1}\]
The cosine of the angle \(\psi\) between \(\mathbf r\) and \(\mathbf v\) is \[ \cos\psi = \frac{\mathbf r\cdot\mathbf v}{r\,v} = \frac{r v_r}{r v} = \frac{v_r}{v}, \tag{2}\] so the sign of \(v_r\) alone determines whether \(\psi\) is less than or greater than \(90^\circ\).
The radial component of velocity is simply \(\dot{r}\) — the time rate of change of the satellite’s distance from Earth.
When the satellite moves away from perigee (the closest point), its distance from Earth is growing until it reaches a maximum at apogee. In this phase, \(\dot{r} > 0\), so the radial velocity \(v_r\) is positive.
When the satellite moves back toward perigee, its distance is shrinking until it reaches a minimum again at perigee. Here, \(\dot{r} < 0\), so \(v_r\) is negative.
We can now conclude:
If \(v_r > 0\), then \(\cos \psi > 0 \;\Rightarrow\; \psi < 90^\circ\). In this case, the velocity has a growing radial component, meaning the satellite is moving away from perigee toward apogee.
If \(v_r < 0\), then \(\cos \psi < 0 \;\Rightarrow\; \psi > 90^\circ\). Here, the velocity has a shrinking radial component, meaning the satellite is moving back toward perigee from apogee.
Thus, a quick test for “toward/away” is simply the sign of \(v_r\).
Full derivation: express \(\cos\psi\) in terms of \(e\) and \(\theta\)
Below we derive an explicit relation \(\cos\psi=\cos\psi(\theta)\) so a reader can compute \(\psi\) (not only its sign).
Orbit geometry
The orbit in polar coordinates (focus at origin) is \[ r(\theta) = \frac{p}{1 + e\cos\theta}, \qquad p = a(1-e^2), \quad 0<e<1. \tag{3}\]
Radial velocity via the chain rule
Because \(r=r(\theta(t))\), \[ v_r \equiv \dot r = \frac{dr}{dt} = \frac{dr}{d\theta}\,\dot\theta. \tag{4}\]
Differentiate \(r(\theta)\) with respect to \(\theta\): \[ \frac{dr}{d\theta} = \frac{d}{d\theta}\!\left(\frac{p}{1+e\cos\theta}\right) = \frac{p\,e\sin\theta}{(1+e\cos\theta)^2}. \tag{5}\]
Specific angular momentum gives \[ h = r^2\dot\theta = \sqrt{\mu p}, \tag{6}\] so \(\dot\theta = h/r^2\). Using \(r=\dfrac{p}{1+e\cos\theta}\) and substituting into Equation 4: \[ \begin{aligned} v_r &= \frac{p\,e\sin\theta}{(1+e\cos\theta)^2}\cdot\frac{h}{r^2} \\ &= \frac{p\,e\sin\theta}{(1+e\cos\theta)^2}\cdot\frac{h(1+e\cos\theta)^2}{p^2} \\ &= \frac{h}{p}\,e\sin\theta = \sqrt{\frac{\mu}{p}}\,e\sin\theta. \end{aligned} \tag{7}\]
Transverse component and speed
From \(h = r v_\theta\), \[ v_\theta = \frac{h}{r} = \sqrt{\frac{\mu}{p}}\,(1+e\cos\theta). \tag{8}\]
Then the speed is \[ \begin{aligned} v &= \sqrt{v_r^2 + v_\theta^2} \\ &= \sqrt{\frac{\mu}{p}\Big(e^2\sin^2\theta + (1+e\cos\theta)^2\Big)} \\ &= \sqrt{\frac{\mu}{p}}\sqrt{1 + 2e\cos\theta + e^2}. \end{aligned} \tag{9}\]
Final formula for \(\cos\psi\)
Using Equation 2, Equation 7 and Equation 9 we get \[ \boxed{% \cos\psi(\theta) = \frac{e\sin\theta}{\sqrt{\,1 + 2e\cos\theta + e^2\,}}% } \tag{10}\]
Why the denominator is always positive (ellipses)
The term under the square root can be rearranged as \[ 1 + 2e\cos\theta + e^2 = (1+e\cos\theta)^2 + e^2\sin^2\theta, \tag{11}\] which is a sum of two non-negative terms. For \(0<e<1\) there is no \(\theta\) making both terms zero simultaneously (that would require \(\cos\theta=-1/e\), impossible). Hence the denominator in Equation 10 is strictly positive for elliptic orbits.
As a simple check, at \(\theta=\pi\) the expression becomes \((1-e)^2>0\).
Conclusion
We can now conclude the following.
Since the velocity vector in an elliptical orbit always has a tangential component, the angle \(\psi\) between \(\mathbf{r}\) and \(\mathbf{v}\) is restricted to the physically meaningful range
\[ 0^\circ < \psi < 180^\circ. \]
The endpoints are excluded, because \(\psi = 0^\circ\) or \(180^\circ\) would imply purely radial motion, which is impossible in a bound Keplerian orbit.
Within this range:
If \(v_r > 0\), then \(\cos\psi > 0 \;\;\Rightarrow\;\; \psi < 90^\circ\) (velocity has a growing radial component → satellite moving away from perigee).
If \(v_r < 0\), then \(\cos\psi < 0 \;\;\Rightarrow\;\; \psi > 90^\circ\) (velocity has a shrinking radial component → satellite moving toward perigee).
We can express this in terms of dot product of the radius and velocity vectors:
\[ \mathbf{r}\cdot\mathbf{v} = r v \cos \psi \]
Since \(r>0\) and \(v>0\), the sign of \(\mathbf{r}\cdot\mathbf{v}\) is determined entirely by \(\cos\psi\):
If \(\mathbf{r}\cdot\mathbf{v} > 0\), then \(0^\circ < \psi < 90^\circ\) (satellite is moving away from perigee).
If \(\mathbf{r}\cdot\mathbf{v} < 0\), then \(90^\circ < \psi < 180^\circ\) (satellite is moving toward perigee).
AI Attribution: Textbooks often state that the dot product of the radius and velocity vectors indicates the direction of a satellite’s motion, but they rarely explain the reasoning in detail. I explored this idea in depth through a long conversation with AI, making detailed notes along the way. The first draft of this post grew out of those notes, and I refined it further with AI’s help.