I've been watching man-made satellites with the naked eye using the "Heavens Above" mobile app. They seem to move incredibly slowly near the horizon, but when they go overhead, they look like they are really cruising. I know they fly about 10x faster than a literal "speeding bullet". Does the slow apparent motion near the horizon mostly mean:

A. They are moving mostly toward or away from me at a near-zero angle of incidence (even at 200-1000 Kilometers overhead)?


B. Their circular orbit keeps their motion mostly perpendicular to me at all times, but their greater distance from me when sighted near the horizon means that the arc they travel is visually much smaller than when they are closest to me (i.e. directly overhead)?

Or something else?

I had an interesting anecdote that relied on the answer being A. But the more I think about it, the more I think the answer is mostly B.

Is there a comprehensible equation for this? If the equation is really hairy, maybe you could include a lay-persons explanation too.


Could I do an experiment where I judge the apparent speed of a satellite that flies nearly parallel to my horizon vs. when it passes directly overhead? I probably need to use the same satellite or pick two that have very similar distances and speeds. Do "arc seconds" have anything to do with this?

  • $\begingroup$ The satellites may be 200 to over 1000 kilometres above the Earth's surface, but maximum (horizon-skimming) line-of-sight distance to a satellite at those altitudes is still much greater than those numbers. $\endgroup$
    – Vikki
    Commented May 10, 2019 at 3:47

6 Answers 6


Here's a geometric construct to back up @uhoh's answer. Start with a satellite in orbit about earth (radius $R$) at height $h$.

enter image description here

The inner circle is the surface, outer is the orbit. Each blue wedge is swept out in equal time by the satellite. Each gold wedge is shows how far you, an observer on the surface, sees it move in that same time. Blowing it up a little:

enter image description here

The satellite on the horizon has a much narrower wedge than the overhead one. This means it is seen to move more slowly. This happens for two reasons: it's further away, and the path isn't perpendicular to your view.

We can make this more exact by calling the (blue) central angle swept in unit time $\Delta \theta$ and (gold) observed angle $\Delta \phi$.

Then overhead:

$\Delta \phi_\rm{overhead} = \frac{(R+h) \Delta \theta}{h}$

On the horizon, you have to take into account both the distance to the orbit, which we'll call $D$, and the relative angle $\theta$:

$\Delta \phi_\rm{horizon} = \frac{(R+h) \Delta \theta \sin{\theta}}{D}$

This could get complicated fast, but note that $\sin{\theta}$ is $D / (R+h)$. Then this simplifies a lot to:

$\Delta \phi_\rm{horizon} = \frac{(R+h) \Delta \theta D / (R+h)}{D} = \Delta \theta $

$ \frac{\Delta \phi_\rm{overhead}}{\Delta \phi_\rm{horizon}} = \frac{R+h}{h}$

So a satellite overhead, not accounting for things like optical illusions or atmospheric refraction, seems to be going a factor $(R+h)/h$ faster than one on the horizon. For a satellite at 600 km, that's a factor of 11; even more if it's a lower orbit.


They are moving mostly toward or away from me at a near-zero angle of incidence (even at 20K Kilometers or 30K Miles overhead)?

Mostly this I believe. But your distance is off. It is very difficult to see high-altitude satellites with the naked eye. Most of the ones you can see are in low-earth orbit between 400 and 1000km altitude.

  • $\begingroup$ The higher the orbit the less the effect (not just because the satellite is moving more slowly) $\endgroup$
    – user20636
    Commented May 4, 2018 at 17:56
  • $\begingroup$ Thanks, yes my distance was off and I have corrected it. Once I did that, I can see that with the earth radius 6,371 km, the satellite 1,000 km above that, that even if I could see the actual horizon, the minimum angle of the satellite would be about 30 degrees. That makes a right triangle with sides roughly 3, 5, 5.8 which would only account for about a 15% speed-up when overhead. $\endgroup$ Commented May 7, 2018 at 14:31

tl;dr: looking at satellites between 300 and 1000 km altitude that happen to pass overhead, they definitely move the fastest when overead and slow way down. At the zenith they move 1.4 and 0.4 degrees per second respectively, and drop a factor of 10 in rate or more as they approach the horizon.

The interesting thing is that the fastest moving thing slows down the quickest, but that's just because it approaches the horizon the quickest.

OK I can't resist leaving a "me too" answer.

The only equation I know is the vis-viva

$$v^2(r) = GM_E\left(\frac{2}{r}-\frac{1}{a}\right)$$

where the Standard Gravitational Parameter $GM_E$ or $\mu$ for Earth is about 3.986E+14 m^3/s^2 (one of the few numbers I know) and $a$ is the semi-major axis.

For a circular orbit $r=a$ and it becomes just:

$$v^2 = GM_E\ / a,$$

and the velocity is just the circumference divided by the period $T$:

$$v = 2 \pi a / T.$$

Square it and set it equal to the previous, and you get:

$$T = 2 \pi \sqrt{a^3 / GM_E},$$

and if you define the rotation angular rate as $\omega = 2 \pi / T$, that becomes

$$ \omega = \sqrt{GM_E/a^3}$$

If I sit on the Earth at $\mathbf{r_{me}} = R \mathbf{\hat{x}}$ and watch a satellite at an altitude $h$ such that it's orbital radius is $R+h$, it's position will be

$$\mathbf{r_{sat}} = (R+h) \left( \mathbf{\hat{x}} \cos(\omega t) + \mathbf{\hat{y}} \sin(\omega t) \right)$$

and the angle between the satellite and the zenith assuming it passes through the zenith will just be

$$\theta = \arctan\left( \frac{y_{me}-y_{sat}}{x_{me}-x_{sat}} \right).$$

I'll switch to Python, most if it is just making the plots:

satellite movement in the sky plot

import numpy as np
import matplotlib.pyplot as plt

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads = 180/pi, pi/180

GMe   = 3.986E+14  # m^3/s^2
R     = 6378. * 1000.   # approx radius of Earth in meters

altitudes  = 1000. * np.arange(300, 1001, 100)  # meters

t  = np.arange(600.)  # 0 to 10 minutes, in seconds

thetas = []
for h in altitudes:

    a = R + h
    omega = np.sqrt(GMe/a**3)
    r_sat = (R + h) * np.array([np.cos(omega*t), np.sin(omega*t)])
    r_me  = R * np.array([1, 0])[:, None] * np.ones_like(t)
    theta = np.arctan2(r_sat[1]-r_me[1], r_sat[0]-r_me[0])
    theta[theta > halfpi] = np.nan

if True:
    fs = 16

    plt.subplot(3, 1, 1)
    for theta in thetas:
        plt.plot(t/60., degs*theta)

    plt.xlabel('minutes', fontsize=fs)
    plt.ylabel('degs from zenith', fontsize=fs)
    plt.text(0.3, 70, '300km')
    plt.text(5.2, 70, '1000km')

    plt.subplot(3, 1, 2)
    for theta in thetas:
        plt.plot(t[1:]/60., degs*(theta[1:] - theta[:-1]))

    plt.xlabel('minutes', fontsize=fs)
    plt.ylabel('degs/sec', fontsize=fs)
    plt.text(0.3, 1.3, '300km')
    plt.text(0.3, 0.2, '1000km')

    plt.subplot(3, 1, 3)
    for theta in thetas:
        plt.plot(degs*theta[1:], degs*(theta[1:] - theta[:-1]))

    plt.xlabel('degs from zenith', fontsize=fs)
    plt.ylabel('degs/sec', fontsize=fs)
    plt.text(30, 1.3,  '300km')
    plt.text(20, 0.16, '1000km')


The satellites you see moving are only at 200km-500km altitude, typically. The slower movement you perceive at the horizon is partly due to the moon illusion described by fred_dot_u and partly due to foreshortening.

Communications satellites at 35000km altitude are geosynchronous; they don't move appreciably relative to an Earth-bound observer.


I'd like to see any relevant equations to back this up, but I suspect that the answer is more one of perception than of mathematics.

I recently viewed a video on YouTube that suggests the interpretation of size and by extrapolation, of movement and speed, is dependent upon other items within view.

The linked video is of the Sydney Opera House, recorded first from directly at the window, with no items in view that are close to the viewer/camera. The famous building appears "normal" sized, as there is no real reference, other than adjacent buildings, roads, etc.

As the camera operator steps away from the window, the frame of the window appears. This provides to the viewer a new reference, which happens to be closer to the opera house than other references. The opera house does not change size in the real world, obviously, but it appears to be much larger in the camera frame.

I've experienced this phenomenon with low altitude moons as well as high altitude moons. Near the horizon, the moon appears large, because there are trees and buildings as a reference, yet above my head, the moon is a tinier circle, subjectively.

I suggest that this phenomenon applies to moving orbital objects. I've observed the International Space Station when it is visible in my area and have noted just as the question is described. Movement seems slow at the first part of the appearance, then it picks up speed and dashes across the sky, slowing near the end as it approaches the opposite horizon.

Once my head is tilted back enough to lose view of the horizon perpendicular to the travel of the ISS, the reference is lost.

For an experiment, using a single satellite, consider to cut a rectangle from a stiff material. Use the frame and a stop watch to time the transit from one point on the frame to the other, when near the low part of the transit, then perform the same test when the satellite is near zenith.

  • 1
    $\begingroup$ The "Moon Illusion" is well known. en.wikipedia.org/wiki/Moon_illusion $\endgroup$ Commented May 4, 2018 at 18:42
  • $\begingroup$ You might note a curiosity regarding the "moon illusion" - it does not occur when viewing it upside down. Its not really reference objects per se, its how we perceive the dome of the sky. $\endgroup$
    – Norm
    Commented May 4, 2018 at 21:55
  • 3
    $\begingroup$ This has very little to do with the Moon effect, which is an illusion. This is a very real phenomenon. $\endgroup$ Commented May 4, 2018 at 22:10
  • $\begingroup$ but the Moon effect will exagerate the perceived slowing, @DavidHammen $\endgroup$
    – user20636
    Commented May 7, 2018 at 1:44

You're right, it's both A and B, the incidence angle and the distance.

Let's start by describing what would happen if you were indeed seeing it traveling at an angle perpendicular to your line of sight.

An object directly above you (90deg from horizon) are a distance away from you equal to its altitude (say, 500km).

But when the same object is near the horizon, let's just say for the sake of argument that it's 10deg above the horizon, then its distance from you is more like six times that - approximately 1/sin(angle above horizon).

That's only an approximation, which doesn't take curvature of the orbit or the horizon curvature into account, but it should be a good enough approximation to illustrate why it appears to move more slowly when it's less elevated in the sky. It appears that way because it's farther away.

In addition to that, orbiting objects overhead certainly would be traveling at an incidence angle near square across your line of sight, and objects near the horizon could very well be traveling in a direction which is a highly acute incidence angle, potentially approaching zero.

  • $\begingroup$ It can't be both A and B because they are mutually exclusive. This answer says it is instead option C, "something else" -- which is the correct answer. $\endgroup$ Commented May 4, 2018 at 23:47
  • 2
    $\begingroup$ They're not mutually exclusive. Something can be both far away AND moving in an oblique direction. $\endgroup$
    – Beanluc
    Commented May 5, 2018 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.