# Why 53º orbits for Starlink?

In the UCS satellite dataset, I see a very modest number of non-Starlink LEO satellites around 53º (see graph below).

Conversely, Starlink satellites are regularly deployed around that inclination.

What advantages does 53º inclination provide SpaceX? Why isn't it more more popular?

• Different but related, and the answer is pptentially helpful here SpaceX's 4,425 satellite constellation - what's the method to the madness?
– uhoh
Mar 8, 2021 at 23:59
• Yeah, pretty much placing them to get coverage of the regions where people are most likely to want/need Starlink over ground-based internet options. Mar 9, 2021 at 15:40

Why 53º orbits for Starlink?

tl;dr: they need some cash ASAP, and to cover most paying customers in the medium term, and some polar inclinations will be added later.

Why isn't it more more popular?

Well right now the only example of real wholesale (okay sic, retail) dumping of spacecrafts into LEO for low-latency internet is Starlink. Until now it's been onesies and twosies or tensies and twentysies, but now someone is basically spraying the sky with little shiny satellites in an unprecedented way.

That may change fairly soon though. There are several other companies looking to fill LEO full of sparklers.1

1imagery refers to both what the sky would look like if you could see in the GHz range, and to how observational astronomers everywhere imagine the sky will look like for a few hours after sunset and an few hours before sunrise to all the telescopes photographing the heavens.

And if you are a radio astronomer then both apply.

The motion of a circular orbit of radius $$r$$, period $$T$$ and angular velocity $$\omega = 2 \pi / T$$ around the equator can be written (in a simplified way) as

$$x(t) = r \sin \omega t$$ $$y(t) = r\cos \omega t$$ $$z(t) = 0$$

If it is inclined at an angle $$i$$ it's motion is

$$x(t) = r \cos i \sin \omega t$$ $$y(t) = r \cos \omega t$$ $$z(t) = r \sin i \sin \omega t$$

The corresponding latitude $$\phi$$ of the orbit's ground track is

$$\phi(t) = \arcsin \frac{z(t)}{r} = \arcsin(\sin i \sin \omega t)$$

Solve for $$t$$:

$$t = \frac{1}{\omega} \arcsin \left( \frac{\sin \phi}{\sin i} \right)$$

The amount of time it spends at each latitude is proportional to $$dt/d\phi$$ which is found from

$$\frac{d\phi(t)}{dt} = \frac{1}{\omega}\frac{\sec i \cos \phi}{\sqrt{1 - \sec^2 i \sin^2 \phiϕ}}$$

from Wolfram alpha.

That looks like the first plot and it's not very helpful because it diverges to infinity. The idea is that inclined orbits spend most of their time above latitudes that are near their inclination.

Let's say you can talk with a Starlink satellite as long as it's 30° above the horizon. A 60° degree half-angle cone from a satellite 400 km above the Earth has a radius of 700 km, which is about +/- 6° in latitude.

If we instead just propagate a spot around the orbit, break up latitude into 1° slices and at each time step score all latitude slices that are within +/- 6° of its ground track, we get a nicer distribution as shown in the second plot.

above: Some simple models of coverage from a constellation having 53° planes.

You can also see the idea in these old plots from several years ago based on SpaceX's early concept for what their constellation might look like.

You can also see how the coverage is flat near the equator and becomes denser near the max latitude, but we have to remember that the coverage below is "blurred" by many degrees.

from SpaceX's 4,425 satellite constellation - what's the method to the madness? (click for larger)

They've refined both the design and the way they will roll it out over years by deploying satellites in groups of sixty many times per years for many years, but basically they want to cover paying customers ASAP, and have targeted North America and perhaps Europe as the likely first group of people who will fork out beaucoup money early on.

SpaceX's Starlink is bleeding money and so they need to get some cash flow quickly, and it seems 53° will do that plus long term cover a lot more people. Later they will deploy "filler" planes for the polar regions.

From Elon Musk tweet (found in the linked video below):

SpaceX needs to pass through a deep chasm of negative cash flow over the next year or so to make Starlink financially viable. Every new satellite constellation in history has gone bankrupt. We hope to be the first that does not.

For the cashflow and cost issues as well as the orbits explained, I really like the video Starlink in the Philippines - What you need to know!

import numpy as np
import matplotlib.pyplot as plt

inc = np.radians(53) # better than doing it yourself it seems

phideg = np.linspace(-90, 90, 18001)
top = np.cos(phi)/np.sin(inc)
bot = np.sqrt(1 - (np.sin(phi)/np.sin(inc))**2)

omega = 2 * np.pi
t = np.linspace(0, 1, 10001)[:-1] # don't count the endpoint twice
latitudes = np.linspace(-90, 91, 1801)
phidegs = np.degrees(np.arcsin(np.sin(inc) * np.sin(omega * t)))
scorez = (np.abs(phidegs - latitudes[:, None]) <= 6)
score = scorez.sum(axis=1)

plt.figure()
plt.imshow(scorez)
plt.show()

plt.figure()

plt.subplot(2, 1, 1)

plt.ylim(0, None)
plt.xlim(-80, 80)
plt.gca().set_yticklabels([])
plt.ylabel('dt/d latitude (arb)')

plt.subplot(2, 1, 2)

plt.plot(latitudes, score)
plt.ylim(0, None)
plt.xlim(-80, 80)
plt.gca().set_yticklabels([])
plt.xlabel('latitude (deg)')
plt.ylabel('score per latitude (arb)')

plt.show()