EE/Ae 157a
Solution #1
Problem 1.
Calculate the orbit period, altitude and inclination for a sun synchronous orbit with a
repeat period of 3 days. We want the satellite to complete 44 orbits in the 3 day cycle.
How wide should the swath be to cover the globe in the 3 day period? Assume the radius
of the Earth is 6380 km.
Solution:
For the orbit to be sun synchronous, we need the precession rate of the orbit to be one
revolution per year. The orbital period can be calculated form the fact that we want the
orbit to complete 251 revolutions in 18 days. Therefore, the period is
𝑇=
3 ∗ 24 ∗ 60 ∗ 60
= 5890.91 𝑠𝑒𝑐 = 98.18 min
44
Now that we know the period, we can solve for the orbit altitude from
1/3
𝑔𝑇 2 𝑅 2
ℎ={
}
4𝜋 2
−𝑅
Using the period and 𝑅 = 6380 km, we find
ℎ = 674.05 𝑘𝑚
With this orbit altitude, we can now solve for the inclination angle from
𝑑Ω
3
cos 𝐼
= − 𝐽2 𝑅 3 √𝑔𝑠
= 1 rev/year
(𝑅 + ℎ)7⁄2
𝑑𝑡
2
The result is
𝐼 = 98.13°
At the equator, the separation between ground tracks is
𝑆=
2𝜋𝑅
= 911 km
44
Therefore, the required minimum swath width is
𝑆𝑊 = 𝑆 sin(𝐼) = 901.2 km
In practice, we will make the swath slightly larger to accommodate distortions due to
local relief. A typical number is about 10% more than the minimum.
Problem 2.
Calculate the orbit altitude and inclination for a sun synchronous orbit with a period of
100.93 minutes. What is the approximate repeat period for this orbit in days? What
would the separation between adjacent orbits be at the equator? What is the minimum
swath width required to cover the equator?
Solution:
The orbit period is only a function of the altitude. Given the period, we can calculate the
altitude from
 gT 2 R 2 
h
2 
 4 
13
R
Using the given period and 𝑅 = 6380 km, we find
ℎ = 805 𝑘𝑚
The nodal precession is
d
3
cos I
  J 2 R3 g s
dt
2
r7 2
with J2 = 0.00108. Setting this value to one revolution per year, we can solve for the
inclination angle. The result is
𝐼 = 98.64°
To find the approximate repeat period, we need to find the integers M and N for which
𝑀 ∗ 24 ∗ 60 ∗ 60
=1
𝑁 ∗ 100.93 ∗ 60
An orbit that completes 214 revolutions in 15 days will nearly satisfy this ratio. Taking
the precession into account, we find that the repeat orbit is indeed nearly 15 days.
The real way, to find the repeat period is to “propagate” the orbit starting at some place
on the surface of the earth. For every time step, we can move the satellite along its orbit
plane by the velocity times the time step. The nadir point is then projected down on the
earth taking into account both the rotation of the earth and the orbit precession. We
repeat this calculation until we find that the nadir point first returns to a previously
covered point on the earth.
Since we are not propagating the orbit, we shall calculate the separation and swath width
required in the usual way. For the chosen orbit, the separation between ground tracks is
187.3 km, and the minimum swath width required is 185.2 km.
Problem 3.
This one is a bit more challenging. We want to
cover the globe with an imager every three days
with a sun synchronous orbit. The camera images
two strips on either side of the nadir line, each with
a width of S. The offset from the nadir line is O.
What should be the relationship between S and O to
ensure that we can cover the entire equator? Next,
calculate the possible orbit altitudes between 500
km and 900 km. What are the required values of S
and the orbit inclination for each orbit?
Nadir Line
O
S
S
Solution:
Placing the orbits along the equator, we can see that if we satisfy the condition that 𝑆 ≥ 𝑂
we will cover the entire equator.
Let us assume we cover N orbits in the 3 day repeat period 𝑇𝑅 . In that case, the orbital
period will be
𝑇=
𝑇𝑅
𝑁
The orbital altitude for such an orbit is
1/3
𝑔𝑇 2 𝑅 2
𝑅+ℎ ={
}
4𝜋 2
We can solve for N to find
𝑔𝑅 2
𝑁 = 𝑇𝑅 √ 2
4𝜋 (𝑅 + ℎ)3
Remember, N must be an integer. Using the values of 500 km and 900 km, we find that
N can be values between 42 and 45. But 42 and 45 will be single day repeat orbits, so the
only two viable values are 43 and 44. The corresponding altitudes are 783 km and 674
km. The swath widths are 466 km and 455 km, and the inclination angles are 98.55 and
98.12 degrees.