BASIC ORBIT MECHANICS

Example Mission Requirements: Spatial and Temporal Scales of Hydrologic Processes Infiltration Percolation Evapotranspiration Lateral Redistribution Runoff Precipitation Intensity Streamflow Day Week Month Year EE/Ae 157a

BASIC ORBITS Circular Orbits Elliptical Orbits: Used most often for earth orbiting remote sensing satellites Nadir trace resembles a sinusoid on planet surface for general case Elliptical Orbits: Used most often for planetary remote sensing Can also be used to increase observation time of certain region on Earth Can provide good balance between global coarse resolution and local high resolution observations EE/Ae 157a

Circular Orbits 𝑅 𝑅+ℎ EE/Ae 157a

CIRCULAR ORBITS Circular orbits balance the inward gravitational force 𝐹 𝑔 and the outward centrifugal force 𝐹 𝑐 : 𝐹 𝑔 =𝑚 𝑔 𝑠 𝑅 𝑅+ℎ 2 𝐹 𝑐 = 𝑚 𝑣 2 𝑅+ℎ When these forces are balanced, the linear velocity of the spacecraft is derived to be 𝐹 𝑔 = 𝐹 𝑐 →𝑣= 𝑔 𝑠 𝑅 2 𝑅+ℎ EE/Ae 157a

Circular Orbits The orbit period is the time it takes the spacecraft to complete one orbit. This is found to be 𝑃 𝑜 = 2𝜋 𝑅+ℎ 𝑣 =2𝜋 𝑅+ℎ 3 𝑔 𝑠 𝑅 2 The point directly under the spacecraft (called the nadir point) also completes one revolution on the surface of the planet during this same time. The velocity at which the nadir point travels along the surface is 𝑣 𝑛 = 2𝜋𝑅 𝑃 𝑜 =𝑣 𝑅 𝑅+ℎ EE/Ae 157a

Planetary Parameters EE/Ae 157a Orbit Altitude = 0.2R   Orbit Altitude = 0.2R Orbit Altitude = 0.5R Planet Radius R (km) Surface Gravity 𝑔 𝑠 (m/sec2) 𝑣 (km/sec) 𝑇 (min) Mercury 2440 3.63 2.7 113 2.4 158 Venus 6050 8.83 6.7 114 5.9 159 Earth 6380 9.81 7.2 111 6.4 155 Moon 1710 1.68 1.55 139 1.38 194 Mars 3395 3.92 3.3 128 3.0 179 Jupiter 71,500 25.9 39.3 229 35.0 320 Saturn 60,000 11.38 23.8 316 21.2 442 EE/Ae 157a

Orbital Velocities EE/Ae 157a

Orbital Periods EE/Ae 157a

ORBIT INCLINATION I EQUATORIAL PLANE ORBITAL EARTH EE/Ae 157a

What is a Day? Because the earth orbits the sun, ORBITAL PLANE SUN Because the earth orbits the sun, it has to move slightly more than one rotation around its axis to have the same point line up with the sun-earth line. This is called a solar day, which is 86400 seconds long. A sidereal day is defined as the time it takes the earth to complete one revolution around its own axis as measured by observing a distant star. It is ~86164 seconds long. EE/Ae 157a

How to calculate Satellite tracks on a planet The satellite moves at a constant velocity in an orbital plane. We shall assume that this plane makes an angle 𝑖 with respect to the equatorial plane. The earth rotates around its polar axis while the satellite moves in its orbit. Relative to an observer on the earth, however, it appears as if the satellite orbit plane is rotating in the opposite direction, but at the same rate that the earth is rotating. Because the earth is not perfectly spherical, i.e. it is flatter at the poles, the orbit will precess around the earth as time elapses. The rate at which the orbit precesses is given by Ω . This can be visualized as our orbital disk rotating slowly around the polar axis of the earth as time goes by. EE/Ae 157a

ORBITAL NODE LONGITUDE The rate of change of the nodal longitude is approximated by: W EARTH VERNAL EQUINOX ORBITAL PLANE SUN 𝑑Ω 𝑑𝑡 =− 3 2 𝐽 2 𝑅 3 𝑔 𝑠 cos 𝐼 𝑅+ℎ 7 2 𝐽 2 =0.00108 EE/Ae 157a

SATELLITE ORBIT PRECESSION EE/Ae 157a

How to calculate Satellite tracks on a planet - 1 For a circular equatorial orbit, we can write the satellite nadir point position as 𝑆 = 𝑟 cos 𝜗 𝑡 𝑟 sin 𝜗 𝑡 0 Here 𝜗 𝑡 is the angle of rotation within the orbital plane measured from some reference starting point. For an orbit inclined by an inclination 𝑖 relative to the equator, the satellite nadir point position is 𝑆 𝑒 = 1 0 0 0 cos 𝑖 − sin 𝑖 0 sin 𝑖 cos 𝑖 𝑟 cos 𝜗 𝑡 𝑟 sin 𝜗 𝑡 0 =𝑟 cos 𝜗 𝑡 cos 𝑖 sin 𝜗 𝑡 sin 𝑖 sin 𝜗 𝑡 =𝑟 cos 𝛾 𝑡 cos 𝛿 𝑡 sin 𝛾 𝑡 cos 𝛿 𝑡 sin 𝛿 𝑡 EE/Ae 157a

How to calculate Satellite tracks on a planet - 2 The latitude and longitude of the nadir point are sin 𝛿 𝑡 = sin 𝑖 sin 𝜗 𝑡 tan 𝛾 𝑡 = cos 𝑖 tan 𝜗 𝑡 Note that the highest latitude that the satellite nadir point will reach is equal to the orbit inclination. These expressions ignore precession of the orbit and rotation of the earth. EE/Ae 157a

How to calculate Satellite tracks on a planet - 3 Let us now assume that the earth is rotating at a rate 𝜔 𝑒 around its axis in a counter-clockwise direction when looking down on the north pole. Then we can model the spacecraft behavior as viewed from the earth by a rotation around the fixed 𝑧 axis of our coordinate system, but rotating from east to west. Therefore, the new version of the satellite position vector becomes 𝑆 𝑠 = cos 𝜔 𝑒 𝑡 sin 𝜔 𝑒 𝑡 0 −sin 𝜔 𝑒 𝑡 cos 𝜔 𝑒 𝑡 0 0 0 1 𝑟 cos 𝜗 𝑡 𝑟 cos 𝑖 sin 𝜗 𝑡 𝑟 sin 𝑖 sin 𝜗 𝑡 =𝑟 cos 𝜔 𝑒 𝑡 cos 𝜗 𝑡 + sin 𝜔 𝑒 𝑡 cos 𝑖 sin 𝜗 𝑡 cos 𝜔 𝑒 𝑡 cos 𝑖 sin 𝜗 𝑡 − sin 𝜔 𝑒 𝑡 cos 𝜗 𝑡 sin 𝑖 sin 𝜗 𝑡 Solving for latitude and longitude, we find sin 𝛿 𝑡 = sin 𝑖 sin 𝜗 𝑡 tan 𝛾 𝑡 + 𝜔 𝑒 𝑡 = cos 𝑖 tan 𝜗 𝑡 EE/Ae 157a

How to calculate Satellite tracks on a planet - 4 To include precession, let us consider just the movements for the earth rotation around its own axis and the precession. The precession direction is defined to be positive when it rotates from west to east, and rotates the orbit plane around the 𝒛−axis of the earth. Therefore, the combination of these two rotations can be written as cos Ω 𝑡 − sin Ω 𝑡 0 sin Ω 𝑡 cos Ω 𝑡 0 0 0 1 cos 𝜔 𝑒 𝑡 sin 𝜔 𝑒 𝑡 0 −sin 𝜔 𝑒 𝑡 cos 𝜔 𝑒 𝑡 0 0 0 1 = cos 𝜔 𝑒 𝑡− Ω 𝑡 sin 𝜔 𝑒 𝑡− Ω 𝑡 0 −sin 𝜔 𝑒 𝑡− Ω 𝑡 cos 𝜔 𝑒 𝑡− Ω 𝑡 0 0 0 1 The resulting expressions for latitude and longitude are sin 𝛿 𝑡 = sin 𝑖 sin 𝜗 𝑡 tan 𝛾 𝑡 + 𝜔 𝑒 𝑡− Ω 𝑡 = cos 𝑖 tan 𝜗 𝑡 EE/Ae 157a

How to calculate Satellite tracks on a planet - 5 At any instant in time, we can now write the expressions for the latitude and longitude of the satellite nadir point as 𝛿 𝑡 = 𝛿 0 + sin −1 sin 𝑖 sin 2𝜋 𝑃 𝑜 𝑡 𝛾 𝑡 = 𝛾 0 + tan −1 cos 𝑖 tan 2𝜋 𝑃 𝑜 𝑡 − 2𝜋 𝑃 𝑒 𝑡+ Ω 𝑡 In these equations we introduced the orbital period of the satellite, 𝑃 𝑜 , and the rotation period of the planet, 𝑃 𝑒 , respectively. Also, 𝛿 0 and 𝛾 0 are the starting latitude and longitude of our orbit tracks. EE/Ae 157a

Repeat Orbits For an orbit to repeat, the satellite ground track has to return to the exact position with the exact same heading as where it started the cycle at the end of the repeat period. This means that the longitude and latitude of the orbit at the end of the period should be the same as the starting values, while the satellite makes an integer number of orbits in its plane, i.e. the repeat orbit has to be an integer multiple of the satellite orbit period. Once we determine the desired repeat period and orbit inclination, we will calculate possible satellite orbit periods with their associated altitudes. EE/Ae 157a

Repeat Orbits- 1 During the repeat period 𝑇 𝑟 , the earth completes 𝑋 revolutions, while the satellite completes 𝑁 orbits. Here 𝑁 has to be an integer to ensure that the starting and ending latitudes are the same. But 𝑋 may not be an integer because of the orbit precession as illustrated below. The earth may have to rotate slightly more than an integer number of times to compensate for the precession. Ω 𝑇 𝑟 Orbit plane at end of period Orbit plane at start of period EE/Ae 157a

Repeat Orbits - 2 Let us assume our starting latitude was zero. After an integer number 𝑁 satellite orbits, the longitude of the nadir point will be 𝛾 𝑇 𝑟 = 𝛾 0 −𝑋2𝜋= 𝛾 0 − 2𝜋 𝑃 𝑒 𝑇 𝑟 + Ω 𝑇 𝑟 We have to return to the starting longitude after 𝑋 rotations of the earth, during which tie our satellite completed exactly 𝑁 orbits. To solve for the repeat period, we therefore have two equations: −𝑋2𝜋= Ω − 2𝜋 𝑃 𝑒 𝑇 𝑟 𝑁 𝑃 𝑜 = 𝑇 𝑟 EE/Ae 157a

Repeat Orbits - 3 Here, 𝑀 is the number of times the earth rotates around its polar axis during the repeat period and 𝑁 is the number of orbits the satellite completes during the repeat period. If we ignore precession, then the repeat period is simply 𝑀 sidereal days. When we take precession into account, we find that 𝑇 𝑟 = 𝑀 𝑃 𝑒 1− Ω 2𝜋 𝑃 𝑒 =𝑁 𝑃 𝑜 We can then calculate the desired orbital period based on how many orbits the satellite has to complete during the repeat period. There may be many options to consider. EE/Ae 157a

Example Let us assume we ignore precession, and choose an orbit that makes 44 orbits in 3 solar days. Then 𝑃 𝑜 = 3 𝑃 𝑠𝑜𝑙𝑎𝑟 44 =5890.9 seconds=98.18 minutes A satellite with this orbital period has to orbit at an altitude of 673.3 km. Let us next assume we want a 60 degree inclination for this orbit. Such an orbit will have a precession rate of approximately – 3.5 degrees per day, or more than 10 degrees in the 3 days we want the orbit to repeat. As a result, this orbit does not repeat in exactly 3 days EE/Ae 157a

Example ignoring precession EE/Ae 157a

𝑀 𝑃 𝑒 1− Ω 2𝜋 𝑃 𝑒 =𝑁 𝑃 𝑜 and inclination = 60 degrees Example - Continued So, what can we do about this? If we really want the repeat period to be 3 solar days, the only option we have is to adjust the inclination. In that case, we have a sun-synchronous orbit, which we will discuss in more detail soon. If we insist to have an inclination of 60 degrees, we need to adjust the orbit altitude so that our orbit repeats after 44 orbits, but now our repeat period may not be 3 solar days. Using the equation on slide 21, we have to find the altitude that satisfies 𝑀 𝑃 𝑒 1− Ω 2𝜋 𝑃 𝑒 =𝑁 𝑃 𝑜 and inclination = 60 degrees Let’s start by realizing that if we pick any altitude, we determine what both the precession rate is (because we fixed the orbit inclination – see page 13) and we fix the orbital period (see slide 6) EE/Ae 157a

Example - Continued Now that the orbital period is determined, the repeat period is also determined because the satellite has to complete 𝑵 orbits during the repeat period. 𝑇 𝑟 =𝑁 𝑃 𝑜 = 𝑀+𝛥 𝑃 𝑒 Here we wrote the number of earth rotations explicitly as an integer plus a fraction of a full revolution. Referring to the figure on slide 21, the orbit will repeat exactly if this fraction of a revolution precisely matches the orbit precession during the repeat period. Δ2𝜋= 𝛺 𝑇 𝑟 We now have to iterate on the orbit altitude until this last equation is true. If the precession is negative, then the delta needs to be relative to the next larger integer EE/Ae 157a

Example - Continued Returning to our original example, we need our orbit to complete 44 times during the repeat period, at an inclination of 60 degrees. For a 60 degree inclination, the precession rate is negative. We therefore expect that we might need just less than 3 sidereal days to complete the 44 orbits. Iterating on the altitude, we find that we need an orbit altitude of 613.9 km, with an orbital period of 96.94 minutes. The precession during the repeat orbit is -10.7 degrees, and the orbit repeat period is 2.97 sidereal days. We find exactly the same result if we start with the equation on slide 21 with 𝑀=3 and the iterate on the altitude until the two halves of that equation match. The repeating tracks are shown on the next slide. EE/Ae 157a

Example with precession- 44 orbit in 3 solar days EE/Ae 157a

Sun Synchronous orbits A sun synchronous orbit passes over any point on the earth at exactly the same mean solar time. This requires the orbit to have a repeat period that is an integer number of solar days. We can accomplish this by picking the precession rate of the orbit to match exactly the rate at which the earth is revolving around the sun. The exact precession rate can be calculated by requiring that 𝑇 𝑟 = 𝑀 𝑃 𝑒 1− Ω 2𝜋 𝑃 𝑒 =𝑀 𝑃 𝑠 Here, 𝑃 𝑠 is the length of the mean solar day, which is equal to 86400 seconds for the earth. The result is Ω =2𝜋 1 𝑃 𝑒 − 1 𝑃 𝑠 EE/Ae 157a

There are many orbit choices The key expression is 𝑇 𝑟 = 𝑀 𝑃 𝑒 1− Ω 2𝜋 𝑃 𝑒 =𝑁 𝑃 𝑜 for exact repeats in an integer number of sidereal days, and 𝑇 𝑟 =𝑀 𝑃 𝑠 =𝑁 𝑃 𝑜 Ω =2𝜋 1 𝑃 𝑒 − 1 𝑃 𝑠 for exact repeats in an integer number of solar days. We might have multiple choices for orbits. Each choice will result in a different coverage pattern during the repeat period. EE/Ae 157a

Can we pic other repeat periods? What if we want a repeat period other than integer numbers of sidereal or solar days? In that case, we need to calculate both the inclination angle and the orbital period. The only thing we can pick is how many orbits the satellite makes during the repeat period, because we always have to complete an integer number of satellite orbits. We then calculate the orbital period from 𝑇 𝑟 =𝑁 𝑃 𝑜 Once we know the orbital period, we can calculate its altitude. We then have to find the precession rate that would give us the exact repeat period from 𝑇 𝑟 =𝑋 𝑃 𝑒 = 𝑀 𝑃 𝑒 1− Ω 2𝜋 𝑃 𝑒 → Ω =2𝜋 1 𝑃 𝑒 − 𝑀 𝑋𝑃 𝑒 Since we already know the orbit altitude from its period, we can use this last expression to calculate the inclination angle. EE/Ae 157a

PERIODIC COVERAGE PATTERNS FOR SUN-SYNCHRONOUS ORBITS EE/Ae 157a

Example: 223 orbits in 16 days EE/Ae 157a

Example: 225 orbits in 16 days EE/Ae 157a

Example: 227 orbits in 16 days EE/Ae 157a

Example: 233 orbits in 16 days (LandSat) EE/Ae 157a

Example: 241 orbits in 16 days EE/Ae 157a

Example: SRTM Orbit EE/Ae 157a

QuikSCAT Orbit: 14 Orbits per day EE/Ae 157a

QUIKSCAT Swaths EE/Ae 157a

Swath Design Orbit tracks are the furthest apart on the equator: EE/Ae 157a

Swath Design - 2 How wide a swath do we need to cover the entire equator? If we have 𝑵 distinct orbits in our repeat period, each swath needs to cover 1/N of the length of the equator to ensure that we get complete coverage of the equator. The fraction of the equator covered by an instrument during a crossing depends on what angle the orbit crosses the equator at. EE/Ae 157a

Swath Design - 3 Taking this geometry into account, we can now state that to a good approximation, if our instrument can cover a swath width of S kilometers, we need a total of 𝑁= 2𝜋𝑅 𝑆 sin 𝐼 orbits to fully cover the equator. The factor 2𝜋𝑅 is the length of the equator. As an example, the LandSAT images are 185 km wide, and the satellites cross the equator at an angle of 98.2 degrees. To cover the equator with a single satellite, we would therefore need 217 equator crossings if we do not allow for any overlap between swaths. To be safe, one would typically use about 10% more swaths to allow for about 10% overlap between swaths. The actual LandSAT orbit makes 233 equator crossings in 16 days. EE/Ae 157a

Special Orbits Geosynchronous orbit has a period equal to the siderial day The satellite orbit period is the same as the rotation period of the planet To an observer on the planet, the satellite traces a figure 8 in the sky. Geostationary orbits are equatorial geosynchronous orbits Sun synchronous orbits provide constant node-to-sun angle The satellite passes over a certain point at exactly the same solar time every day EE/Ae 157a

CIRCULAR GEOSYNCHRONOUS ORBIT TRACE EE/Ae 157a

ELLIPTICAL ORBITS The orbit is defined by: EE/Ae 157a

ELLIPTICAL GEOSYNCHRONOUS ORBIT TRACE EE/Ae 157a

ELLIPTICAL ORBIT GROUND TRACE EE/Ae 157a

ORBIT SELECTION Minimize Earth atmospheric drag --> h > 200 km Global coverage --> polar or near-polar orbit Constant illumination geometry --> sun-synchronous orbit Thermal inertia observations --> day and night pass over same area Minimize radar sensor power --> low altitude Minimize gravity anomalies perturbation --> high altitude Measure gravity anomalies --> low altitude Continuous monitoring --> geostationary or geosynchronous orbit EE/Ae 157a