Earth Geometry Viewed from Space

The most common problem in space mission geometry is to determine the relative geometry of objects on the Earth's surface as seen from the spacecraft One example is to use the given coordinates of a target on the Earth to determine its coordinates in the spacecraft field of view. Another is to determine the intercept point on the surface of the Earth corresponding to a given direction in spacecraft coordinates.

To begin, we determine p, the angular radius of the spherical Earth as seen from the spacecraft, and A0, the angular radius measured at the center of the Earth of the region seen by the spacecraft (see Fig. 5-11). Because we have assumed a spherical Earth, the line from the spacecraft to the Earth's horizon is perpendicular to the Earth's radius, and therefore sin p = cos A0 = ^ (5-16)

Re+H

where RE is the radius of the Earth and if is the altitude of the satellite.

Fig. 5-11. Relationship Between Geometry as Viewed from the Spacecraft and from the Center of the Earth. See also Fig. 5-12.

Thus, the Earth forms a small circle of radius p on the spacecraft sky, and the spacecraft sees the area within a small circle of radius A0 on the surface of the Earth. The distance, Dmax, to the horizon is given by (see Fig. 5-13 below):

The spherical-Earth approximation is adequate for most mission geometry applications. However, for precise work, we must apply a correction for oblateness, as explained in detail by Liu [1978] or Collins [1992]. The Earth's oblateness has two distinct effects on the shape of the Earth as seen from space. First, the Earth appears somewhat oblate rather than round, and second, the center of the visible oblate Earth is displaced from the true geometric center of the Earth. For all remaining computations in this section, we will use spherical coordinates both on the Earth and in the spacecraft frame. Computationally, we can treat oblateness and surface irregularities as simply the target's altitude above or below a purely spherical Earth. That the Earth's real surface is both irregular and oblate is immaterial to the computation, and, therefore, the results are exact

We wish to find the angular relationships between a target, P, on the surface of the Earth, and a spacecraft with subsatellite point, SSP, also on the surface of the Earth, as shown in Fig. 5-12. We assume that the subsatellite point's latitude, Latssp and longitude, Longssp, are known. Depending on the application, we wish to solve one of two problems: (1) given the coordinates of a target on the Earth, find its coordinates viewed by the spacecraft, or (2) given the coordinates of a direction relative to the spacecraft, find the coordinates of the intercept on the surface of the Earth. In both cases, we determine the relative angles between SSP and P on the Earth's surface and then transform these angles into spacecraft coordinates.

Given the coordinates of the subsatellite point (Longggp, LatSSP) and target (Longp, Latp), arid defining AL = | LongSSp - Longp |, we wish to find the azimuth, <p£, measured eastward from north, and angular distance, A, from the subsatellite point to the target (See Fig. 5-12.) These are given by cos A = sin Latssp s'n Latp + cos Latssp cos Latp cos AL (A < 180 deg) (5-19)

cos 0E = (sin Latp - cos A sin Latssp )/(sin A cos LatSsp ) (5-20)

where <P£ < 180 deg if P is east of SSP and 0E> 180 deg if P is west of SSP.

Fig. 5-12. Relationship Between Target and Subsatellite Point on the Earth's Surface.

Alternatively, given the position of the subsatellite point (Longssp, Latssp) and the position of the target relative to this point (&£, A), we want to determine the geographic coordinates of the target {Longp, Latp):

cos Latp = cos A sin LatSSP + sin A cos Latssp cos <PE (Latp <180 deg) (5-21)

cos AL = (cos A - sin Latssp sin Latp)! (cos Lot ssp cos Latp) (5-22)

where Latp = 90 deg - Latp and P is east of SSP if <Z>£ < 180 deg and west of SSP if &E > 180 deg.

We now wish to transform the coordinates on the Earth's surface to coordinates as seen from the spacecraft By symmetry, the azimuth of the target relative to north is the same as viewed from either the spacecraft or the Earth. That is,

True Outer Horizon

True Outer Horizon

SubsatelBte Point, SSP

Fig. 5-13. Definition of Angular Relationships Between Satellite, Target, and Earth Center.

SubsatelBte Point, SSP

Fig. 5-13. Definition of Angular Relationships Between Satellite, Target, and Earth Center.

Generally, then, the only problem is to find the relationship between the nadir angle, t], measured at the spacecraft from the subsatellite point (= nadir) to the target; the Earth central angle, A, measured at the center of the Earth from the subsatellite point to the target; and the grazing angle or spacecraft elevation angle, £, measured at the target between the spacecraft and the local horizontal. Figure 5-13 defines these angles and related distances. First, we find the angular radius of the Earth, p, from sinp =cosAq=

which is the same as Eq. (5-16). Next, if A is known, we find rj from tan 77= sinpsinA 1-sinpcosA

Or, if e is known, we find r) from sin 77 = cos esin p Finally, the remaining angle and side are obtained from ij + A+e=90deg D = Re (sin A/sin rf)

Figure 5-14 summarizes the process of transforming between spacecraft coordinates and Earth coordinates.

As an example, consider a satellite at an altitude of 1,000 km. From Eq. (5-16), the angular radius of the Earth p = 59.8 deg. From Eqs. (5-17) and (5-18), the horizon is 30.2 deg in Earth central angle from the subsatellite point and is at la line-of-sight

First, compute the angular radius of Earth, p slnp = cos^o = RE/(RE + H) (5-24) To compute spacecraft viewing angles given the subsatellite point at (Longssp. L&ssp) and target at (LongP, Latp), and Ai. b | Longssp ~ Longp \

cos A = sin Lafsspsin Latp-t-cos Lafsspcos Latpcos Ai (A <180 deg) (5-19)

cos&E= (sin Latp - cos A sin Latssp) / (sin A cos Latssp) (5-20)

tan ?} = sin p sin A/(1-sin p cos A) (5-25) To compute coordinates on the Earth given the subsatellite point at (Longssp, Latssp) and target direction (@b 7J):

cos Latp' = cos A sin Latgsp + sin A cos Latssp cos Oe (Latp' < 180 deg) (5-21)

cos A L = (cos A - sin Latssp sin Latp)/(cos Latssp cos Latp) (5-22)

Fig. 5-14. Summary of the Process of Transforming Between Spacecraft Viewing Angles and Earth Coordinates. Equation numbers are listed In the figure and variables are as defined in Figs. 5-11 and 5-12.

distance of 3,709 km from the satellite. We will assume a ground station at Hawaii (Latp = 22 deg, Longp = 200 deg) and a subsatellite point at LatSSP =10 deg, Longssp = 185 deg. From Eqs. (5-19) and (5-20), the ground station is a distance A = 18.7 deg from the subsatellite point, and has an azimuth relative to north = 48.3 deg. Using Eqs. (5-25) and (5-28) to transform into spacecraft coordinates, we find that from the spacecraft the target is 56.8 deg up from nadir (rf) at a line of sight distance, D, of2,444 km. From Eq. (5-27), the elevation of the spacecraft as seen from the ground station is 14.5 deg. The substantial foreshortening at the horizon can be seen in that at £ = 14.5 deg we are nearly half way from the horizon to the subsatellite point (A = 18.7 deg vs. 30.2 deg at the horizon).

Using these equations, we can construct Fig. 5-15, which shows the Earth as seen from 1,000 km over Mexico's Yucatan Peninsula in the Gulf of Mexico. The left side shows the geometry as seen on the surface of the Earth. The right side shows the geometry as seen by the spacecraft projected onto the spacecraft-centered celestial sphere. As computed above, the maximum Earth central angle will be approximately 30 deg from this altitude such that the spacecraft can see from northwestern South America to Maine on the East Coast of the U.S. and Los Angeles on the West Coast. The angular radius of the Earth as seen from the spacecraft will be 90 - 30 = 60 deg as shown in Fig. 5-15B. Because the spacecraft is over 20 North latitude, the direction to nadir in spacecraft coordinates will be 20 deg south of the celestial equator. (The direction from the spacecraft to the Earth's center is exactly opposite the direction from the Earth's center to the spacecraft)

Even after staring at it a bit, the view from the spacecraft in Rg. 5-15B looks strange. First, recall that we are looking at the spacecraft-centered celestial sphere from the outside. The spacecraft is at the center of the sphere. Therefore, the view for us is reversed from right-to-left as seen by the spacecraft so that the Atlantic is on the left and the Pacific on the right Nonetheless, there still appear to be distortions in the view. Mexico has an odd shape and South America has almost disappeared. All of this

Equatoi

A. Geometry on the Earth's Surface (SSP=SubsaJetKe Point)

Equatoi

A. Geometry on the Earth's Surface (SSP=SubsaJetKe Point)

B. Geometry Seen on the Spacecraft Centered

A'. Region on the Earth Seen by the 35 mm

Camera Fran» Shown in (ff) B\ Reld of View of a 35 mm Camera with a Normal Lens Looking Along the East Coast of the US.

B". Enlargement of the 35 mm Frame Showing the Region from Georgia to Massachusetts.

Fig. 5-15. Viewing Geometry for a Satellite at 1,000 km over the Yucatan Peninsula at 90 deg W longitude and 20 deg N latitude. See text for discussion. [Copyright by Microcosm; reproduced by permission.]

is due to the very strong foreshortening at the edge of the Earth's disk. Notice for example that Jacksonville, FL, is about halfway from the subsatellite point to the horizon. This means that only l/4th of the area seen by the spacecraft is closer to the subsatellite point than Jacksonville. Nonetheless, as seen from the perspective of the spacecraft, Jacksonville is 54 deg from nadir, i.e., 90% of the way to the horizon with 3/4ths of the visible area beyond it

The rectangle in the upper left of Fig. 5- 15B is the field of view of a 35 mm camera with a 50 mm focal length lens (a normal lens that is neither wide angle nor telephoto). The cameraperson on our spacecraft has photographed Florida and the eastern seaboard of the US to approximately Maine The region seen on the Earth is shown in Fig. 5-15A and 5-15B' and an enlargement of a portion of the photo from Georgia to Maine is shown in Fig. 5-15B". Note the dramatic foreshortening as Long Island and Cape Cod become little more than horizontal lines, even though they are some distance from the horizon. This distortion does not come from the plotting style, but is what the spacecraft sees. We see the same effect standing on a hilltop or a mountain. (In a sense, the spacecraft is simply a very tall mountain.) Most of our angular field of view is taken up by the field or mountain top we are standing on. For our satellite, most of what is seen is the Yucatan and Gulf of Mexico directly below. There is lots of real estate at the horizon, but it appears very compressed. From the spacecraft, 1 can point an antenna at Long Island, but I can not map it We must keep this picture in mind whenever we assess a spacecraft's fields of view or measurement needs.

Thus far we have considered spacecraft geometry only from the point of view of a spacecraft fixed over one point on the Earth. In fact, of course, the spacecraft is traveling at high velocity. Figure 5-16A shows the path of the subsatellite point over the Earth's surface, called the satellite's ground trace or ground track. Locally, the ground trace is very nearly the arc of a great circle. However, because of the Earth's rotation, the spacecraft moves over the Earth's surface in a spiral pattern with a displacement at successive equator crossings directly proportional to the orbit period. For a satellite in a circular orbit at inclination i, the subsatellite latitude, 8S, and longitude, Ls, relative to the ascending node are sin 8S = sin i sin (a t) (5-29)

where t is the time since the satellite crossed the equator northbound, coE= 0.004 178 07 deg/s is the rotational velocity of the Earth on its axis, and w is the satellite's angular velocity. For a satellite in a circular orbit, (o in deg/s is related to the period, P, in minutes by m= 6/P < 0.071 deg/s (5-31)

where 0.071 deg/s is the maximum angular velocity of a spacecraft in a circular orbit Similarly, the ground track velocity, Vg, is

where RE = 6,378 km is the equatorial radius of the Earth. For additional information on the satellite ground trace and coverage, taking into account the rotation of the Earth, see Chap. 8 of Wertz [2001].

Fig. 5-16B shows the swath coverage for a satellite in low-Earth orbit The swath is the area on the surface of the Earth around the ground trace that the satellite can observe as it passes overhead. From the formulas for stationary geometry in Eqs. (5-24) to (5-27), we can compute the width of the swath in terms of the Earth central angle, A. Neglecting the Earth's rotation, the area coverage rate, ACR, of a spacecraft will be

Ground Track

Ground Track

A. Satellite Ground Track. B. Swath Coverage for Satellite Ground Track

Fig. 5-16. Path of a Satellite Over the Earth's Surface. A swath which goes from horizon to horizon will cover a very large area, although we will see most of this area at very shallow elevation angles near the horizon.

A. Satellite Ground Track. B. Swath Coverage for Satellite Ground Track

Fig. 5-16. Path of a Satellite Over the Earth's Surface. A swath which goes from horizon to horizon will cover a very large area, although we will see most of this area at very shallow elevation angles near the horizon.

where Xower is the effective outer horizon, Ximeris the inner horizon, the area on the Earth's surface is in steradians, and P is the orbital period of the satellite. The plus sign applies to horizons on opposite sides of the ground trace and the minus sign to both horizons on one side, that is, when the spacecraft looks exclusively left or right For a swath of width 2A symmetric about the ground trace, this reduces to

Alternatively, this can be expressed in tenns of the limiting grazing angle (or elevation angle), e, and angular radius of the Earth, p, as

ACR = (4n/P) cos (e + arc sin (cos e sin p)) (5-35)

Because the curvature of the Earth's surface strongly affects the ACR, Eqs. (5-33) to (5-35) are not equal to the length of the arc between the effective horizons times either the velocity of the spacecraft or the velocity of the subsatellite point

53 Apparent Motion of Satellites for an Observer on the Earth

Even for satellites in perfectly circular orbits, the apparent motion of a satellite across the sky for an observer on die Earth's surface is not a simple geometrical figure. If the observer is in the orbit plane, then the apparent path of the satellite will be a great circle going directly overhead. If the observer is somewhat outside of the orbit plane, then the instantaneous orbit will be a large circle in three-dimensional space viewed from somewhat outside the plane of the circle and projected onto the observer's celestial sphere.

Because the apparent satellite path is not a simple geometrical figure, it is best computed using a simulation program. Available commercial programs include Satellite Tool Kit (1990), Orbit View and Orbit Workbench (1991), Orbit II Plus (1991), and MicroGLOBE (1990), which generated the figures in this chapter. These programs also work with elliptical orbits, so they are convenient—along with the appropriate formulas from this chapter—for evaluating specific orbit geometry. Unfortunately, a simulation does not provide the desired physical insight into the apparent motion of satellites. Neither does it provide a rapid method of evaluating geometry in a general case, as is most appropriate when first designing a mission. For these problems, we are interested in either bounding or approximating the apparent motion of satellites rather than in computing it precisely. After all, the details of a particular pass will depend greatly on the individual geometrical conditions. Approximate analytic formulas are provided by Wertz [1981,2001]. For mission design, the circular orbit formulas provided below for satellites in low-Earth orbit and geosynchronous orbit work well.

53.1 Satellites in Circular Low-Earth Orbit

We assume a satellite is in a circular low-Earth orbit passing near a target or ground station. We also assume that the orbit is low enough that we can ignore the Earth's rotation in the relatively brief period for which the satellite passes overhead.* We wish to determine the characteristics of the apparent satellite motion as seen from the ground station. Throughout this section we use the notation adopted in Sec. 5.2. Figure 5-17 shows the geometry. The small circle centered on the ground station represents the subsatellite points at which the spacecraft elevation, £, seen by the ground station is greater than some minimum e^. The nature of the communication or observation will determine the value of e^. For communications, the satellite typically must be more than 5 deg above the horizon, so e^ = 5 deg. The size of this circle of accessibility strongly depends on the value of c^,, as emphasized in the discussion ofFig. 5-15. In Fig. 5-17 we have assumed a satellite altitude of 1,000 km. The dashed circle surrounding the ground station is at e^ = 0 deg (that is, the satellite's true outer horizon), and the solid circle represents e^ = 5 deg. In practice we typically select a specific value of e^ and use that number. However, you should remain aware that many of the computed parameters are extremely sensitive to this value.

Fig. 5-17. Geometry of Satellite Ground Track Relative to an Observer on the Earth's Surface.

* See Chap. 9 of Wertz [2001] for a more accurate approximation which takes the Earth's rotation into account.

Given a value of £„,„, we can define the maximum Earth central angle, A^p the maximum nadir angle, rjnca, measured at the satellite from nadir to the ground station, and the maximum range, Dmax, at which the satellite will still be in view. These parameters as determined by applying Eqs. (5-26a) to (5-28) are given by:

SI" Timor where p is the angular radius of the Earth as seen from the satellite, that is, sin p = RE/(RE+H). We call the small circle of radius centered on the target the effective horizon, corresponding in our example to = 5 deg, to distinguish it from the true or geometrical horizon for which emin = 0 deg. Whenever the subsateJlite point lies within the effective horizon around die target or ground station, then communications or observations are possible. The duration, T, of this contact and the maximum elevation angle, f^^, of the satellite depends on how close the ground station is to the satellite's ground track on any given orbit pass.

As described in Chap. 6, the plane of a spacecraft's orbit and, therefore, the ground track, is normally defined by the inclination, i, and either the right ascension, Q or longitude, Lnoje, of the ascending node. Except for orbit perturbations, Q, which is defined relative to the stars, remains fixed in inertial space while the Earth rotates under the orbit On the other hand, L„0i/eis defined relative to the Earth's surface and, therefore, increases by 360 deg in 1,436 min, which is the rotation period of the Earth relative to the stars. (Again, orbit perturbations affect the exact rotation rate.) Because of this orbit rotation relative to the Earth, it is convenient to speak of the instantaneous ascending node which is Ln0de evaluated at the time of an observation or passage over a ground station. For purposes of geometry it is also often appropriate to work in terms of the instantaneous orbit pole, or the pole of the orbit plane at the time of the observation. The coordinates of this pole are latpote= 90 deg-i (5-39)

longpole =W-90deg (5-40)

A satellite passes directly over a target or ground station (identified by the subscript gs) on the Earth's surface if and only if sin (/onggj- Lnode) = tan lat^ / tan i (5-41)

There are two valid solutions to the above equation corresponding to the satellite passing over the ground station on the northbound leg of the orbit or on the southbound leg. To determine when after crossing the equator the satellite passes over the ground station for a circular orbit, we can determine n, the arc length along the instantaneous ground track from the ascending node to the ground station, from sin// = sin latgsl sin i (5-42)

Again, the two valid solutions correspond to the northbound and southbound passes.

Figure 5-17 defines the parameters of the satellite's pass overhead in terms ofAm/n, the minimum Earth central angle between the satellite's ground track and the ground station. This is 90 deg minus the angular distance measured at the center of the Earth from the ground station to the instantaneous orbit pole at the time of contact If we know the latitude and longitude of the orbit pole and ground station, gs, then the value of Kir,is sin A^/n = sin latpoie sin latgs + cos latpo\e cos latgg cos (¿Jong) (5-43)

where Along is the longitude difference between gs and the orbit pole. At the point of closest approach, we can compute the minimum nadir angle, rjml„,maximum elevation angle, e^, and minimum range, Z>min as t3" Vmin= sinpsin* (5-44)

l-sinpcosAm;(J

At the point of closest approach, the satellite is moving perpendicular to the line of sight to the ground station. Thus, the maximum angular rate of the satellite as seen from the ground station, Qmax, will be umax '

where Vsat is the orbital velocity of the satellite, and P is the orbit period.

Finally, it is convenient to compute the total azimuth range, A<j>, which the satellite covers as seen by the ground station, the total time in view, T, and die azimuth, <j>center at the center of the viewing arc at which the elevation angle is a maximum:

180 deg J ^cosAm/n where the arc cos is in degrees. Center ¡s related to <j>poie, the azimuth to the direction to the projection of the orbit pole onto the ground by teener = 180 deg - ^ (5-50)

cos <ppoie = (sin latppie — sin A^/^ sin latgs) / (cos Am/n cos latgs) (5-51 )

where <L0[e < 180 deg if the orbit pole is east of the ground station and > 180 deg if the oroit pole is west of the ground station. The maximum time in view, Tmar> occurs when the satellite passes overhead and A^ = 0. Eq. (5-49) then reduces to:

If satellite passes are approximately evenly distributed in off-ground track angle, then the average pass duration is about 80% of Tmax and 86% or more of the passes will be longer than half Tmwr

Table 5-4 summarizes the computations for ground station coverage and provides a worked example. Note that as indicated above, J is particularly sensitive to e^. If, for example, we assume a mountain-top ground station with 2 deg, then the time in view increases by 15% to 14.27 min. Figure 5-18 shows samples of several ground tracks for satellites in a 1,000 km orbit

TABLE 5-4. Summary of Computations for Ground Station Pass Parameters. We assume the following parameters: orbit pole at lat^=61.5 deg, long^ = 100 deg; Hawaii ground station at lafo = 22 deg, longgs = 200 deg; minimum allowable elevation angle = 5 deg. The result is a typical pass time-in-view of about 12 min.

TABLE 5-4. Summary of Computations for Ground Station Pass Parameters. We assume the following parameters: orbit pole at lat^=61.5 deg, long^ = 100 deg; Hawaii ground station at lafo = 22 deg, longgs = 200 deg; minimum allowable elevation angle = 5 deg. The result is a typical pass time-in-view of about 12 min.

Parameter

Formula

Eq. No.

Example

Earth Angular Radius, p

sin p = RE/{RE + H)

0 -1

Post a comment