Orbital Maneuvers - Michigan Tech IT Support Centerpages.mtu.edu/~ooabdelk/Space Mechanics Book/Ch 9/Space... · Web viewSpacecraft is not inserted in an orbit to stay forever! A

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9 Orbital Maneuvers

Philosophy is such an impertinently litigious lady that a man had as good be engaged in lawsuits as to have to do with her.

Isaac Newton in a letter to his friend Edmund Halley, June 20, 1687

9.1 Introduction9.1.1 Orbital Energy

Spacecraft is not inserted in an orbit to stay forever! A spacecraft may need to change its orbit once or more during its life time due to many reasons. A launch vehicle may insert a geostationary (GEO) satellite into an initial low Earth orbit (LEO) which is much lower than the final operational orbit. Then, the satellite should transfer from the initial orbit to its final orbit. Another need may arise if a surveillance satellite has to change its orbit in order to track a new target. Interplanetary missions usually require many orbit transfers until the spacecraft is inserted into the operational orbit or to use the same spacecraft to accomplish more than one mission. At the satellite end of life (EOF), the satellite may be kicked out of its orbit whether to reenter the Earth’s atmosphere or to rest in a graveyard orbit.

Any analysis of orbital maneuvers, i.e., the transfer of a satellite from one

orbit to another by means of a change in velocity, logically begins with the

energy as

V 2=μ (2r−1a) ( 9-1)

99

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http://en.wikipedia.org/wiki/Universe

http://en.wikipedia.org/wiki/Classical_mechanics

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http://en.wikipedia.org/wiki/Law_of_universal_gravitation

C H A P T E R 9 O R B I T A L M A N E U V E R S

Where V is the magnitude of the orbital velocity at some point, r the magnitude of the radius from the focus to that point, a the semimajor axis

of the orbit, and μ the gravitational constant of the attracting body. Fig. illustrates r, V, and a.

Equation can be rearranged asV 2

2− μr=−μ2a

( 9-2 )

Where it is evident that

Note that total energy/satellite mass is dependent only on a. as a increases, energy increases.

Figure 9- 1. Basic orbital maneuver.

C H A P T E R 9 O R B I T A L M A N E U V E R S 3

Figure 9-2.Conservation of energy relates r, V and a.

9.2 Basic Orbital ManeuversOrbital maneuvers are based on the principle that an orbit is uniquely determined by the position and vector at any point. Conversely, changing the velocity vector at any point instantly transforms the trajectory to a new one corresponding to the new velocity vector. Any conic orbit can be transformed into another conic orbit by changing the spacecraft velocity vector.

9.2.1 Delta–V Budget

Orbital transfers are usually achieved using the propulsion system onboard the spacecraft. Since the propellant mass on board is limited, it is very crucial for mission planning to estimate the propellant required for every transfer. The overall need for propulsion is usually expressed in terms of spacecraft total velocity change, or DV (Delta-V) budget. We assume the propulsion is applied impulsively, i.e. the velocity change will be acquired instantaneously. This assumption is reasonably valid for high-thrust propulsion.


Figure 9-3. Delta-V Budget.

From rocket theory shown in Error: Reference source not found, we can express the force produce by the impulsive thrust as:

F= ˙mνe=m I sp g0

M dVdt

=−dMdt

νe

dVνe

=−dMM

∆V=−ν e ln(M f

M i)=νe ln ( M i

M f)

( 9-3 )

where I sp=specific impulse = thrust/rate of fuel consumption. The spacecraft ‘sinitial and final mass, and the propellant mass are:

M i= spacecraft initial mass M f= spacecraft final mass M p=propellant mass used g0=9.81m/s²

Figure 9-4. Impulsive thrust produced based on rocket theory.


∆V=I sp g0 ln (1+M p

M f)

M p=M f [exp( ∆VI sp g0 )−1] ¿M i[1−exp(−∆V

I sp g0 )]( 9-4)

9.3 Satellite LaunchHigh-altitudes (above 200 km) may be achieved through two burns separated by coasting phase. The first burn is nearly vertical and places the satellite into an elliptic orbit with apogee at the final orbit radius. The satellite then coast (no burn) until it reaches the apogee. A second burn can be used to insert the satellite into its final LEO orbit.

9.4 Coplanar ManeuversWhen a satellite is launched, it can be placed into desired orbit through:

1. Directly from launch,

2. A booster at particular point to transfer into another orbit.

In section 9.3, the method (2) has been introduced, which is known as orbit maneuver. Orbit maneuver had its roots in the classical formulas and dynamics of Astrodynamics from several centuries ago. However, the application of orbit maneuver did not occur until after the launch of Sputnik in 1957.

Orbit maneuver is based on the fundamental principle that an orbit is uniquely determined by the position and velocity at any point. Therefore, changing the velocity vector at any point instantly transforms the trajectory to correspond to the new velocity vector. Thus, the orbit of a satellite is changed.

Coplanar maneuver only involves the change of semimajor axis and eccentricity of the orbit without changing the orbit plane. In this section, four kind of coplanar maneuvers are introduced:

i. Tangential-Orbit Maneuver,

ii. Non-tangential Orbit Maneuver,

Figure 9-5. Satellite launch.


iii. Hohmann Transfer,

iv. Bielliptic Orbit Transfer.

9.4.1 Tangential-Orbit Maneuver

Tangential-orbit maneuver occurs at the point where the velocity vector of spacecraft is tangent to its position vector, typically at perigee point.

EXAMPLE 9-1Determine the ∆V required to transfer from a circular orbit into elliptic orbit.

SOLUTION

The ∆V between two orbit can be shown as follow:

V ∘=√ μR,V p=√ 2μR − μ

a ( 9-5)

Figure 9-6. Single coplanar maneuver.

Figure 9-6 shows a typical tangential orbit maneuver at perigee point. Using the equation 9-5, the ∆V required is,

∆V=√ 2 μR −μa−√ μ

R

DVVcircVp

DV


9.4.2 Non-Tangential Coplanar Maneuver

The orbit maneuver does not limited only at apogee and perigee point. If condition is allowed, the satellite able to perform the orbit maneuvers at any point.

Figure 9- 1 shows the ∆V vector required for a non-tangential orbit maneuver, where α is the difference angle between the flight path angle of V1 and V2.

∆V=√v12+v22−2v1 v2 cos∝∝=∅ 1−∅ 2

( 9-6 )

9.4.3 Hohmann Transfer

The Hohmann’s transfer is the minimum two-impulse transfer between coplanar circular orbits. It can be used to transfer a satellite between two nonintersecting orbits (Walters Hohmann 1925).

The fundamental of the Hohmann’s transfer is a simple maneuver. This maneuver employs an intermediate elliptic orbit which is tangent to both initial and final orbits at their apsides. To accomplish the transfer, two burns are needed. The first burn will insert the spacecraft into the transfer orbit, where it will coast from periapsis to apoapsis. At apoapsis, the second burn is applied to insert spacecraft into final orbit.

Error: Reference source not found represents a Hohmann’s transfer from a circular orbit into another circular orbit. A tangential ΔV1 is applied to the circular orbit velocity. The magnitude of ΔV1 is determined by the requirement that the apogee radius of the resulting transfer ellipse must equal the radius of the final circular orbit. When the satellite reaches apogee of the transfer orbit, another ΔV must be added or the satellite will remain in the transfer ellipse. This ΔV is the difference between the apogee velocity in the transfer orbit and the circular orbit velocity in the final orbit. After ΔV2 has been applied, the satellite is in the final orbit, and the transfer has been completed.


∆V 1=V p ,t−V 1=√ 2μr1 − μa−√ μ

r 1 (9-7)

∆V 2=V 2−V a , t=√ μr2

−√ 2 μr2 −μa (9-8)

TOF=12Pt=π √ a3μ (9-9)

r p , t=r1 , ra , t=r2 (9-10)

EXAMPLE 9-2Determine the total ∆V required for Hohmann transfer to transfer from a LEO with hinitial = 191.344 km into GEO.

SOLUTION

The initial and final radius is,rinitial = 191.344 + 6378.145 = 6569.489 km

rfinal = 42164.215 km

At first impulse, the delta-v required is,

Figure 9-7. Hohmann transfer.


∆V 1=√ 2μr initial− μa−√ μ

r initial

where,

a=rinitial+rfinal

2=24366.852 km

Thus, ∆V 1=2.4571km / secFor the second impulse, the delta-v required is,

∆V 2=√ μr final

−√ 2μr final−μa=1.4782 km /sec

The total delta-v require is,∆V TOTAL=2.4571+1.4782=3.9353 km/sec

EXAMPLE 9-3Two geocentric elliptical orbits have common apse lines and their perigees are on the same side of the Earth. The first orbit has a perigee radius of r p=7000 km and e=0.3, whereas for the second orbit r p=32000 km and e=0.5.

a. Find the minimum total delta-v and the time of flight for a transfer from the perigee of the inner orbit to the apogee of the outer orbit.

b. Do part (a) for a transfer from the apogee of the inner orbit to the perigee of the outer orbit.

SOLUTION

a. For 1st orbit: P1=r p1 (1+e1 )=9100 km

a1=P11−e1

2=10000 km

v∫¿p1=√2μr p1 − μ

a1=8.6038 ¿ km/sec

For 2nd orbit: P2=48000 km & a2=64000 km

a=ra+r p2

⇒ ra2=96000 km


v fina2=√ 2μr a2 − μa2

=1.4408 km/sec

For transient orbit:r pt=r p1=7000 km & rat=ra2=96000 km

∴at=51500 km

v transp1=√ 2μr p1 − μa t

=10.3027 km/sec

v transa2=√ 2μra2 − μat

=0.7512 km/sec

v1=v transp1−v∫¿p1=1.6989¿ km/sec

v2=v fina2−v transa2=0.6896 km/sec∴ v total=2.3885 km/sec

ntrans=√ μa trans3 =5.402×10−5 sec−1

T trans=2πntrans

=1.1631×105 sec

Time of flight, TOF:

TOF=T trans

2=2.3262×105 sec ¿16.1544hr

b. For 1st orbit:ra1=a1 (1+e1 )=13000 km

v∫¿a1=√ 2μra1 − μ

a1=4.6328¿ km/sec

For 2nd orbit:

v finp2=√ 2μr p2

− μa2

=4.3225 km/sec

For transient orbit:r pt=ra1=13000 km & rat=r p2=32000 km

∴at=22500 km

v transa1=√ 2μra1 − μat

=6.6036 km/sec

v transp2=√ 2 μr p2 − μat

=2.6827 km/sec

C H A P T E R 9 O R B I T A L M A N E U V E R S 1 1

v1=vtransa1−v∫¿a1=1.9708 ¿ km/sec v2=v fin p2

−v trans p2=1.6398 km/sec∴ v total=3.6106 km/secT trans=3.3588×10

4 secTime of flight, TOF:

TOF=4.665 hr

EXAMPLE 9-4

A spacecraft is in a 300 km circular earth orbit. Calculate the transfer orbit time for a Hohmann transfer to a 3000 km coplanar circular Earth orbit.

SOLUTION

For initial orbit, 1: r1=6678.145 km ⇒ v1=7.7258 km/sec

For final orbit, 3: r3=9378.145 km ⇒ v3=6.5194 km/sec

For elliptical transient orbit, 2:r2 p=ri=6678.145 kmr2a=rf=9378.145 km

a2=r 2a+r2 p2

=8028.1 km

Transfer orbit time T 2

n2=√ μa23=8.7771×10

−4 sec−1

∴T 2=( 2πnt )2

=0.9943 hr

Figure 9-8. Bielliptic Transfer.


9.4.4 Bi-elliptic Transfer

Another type of orbit transfer that based on Hohmann transfer, which is called Bi-elliptic Transfer involves series of two Hohmann transfer.

The bi-elliptic transfer requires total of three impulse burn with two transfer orbit. The first burn is injected to insert the spacecraft into first transfer orbit at periapsis. When the spacecraft coasts to the apoapsis of the first transfer orbit, second impulse is injected to insert the spacecraft into second transfer orbit. Then, the spacecraft orbits along the transfer orbit to new apoapsis point. Finally, another impulse is injected to insert the spacecraft into the destination orbit. Figure 9-8 illustrates a bi-elliptic transfer between two circular orbits.

The bi-elliptic transfer requires much longer transfer time compared to the Hohmann Transfer. However, bi-elliptic is more efficient for long distance orbit transfer. Fig. 9-10 shows the cost comparison between Hohmann and Bi-elliptic Transfer. R is the ratio of final to initial radius for both orbits, where R* is the ratio of apogee radius of transfer orbit to initial orbit in bi-elliptic orbit. For R < 11.94, Hohmann transfer requires less cost than bi-elliptic transfer. For R > 15.58, bi-elliptic transfer performs better.


Figure 9-9.Delta-v Cost Comparison between Hohmann and Bi-elliptic Transfer.

EXAMPLE 9-5Determine the total ∆V required and time of flight for a bi-elliptic transfer with given orbit properties:

Initial orbit, hinitial = 191.344 km

Apogee altitude of transfer orbit, hapog = 503873 km

Final orbit, hfinal = 376310 kmSOLUTION

The initial, transfer orbit apogee and final radius are,rinitial = 191.344 + 6378.145 = 6569.489 km

rtrans = 503873 + 6378.145 = 510251.145 km

rfinal = 376310 + 6378.145 = 382688.145 km

And the semimajor axis for both transfer orbits are,

a1=r initial+rtrans

2=258410.317 km

a2=r final+rtrans

2=446469.645 km

At first impulse, the delta-v required is,

∆V 1=√ 2μrinitial

− μa1

−√ μr initial

=3.156 km /sec

At the second impulse, the delta-v required is,


∆V 2=√ 2 μrtrans− μa2

−√ 2μr trans− μa1

=0.677 km /sec

At the third impulse, the delta-v required is,

∆V 3=√ μr final

−√ 2 μr final− μa2

=−0.0705 km / sec

The total delta-v require is,∆V TOTAL=3.156+0.677+0.0705=3.9035 km /sec

The time of flight is,

TOF=π ×(√ a13μ +√ a23μ )=2138113.26 sec=593.92hr

EXAMPLE 9-6

A spacecraft is in a 300 km circular Earth orbit. Calculatea. The total delta-v required for the bi-elliptical transfer to a 3000

km altitude coplanar circular orbit shown, andb. Compare the total transfer time with the Hohmann’s transfer

time in Example 9-4.

SOLUTION

a. For initial orbit, 1: r1=6678.145 km ⇒ v1=7.7258 km/sec

For final orbit, 4: r 4=9378.145 km ⇒ v4=6.5194 km/sec

For first elliptical transient orbit, 2:e2=¿0.3

r2 A=r1=6678.145 km

a2=r2 A

1−e2=9540.21 km

r2B=rB=a2 (1+e2 )=12402.26 km


v2A=√ 2μr2A − μa2

=8.8087 km/sec

v2B=√ 2μr 2B − μa2

=4.74316 km/sec

For second elliptical transient orbit, 3:r3C=r4=9378.145 kmr3B=rB=12402.26 km

a3=r3C+r3B2

=10890.2 km

v3C=√ 2 μr3C − μa3

=6.9573 km/sec

v3 B=√ 2μr3B − μ

a3=5.26088 km/sec

v A=v2 A−v1=1.0829 km/sec

vB=v3B−v2B=0.51772 km/sec vC=v3C−v4=0.43793 km/sec

∴ v total=| vA|+| vB|+| vC|=2.03855 km/sec

b. Total transfer time, T total

T total=T22

+T 32

where,

T 2=2π

√ μa23

=9.2736×103 sec

T 3=2π

√ μa33

=1.131×104 sec

∴T TOTAL=1.02918×104 sec ¿2.859 hr

From Example 9-4, Hohmann Transfer only requires 0.9943 hr to transfer the spacecraft into another orbit. However, the bi-elliptic requires 3 times longer of transfer time to transfer the spacecraft into another orbit.


9.4.5 General Coplanar Transfer between Circular Orbits

Transfer between circular coplanar orbits only requires that the transfer orbit intersect or at least be tangent to both of the circular orbits. It is obvious that the periapsis radius of the transfer orbit must be equal to or less than the radius of the inner orbit and the apoapsis radius must be equal to or exceed the radius of the outer orbit if the transfer orbit is to touch both circular orbits. This condition can be expressed mathematically in Figure 9-10.

Figure 9-10.General coplanar transfer between circular orbits.

9.4.6 Phasing Maneuver

Most coplanar maneuver involves change of orbit size and shape. However, in some situation, the spacecraft required to change its position at a given time. Especially for the spacecraft rendezous case where the interceptor spacecraft required to intercept (or meet) the target spacecraft when it is behind or ahead of the target spacecraft in the orbit.


Figure 9-11. Phasing Maneuver.

Figure 9-11 shows an illustration of phasing maneuver. If the interceptor is behind the target spacecraft, then the phasing orbit required to be smaller than the original orbit, and vice versa.

Given that the phase angle (or difference of two true anomaly) between two spacecraft is ∆θ. Then, the one orbit period required by the phasing orbit is:

τ phase=2π−∆θn tgt

(9-11)

where ntgt is the mean motion of target spacecraft (or the original orbit). Then, we can determine the semimajor axis for the phasing orbit, that is,

τ phase=2 πnphase

=2π √ aphase3

μ

a phase¿( τ phase √μ2π )2/3

(9-12)


EXAMPLE 9-7Determine the semimajor axis of the phasing orbit, given that the position of target and interceptor spacecraft are:rtgt=4 J DU r∫ ¿=2 J +2√3 K DU¿

SOLUTION

First, the phase angle between spacecraft is,∆θ=cos−1¿

The mean motion of the original orbit is,

ntgt=√ μr3

=0.125 rad /TU

Then, the one orbit period required for the phasing orbit is,

τ phase=2π−∆θn tgt

=41.888TU

The semimajor axis for the phasing orbit is,

a phase¿( τ phase√μ2π )2/3

=3.5422DU

9.5Out-of-Plane Orbit Maneuvers

A velocity change which lies in the plane of the orbit can change its size or shape, or rotate the line of apsides. To change the orientation of the orbit plane in space, the ∆V impulse-vector inserted to the spacecraft should not parallel to the spacecraft velocity vector.

9.5.1 Simple Plane Change

Orbital maneuvers are characterized by a change in orbital velocity. If a velocity vector increment, ΔV that is perpendicular to a satellite velocity vector, V1 is added, then its results a new satellite velocity vector, V2. The perpendicular ΔV does not change the speed and flight-path angle of the satellite, but only the inclination of the orbit. The maneuver is called simple plane change (see Figure 9-12).


For a circular orbit spacecraft that performs the simple plane change through an angle θ, the semimajor axis, a and eccentricity, e are remain the same. Thus, the velocity of spacecraft at before and after the plane change

are equal, V 1=V 2 . Using the velocity vector triangle illustration in Figure 9-13, the delta-v required is,

∆V=2V sin θ2

(9-13)

EXAMPLE 9-8Determine the ∆V required for a satellite to change its orbit plane from inclination 10° to inclination 25° at altitude 600km.

SOLUTION

The radius of the orbit is,r = 600 + 6378.145 = 6978.145 kmThe delta-v required for the plane change is,

∆V=2×√ μr × sin( 25−102 )=1.973 km / sec

9.5.2 General Plane Change Maneuver

In general, plane change maneuver involves the inclination and RAAN change while the size and shape of orbit remain the same. The change of the RAAN in plane change maneuver results that both orbit do not intersect

Figure 9-13.Velocity vector triangle for circular orbit plane change.

Figure 9-12.Simple plane change.

Figure 9-14.General Plane Change Maneuver. Figure 9-15.Argument of latitude of intersection point.


at the original RAAN location. Figure 9-14 shows the example of general plane change maneuver.

Nodes 1 and 2 are the direction of RAAN for both initial and final orbits respectively, where else the e is the eccentricity vector (also known as argument of perigee) for the initial orbit.

The delta-v required for the general plane change maneuver is shown in equation 9-17. The α angle is determined using equation 9-14, and ALa is the argument of latitude of intersection point, that is shown in Figure 9-15.

cos α=cos iinitialcos ifinal+sin iinitial sin ifinal cos (ΔΩ ) (9-14)

sin ALa=sin ifi nalsin (∆Ω )

sinα (9-15)

ALa=θ+ω (9-16)


∆V=2Vsin(α2 ) (9-17)

EXAMPLE 9-9Compute ΔV required to change the right ascension of the ascending node of the following orbit to 100o West: rp1 = 1.1DU, e1 = 0.1, i = 45˚ , Ω = 40˚ West, ω = 10˚

SOLUTION

The orbit of satellite is transfer to Ω = 100˚ West.The ΔV required is given by,

∆V=2Vsin( α2 )where α is,

cos α=cos iinitialcos ifinal+sin iinitial sin ifinal cos (ΔΩ )ΔΩ=−40 °+100 °=60 °

α=41.4 °

The speed of satellite at that particular point is,

V=√ 2μr −μa

where,a = 1.22DU

And r can be obtain through,

sin ALa=sin ifinal sin (∆Ω )

sinαALa = 67.8˚

θ = ALa – ω = 57.8˚

r=r p1 (1+e )1+ecosθ

=1.14878DU

Thus,V = 0.96 DU/TU

ΔV = 0.679 DU/TU


9.5.3 Combined Maneuver

Frequently, the spacecraft orbit needs to be raised as well as titled. Two orbital transfers may then be applied:

-A coplanar maneuver to raise the orbit (change radius), then

-A plane change to tilt the orbit.

However, performing two separates orbit maneuvers is fuel inefficient because number of burns increased. Also, the time required for spacecraft to arrive at final orbit is much longer. Therefore, as an alternative, these two maneuvers can be combined in one maneuver to perform both tasks in one burn which is more economic (require less fuel) and faster.

There are a few type of combined maneuver available in study. In this section, we will introduce the minimum inclination maneuver.

Figure 9-16 shows the minimum inclination maneuver for a spacecraft. Both initial and final velocity of plane change maneuver contains the Hohmann transfer’s contribution. The change of inclination between initial, transfer and final orbit is chosen in the way such that the required cost is minimum. Here, a scaling term, s is introduced to determine change of inclination required between orbits.

∆ iinitial=s ∆ i

∆ ifinal=(1−s)∆ i (9-18)


Figure 9-16.Orbit transfer of a spacecraft using combined maneuver.

The total delta-v that required for the combined maneuver is,

∆V=√V transa2 +V initial

2 −2V initialV transacos (s ∆ i)

+√V transb2 +V final

2 −2V finalV transbcos ( (1−s )∆i ) (9-19)

Now, the optimum scaling, s is required to determine to produce minimum cost. Then, we taked ∆V /ds=0

s≈ 1∆ itan−1[ sin (∆i )

V initialV transa

V finalV transb

cos (∆ i ) ] (9-18)

For circular initial and final orbits:

V initialV transa

V finalV transb

=√R3 , R=r finalrinitial

(9-19)


EXAMPLE 9-10

Calculate the total delta-v required for a spacecraft to transfer from an orbit, r1 = 1.02 DU to r2 = 2.33 DU with the change of inclination ∆i = 10°.

SOLUTION

We have the initial and final radius, r1 and r2. Then semimajor axis for the transfer orbit is,

a=r1+r22

=1.675DU

The velocities at each location are:

V initial=√ μr1

=0.9901DU /TU

V transa=√ 2μr1 −μa=1.1678DU /TU

V transb=√ 2μr2 − μa=0.5512DU /TU

V final=√ μr2

=0.6551DU /TU

Then, we need to determine the scaling, s, that is:

s= 1Δi tan

−1[ sin (Δi )R3/2+cos (Δi ) ]=0.224105

Therefore, the total delta-v is,∆V=0.3464DU /TU

9.6 Problems1. A space vehicle in a circular orbit at an altitude of 0.0784 DU above the

Earth executes a Hohmann transfer to a 0.1568 DU circular orbit. Calculate the total delta-v requirement.

2. Calculate the total delta-v required for a Hohmann transfer from a circular orbit of radius 4 r to a circular orbit of radius16 r.


3. Determine the total time of flight for Hohmann and bi-elliptic transfer with given orbit properties:

i. Initial orbit, hinitial = 200 km,ii. Apogee altitude of transfer orbit for bi-elliptic transfer, hapog = 35000

km,iii. Final orbit, hfinal = 30000 km.

4. Determine the total ∆V required for a bi-elliptic transfer with given orbit properties:

i. Initial orbit, r1 = 1.013 DU,ii. Apogee altitude of transfer orbit, r2 = 3.021 DU,iii. Final orbit, r3 = 2.601 DU.

i.Determine the

total ∆V required and time-of-flight for a bi-elliptic

transfer to place a spacecraft from r1 = 6578.145 km into GEO. Given that, the apogee radius of transfer orbit is, r2 = 46378.145 km.

ii. Given that the target and interceptor spacecraft are orbiting around the Earth in equatorial orbit. Determine the semimajor axis of the phasing orbit. Both spacecrafts’ positions at that time are:

rtgt=12546.387 I+10527.667 J kmr∫ ¿=16129.324 I +2844.035 J km¿

iii. Given that the target and interceptor spacecraft are in an equatorial orbit with semimajor axis of 15000 km and eccentricity 0.1. Determine the semimajor axis of the phasing orbit if the distances of target and interceptor spacecraft to the Earth center are 13,574.4216 km and 13,615.9737 km respectively at the time. (Assume that the true anomaly of both spacecraft are in first quadrant.)

iv. Determine the ∆V required for a satellite to change its orbit plane from equatorial orbit to an orbit with inclination 10° at altitude 400km.


v. Determine the total ∆V required for a satellite to change its orbit plane at inclination 5° to GEO.

vi. Compute ΔV required to change from following orbit to the right ascension of the ascending node at 35˚ and inclination at 15˚:

rp1 = 1.08 DU, e1 = 0.05, i = 20˚ , Ω = 20˚, ω = 5˚

vii. Compute ΔV required to change from following orbit to the right ascension of the ascending node at 30˚ and inclination at 25˚:

Altitude, h = 300 km, e1 = 0, i = 10˚ , Ω = 25˚

viii. Calculate the total delta-v required for a spacecraft to transfer from an orbit with altitude, h1 = 400 km to geosynchronous orbit with the change of inclination ∆i = 25°.

ix. Calculate the total delta-v required for a spacecraft to transfer from an orbit, r1 = 1.157 DU to r2 = 4.136 DU with the change of inclination ∆i = 20° using:a. Hohmann transfer followed by simple plane change.b. Combined Maneuver.

9.7 ReferencesChobotov, V. (2002). Orbital Mechanics. Reston, Virginia, American Institute of Aeronautics and Astronautics, Inc.

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Orbital Maneuvers - Michigan Tech IT Support Centerpages.mtu.edu/~ooabdelk/Space Mechanics Book/Ch 9/Space... · Web viewSpacecraft is not inserted in an orbit to stay forever! A