EXAMPLE PROBLEMS.docx
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EXAMPLE PROBLEMS.docx
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EXAMPLEPROBLEMS
EXAMPLEPROBLEMS
PROBLEM1.1
CalculatethevelocityofanartificialsatelliteorbitingtheEarthinacircularorbit
atanaltitudeof200kmabovetheEarth'ssurface.
SOLUTION,
FromBasicsConstants,
RadiusofEarth=6,378.140km
GMofEarth=3.986005x1014m3/s2
Given:
r=(6,378.14+200)x1,000=6,578,140m
Equation(1.6),
v=SQRT[GM/r]
v=SQRT[3.986005x1014/6,578,140]
v=7,784m/s
PROBLEM1.2
Calculatetheperiodofrevolutionforthesatelliteinproblem1.1.
SOLUTION,
Given:
r=6,578,140m
Equation(1.9),
P2=4x
2xr3/GM
P=SQRT[4x
2xr3/GM]
P=SQRT[4x
2x6,578,1403/3.986005x1014]
P=5,310s
PROBLEM1.3
CalculatetheradiusoforbitforaEarthsatelliteinageosynchronousorbit,wherethe
Earth'srotationalperiodis86,164.1seconds.
SOLUTION,
Given:
P=86,164.1s
Equation(1.9),
P2=4x
2xr3/GM
r=[P2xGM/(4x
2)]1/3
r=[86,164.12x3.986005x1014/(4x
2)]1/3
r=42,164,170m
PROBLEM1.4
AnartificialEarthsatelliteisinanellipticalorbitwhichbringsittoanaltitudeof
250kmatperigeeandouttoanaltitudeof500kmatapogee.Calculatethevelocityof
thesatelliteatbothperigeeandapogee.
SOLUTION,
Given:
Rp=(6,378.14+250)x1,000=6,628,140m
Ra=(6,378.14+500)x1,000=6,878,140m
Equations(1.16)and(1.17),
Vp=SQRT[2xGMxRa/(Rpx(Ra+Rp))]
Vp=SQRT[2x3.986005x1014x6,878,140/(6,628,140x(6,878,140+6,628,140))]
Vp=7,826m/s
Va=SQRT[2xGMxRp/(Rax(Ra+Rp))]
Va=SQRT[2x3.986005x1014x6,628,140/(6,878,140x(6,878,140+6,628,140))]
Va=7,542m/s
PROBLEM1.5
AsatelliteinEarthorbitpassesthroughitsperigeepointatanaltitudeof200km
abovetheEarth'ssurfaceandatavelocityof7,850m/s.Calculatetheapogeealtitude
ofthesatellite.
SOLUTION,
Given:
Rp=(6,378.14+200)x1,000=6,578,140m
Vp=7,850m/s
Equation(1.18),
Ra=Rp/[2xGM/(RpxVp2)-1]
Ra=6,578,140/[2x3.986005x1014/(6,578,140x7,8502)-1]
Ra=6,805,140m
Altitude@apogee=6,805,140/1,000-6,378.14=427.0km
PROBLEM1.6
Calculatetheeccentricityoftheorbitforthesatelliteinproblem1.5.
SOLUTION,
Given:
Rp=6,578,140m
Vp=7,850m/s
Equation(1.20),
e=RpxVp2/GM-1
e=6,578,140x7,8502/3.986005x1014-1
e=0.01696
PROBLEM1.7
AsatelliteinEarthorbithasasemi-majoraxisof6,700kmandaneccentricityof0.01.
Calculatethesatellite'saltitudeatbothperigeeandapogee.
SOLUTION,
Given:
a=6,700km
e=0.01
Equation(1.21)and(1.22),
Rp=ax(1-e)
Rp=6,700x(1-.01)
Rp=6,633km
Altitude@perigee=6,633-6,378.14=254.9km
Ra=ax(1+e)
Ra=6,700x(1+.01)
Ra=6,767km
Altitude@apogee=6,767-6,378.14=388.9km
PROBLEM1.8
AsatelliteislaunchedintoEarthorbitwhereitslaunchvehicleburnsoutatan
altitudeof250km.Atburnoutthesatellite'svelocityis7,900m/swith
equalto89degrees.Calculatethesatellite'saltitudeatperigeeandapogee.
SOLUTION,
Given:
r1=(6,378.14+250)x1,000=6,628,140m
v1=7,900m/s
=89o
Equation(1.26),
(Rp/r1)1,2=(-C+/-SQRT[C2-4x(1-C)x-sin2
])/(2x(1-C))
whereC=2xGM/(r1xv12)
C=2x3.986005x1014/(6,628,140x7,9002)
C=1.92718
(Rp/r1)1,2=(-1.92718+/-SQRT[1.927182-4x-0.92718x-sin2(89)])/(2x-0.92718)
(Rp/r1)1,2=0.996019and1.08252
PerigeeRadius,Rp=Rp1=r1x(Rp/r1)1
Rp=6,628,140x0.996019
Rp=6,601,750m
Altitude@perigee=6,601,750/1,000-6,378.14=223.6km
ApogeeRadius,Ra=Rp2=r1x(Rp/r1)2
Ra=6,628,140x1.08252
Ra=7,175,090m
Altitude@agogee=7,175,090/1,000-6,378.14=797.0km
PROBLEM1.9
Calculatetheeccentricityoftheorbitforthesatelliteinproblem1.8.
SOLUTION,
Given:
r1=6,628,140m
v1=7,900m/s
=89o
Equation(1.27),
e=SQRT[(r1xv12/GM-1)2xsin2
+cos2
]
e=SQRT[(6,628,140x7,9002/3.986005x1014-1)2xsin2(89)+cos2(89)]
e=0.04162
PROBLEM1.10
Calculatetheangle
fromperigeepointtolaunchpointforthesatellite
inproblem1.8.
SOLUTION,
Given:
r1=6,628,140m
v1=7,900m/s
=89o
Equation(1.28),
tan
=(r1xv12/GM)xsin
xcos
/[(r1xv12/GM)xsin2
-1]
tan
=(6,628,140x7,9002/3.986005x1014)xsin(89)xcos(89)
/[(6,628,140x7,9002/3.986005x1014)xsin2(89)-1]
tan
=0.48329
=arctan(0.48329)
=25.79o
PROBLEM1.11
Asatelliteisinanorbitwithasemi-majoraxisof7,500kmandaneccentricity
of0.1.Calculatethetimeittakestomovefromaposition30degreespastperigee
to90degreespastperigee.
SOLUTION,
Given:
a=7,500x1,000=7,500,000m
e=0.1
tO=0
vO=30degx
/180=0.52360radians
v=90degx
/180=1.57080radians
Equation(1.31),
cosE=(e+cosv)/(1+ecosv)
Eo=arccos[(0.1+cos(0.52360))/(1+0.1xcos(0.52360))]
Eo=0.47557radians
E=arccos[(0.1+cos(1.57080))/(1+0.1xcos(1.57080))]
E=1.47063radians
Equation(1.32),
M=E-exsinE
Mo=0.47557-0.1xsin(0.47557)
Mo=0.42978radians
M=1.47063-0.1xsin(1.47063)
M=1.37113radians
Equation(1.30),
n=SQRT[GM/a3]
n=SQRT[3.986005x1014/7,500,0003]
n=0.00097202rad/s
Equation(1.29),
M-Mo=nx(t-tO)
t=tO+(M-Mo)/n
t=0+(1.37113-0.42978)/0.00097202
t=968.4s
PROBLEM1.12
Thesatelliteinproblem1.11hasatrueanomalyof90degrees.Whatwillbethe
satellite'sposition,i.e.it'strueanomaly,20minuteslater?
SOLUTION,
Given:
a=7,500,000m
e=0.1
tO=0
t=20x60=1,200s
vO=90x
/180=1.57080rad
Fromproblem1.11,
Mo=1.37113rad
n=0.00097202rad/s
Equation(1.29),
M-Mo=nx(t-tO)
M=Mo+nx(t-tO)
M=1.37113+0.00097202x(1,200-0)
M=2.53755
METHOD#1,LowAccuracy:
Equation(1.33),
v~M+2xexsinM+1.25xe2xsin2M
v~2.53755+2x0.1xsin(2.53755)+1.25x0.12xsin(2x2.53755)
v~2.63946=151.2degrees
METHOD#2,HighAccuracy:
Equation(1.32),
M=E-exsinE
2.53755=E-0.1xsinE
Byiteration,E=2.58996radians
Equation(1.31),
cosE=(e+cosv)/(1+ecosv)
Rearrangingvariablesgives,
cosv=(cosE-e)/(1-ecosE)
v=arccos[(cos(2.58996)-0.1)/(1-0.1xcos(2.58996)]
v=2.64034=151.3degrees
PROBLEM1.13
Forthesatelliteinproblems1.11and1.12,calculatethelengthofitsposition
vector,itsflight-pathangle,anditsvelocitywhenthesatellite'strueanomaly
is225degrees.
SOLUTION,
Given:
a=7,500,000m
e=0.1
v=225degrees
Equations(1.34)and(1.35),
r=ax(1-e2)/(1+excosv)
r=7,500,000x(1-0.12)/(1+0.1xcos(225))
r=7,989,977m
=arctan[exsinv/(1+excosv)]
=arctan[0.1xsin(225)/(1+0.1xcos(225))]
=-4.351degrees
Equation(1.36),
v=SQRT[GMxax(1-e2)]/(rxcos
)
v=SQRT[3.986005x1014x7,500,000x(1-0.12)]/(7,989,977xcos(-4.351))
v=6,828m/s
PROBLEM1.14
Calculatetheperturbationsinlongitudeoftheascendingnodeandargumentof
perigeecausedbytheMoonandSunfortheInternationalSpaceStationorbiting
atanaltitudeof400km,aninclinationof51.6degrees,andwithanorbital
periodof92.6minutes.
SOLUTION,
Given:
i=51.6degrees
n=1436/92.6=15.5revolutions/day
Equations(1.37)through(1.40),
_moon=-0.00338xcos(i)/n
_moon=-0.00338xcos(51.6)/15.5
_moon=-0.000135deg/day
_sun=-0.00154xcos(i)/n
_sun=-0.00154xcos(51.6)/15.5
_sun=-0.0000617deg/day
_moon=0.00169x(4-5xsin2i)/n
_moon=0.00169x(4-5xsin251.6)/15.5
_moon=0.000101deg/day
_sun=0.00077x(4-5xsin2i)/n
_sun=0.00077x(4-5xsin251.6)/15.5
_sun=0.000046deg/day
PROBLEM1.15
Asatelliteisinanorbitwithasemi-majoraxisof7,500km,aninclination
of28.5degrees,andaneccentricityof0.1.CalculatetheJ2perturbationsin
longitudeoftheascendingnodeandargumentofperigee.
SOLUTION,
Given:
a=7,500km
i=28.5degrees
e=0.1
Equations(1.41)and(1.42),
_J2=-2.06474x1014xa-7/2x(cosi)x(1-e2)-2
_J2=-2.06474x1014x(7,500)-7/2x(cos28.5)x(1-(0.1)2)-2
_J2=-5.067deg/day
_J2=1.03237x1014xa-7/2x(4-5xsin2i)x(1-e2)-2
_J2=1.03237x1014x(7,500)-7/2x(4-5xsin228.5)x(1-(0.1)2)-2
_J2=8.250deg/day
PROBLEM1.16
AsatelliteisinacircularEarthorbitatanaltitudeof400km.Thesatellite
hasacylindricalshape2mindiameterby4mlongandhasamassof1,000kg.The
satelliteistravelingwithitslongaxisperpendiculartothevelocityvectorand
it'sdragcoefficientis2.67.Calculatetheperturbationsduetoatmosphericdrag
andestimatethesatellite'slifetime.
SOLUTION,
Given:
a=(6,378.14+400)x1,000=6,778,140m
A=2x4=8m2
m=1,000kg
CD=2.67
FromAtmosphereProperties,
=2.62x10-12kg/m3
H=58.2km
Equation(1.6),
V=SQRT[GM/a]
V=SQRT[3.986005x1014/6,778,140]
V=7,669m/s
Equations(1.46)through(1.48),
arev=(-2x
xCDxAx
xa2)/m
arev=(-2x
x2.67x8x2.62x10-12x6,778,1402)/1,000
arev=-16.2m
Prev=(-6x
2xCDxAx
xa2)/(mxV)
Prev=(-6x
2x2.67x8x2.62x10-12x6,778,1402)/(1,000x7,669)
Prev=-0.0199s
Vrev=(
xCDxAx
xaxV)/m
Vrev=(
x2.67x8x2.62x10-12x6,778,140x7,669)/1,000
Vrev=0.00914m/s
Equation(1.49),
L~-H/
arev
L~-(58.2x1,000)/-16.
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