AOE 4134
Astrodynamics
HW Set 4, Problem 2
October 7, 1998
•We first consider the heliocentric orbit from a radius of 1 AU (Earth) to a radius of
5.2 AU (Jupiter). We consider the Hohmann case.
E=v2
p
2−µ
rp
=µ
2at
.
•For the Hohmann case 2at=r1+r2and this leads to
vp=s2µr2
r1(r1+r2)=1.2953SU.
•Since circular speed at the Earth’s mean solar distance is 1SUwe get
∆v1=vp−vc1=.2953 SU=8.791 km/s.
•From the Patched Conic approximation we interpret this ∆v1as the hyperbolic escape
speed in a geocentric hyperbola. Again, an energy analysis is applied
E=v2
p
2−µ⊕
rp
=v2
∞
2→vp=v2
∞+2µ⊕
rp1/2
.
The numerical values are rp=1.0871 DU⊕(300 nmi orbit) and v∞=8.791 km/s=
1.1124 SU⊕, leading to vp=.1.7542 SU⊕. Since we begin in a circular parking orbit
∆vp=vp−vc=.7951 SU⊕=6.283km/s
•Finally we consider the time-of-flight and required phase angle. Since the Hohmann
transfer uses precisely one-half the ellipse we have TOF =π√a
3= 17.15TU
.
•Jupiter’s angular rate is .0843/T U, so that during the TOF Jupiter will traverse a Sun
central angle of 1.446 radians, while the spacecraft will traverse π. Thus, at insertion
into the heliocentric orbit Jupiter must lead by 1.695 ≈97.1o.
1
