University of California, Berkeley
Department of Mechanical Engineering
ME 170, Spring 2017
Homework 7
Problem 1
Recall that a particle moving under the influence of a central force has a constant areal
velocity
h
AË™ =
,
(1)
2m
where h/m is the angular momentum per unit mass. Also,
 h
m 2
= GM `, (2) where ` is the semilatus rectum of the orbit. For an elliptical orbit, with semimajor axis a,
semiminor axis b, and eccentricity ε < 1,
` = a(1 − ε2 ), b2 = a2 (1 − ε2 ). (3) (a) The orbital period τ is given by
Ï„=
Deduce that
Ï„2 = A(Ï„ )
.
AË™ (4) 4Ï€ 2 3
a,
GM (5) which is a statement of Kepler’s third law.
(b) Examine how well (5) holds for the solar system. Complete the table below by obtaining
the period τ and semimajor axis a of each planet’s orbit from the given data sheet. Then
calculate τ 2 /a3 and compare it with 4π 2 /GMS (MS = 1.9884 × 1030 kg, G = 6.674 × 10−11
m3 /(kg · s2 )).
Planet Orbital Period Ï„ (s) Semimajor Axis a (m) Ï„ 2 /a3
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
1 % error (c) Repeat Part (b) for Jupiter’s four largest (Galilean) moons. The mass of Jupiter is
MJ = 1.8985 × 1027 kg.
Moon
Orbital Period Ï„ (s) Semimajor Axis a (m) Ï„ 2 /a3
Io
Europa
Ganymede
Callisto % error Problem 2
In 2008, astronomers discovered extrasolar planets orbiting the young star HR 8799, which
is located 129 light years away from earth. The mass of the star is
MH = 1.56 MS .
For the four planets in the system, the semi-major orbital axes are
HR 8799 a :
HR 8799 b :
HR 8799 c :
HR 8799 d : 68.0
42.9
27.0
14.5 au
au
au
au, where 1 au = 149.598 × 109 m (1 light year = 63241 au). Use Kepler’s third law,
Ï„2
4Ï€ 2
=
a3
GMH s2 /m3 (6) to calculate the periods of these planets in years (1 year = 365.25 days = 31.5576 ×106 s).
Problem 3
Consider a satellite orbiting the earth in a circular orbit O1 of altitude 6000 km. Take the
earth’s mean radius R to be 6378 km and the gravitational parameter GME to be 398.6×1012
m3 /s2 . Also, recall that for Kepler orbits, the specific angular momentum h/m of the motion
is related to the semilatus rectum ` of the orbit by
 2
h
= GME `,
(7)
m
while the specific energy E/m of the motion is related to the semilatus rectum l and the
orbital eccentricity ε by
2 (E/m)
ε2 − 1
=
.
(8)
GME
`
2 (a) Show that for a circular orbit of radius r0 , the satellite speed v0 satisfies the relation
v02 = GME
,
r0 (9) and the specific energy is given by
E
1 GME
=−
.
m
2 r0 (10) (b) Calculate the quantities h1 /m and E1 /m for the circular orbit O1 .
(c) Suppose that at a point A on the orbit O1 the speed of the satellite is increased due to
a tangential impulsive thrust by an amount
∆vA = 660 m/s. (11) Let vA0 = vA + ∆vA . Calculate the dynamical quantities h2 /m and E2 /m for the new orbit
O2 . Show that it is elliptical. Denote its apogee by B. Calculate the semilatus rectum `2
and the eccentricity ε2 of O2 . Also, calculate the semimajor and semiminor axes of the orbit,
as well as the distances rp2 and ra2 to perigee and apogee, respectively. Using conservation
of angular momentum, calculate the speed vB of the satellite at apogee. Sketch the orbits
O1 and O2 .
(d) Next, let the speed of the satellite be impulsively decreased at apogee by 200 m/s:
∆vB = −200 m/s. (12) Denote the new speed of the satellite by vB0 , and the new orbit by O3 . Determine the orbital
parameters for O3 ; use a subscript to identify them. Add the new orbit to your sketch.
Denote its perigee by C. Calculate the satellite’s speed vC at perigee.
(e) Argue that by reversing the increments (12) and (11), at B and A, the satellite could be
returned to its original circular orbit O1 at a speed-increment cost of ∆v = 860 m/s.
(f ) As an alternative way to return to O1 , a Hohmann transfer semiellipse O4 may be
constructed with perigee at C and apogee at a point D that lies on the circle O1 and is
diametrically opposite to A. Thus,
r4p = rC = r3p , r4a = rD = 12.378 × 106 m. (13) Calculate the quantities a4 , ε4 , `4 , and b4 for the transfer orbit. Then, use Eqns. (7) and (8)
to determine h4 /m and E4 /m.
(g) Use the value h4 /m to calculate the satellite’s speed vC0 in the orbit O4 , after the impulse
at C. Likewise, calculate the speed vD which it has at apogee D, before the final impulse
that returns it to the circular orbit O4 .
3 (h) Sum up the absolute values of the speed increments in Part (g) and compare the cost
to that in Part (e).
Problem 4
Suppose that an intercontinental ballistic missile is launched from the earth’s surface with
a speed v0 = 6.7 km/s and a flight-path angle φ0 = 20◦ . The radius of the earth is 6378 km.
(a) Use the initial data to determine the dynamical constants h/m and E/m of the missile’s
orbit.
(b) Apply Eqns. (7) and (8) to calculate the semilatus rectum and eccentricity of the orbit.
(c) Calculate the semimajor and semiminor axes.
(d) Find the apogee and perigee.
(e) Calculate the speed of the missile at apogee.
(f ) Recall that the orbit is described by the equation
r= `
,
1 + ε cos θ (14) where the angle θ is the true anomaly. Calculate the value θ0 of θ at launch.
(g) Calculate the maximum altitude and range of the missile.
(h) Sketch the missile’s orbit in relation to the earth.
Problem 5
Consider an attractive central force field of the type
F = −f (r)er , f > 0. (15) The angular momentum and energy integrals are given by
h = mr2 θ˙ (> 0) (16) and


1
2
2Ë™
m r˙ + r θ + V = E.
2
Let 4 (17) u= 1
r (18) and note that
 dr du Ë™
h du
rË™ =
θ=−
,
du dθ
m dθ h
r¨ = −
m 2 u2 d2 u
.
dθ2 (19) The equation of motion for r, namely

r¨ − h
m 2 1
f
= 0,
+
3
r
m (20) may then be expressed as
f /m 1
d2 u
+u−
= 0.
2
dθ
(h/m)2 u2 (21) Suppose that the law of attraction is that of inverse cube, i.e.,
µ
f
= 3,
m
r (µ > 0). (22) (a) Calculate the corresponding potential energy function V (r), taking V → 0 as r → ∞.
(b) Deduce that
 
h 2
1
r¨ − ( ) − µ 3 = 0,
m
r


µ
d2 u
+ 1−
u = 0,
dθ2
(h/m)2 (23)
(24) and
 du
dθ 2  
µ
2E/m
2
+ 1−
u
=
.
(h/m)2
(h/m)2 (25) (c) First, consider the case
 h
m 2
− µ = 0. (26) Deduce that
r¨ = 0, d2 u
= 0,
dθ2
5 (27) and show that the orbit must be of the form
1
= u = Aθ + B,
r (28) where A and B are constants of integration. Further, argue that
E
1 h
= ( )2 A2 ≥ 0.
m
2 m (29) What is the special case E = 0? If E > 0, prove that the orbit cannot have any apsis. Sketch
the particular orbit
r= 1
.
θ
1 + 10 (30) (d) Next, consider the case
 h
m 2
− µ > 0. (31) d2 u
+ ω 2 u = 0.
dθ2 (32) Equation (24) is then if the form Deduce that the orbits are described by
1
= u = A cos ωθ + B sin ωθ = C cos(ωθ + ψ),
r (33) where A, B, C, ψ are constants. Relate C to the dynamical constants and deduce that
E > 0 for this case. Also, show that there is only one apsidal distance, which is given by
1
(h/m)2 − µ
=
.
r
E/m (34) 1
= u = cos 4θ,
r (35) For the particular case Find the values of θ at which the apsides occur and also solve for the apsidal distance. Sketch
the orbits.
(e) Finally, consider the case
6 d2 u
− µ < 0.
dθ2 (36) d2 u
− q 2 θ = 0.
2
dθ (37) 1
= u = Aeqθ + Be−qθ ,
r (38) Equation (24) is then of the form Show that the orbits are of the form where A and B are constants.
(For the inverse cube law, the orbits are known collectively as Cotes’s spirals.) 7

 

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