Introduction to Celestial Mechanics:
Solutions to Exercises
RICHARD FITZPATRICK
University of Texas at Austin
,
,1 Newtonian mechanics
Solutions to exercises
1.1 Consider the motion of a point object whose position vector is r = (x, y, z) in one
frame of reference, and r = (x , y , z ) in another. Let the origin of the latter frame
be moving at the instantaneous velocity u(t) = [u x (t), uy (t), uz (t)] with respect to that
of the former, and let both origins coincide at t = 0. The appropriate generalization
of the Galilean coordinate transform (1.1)–(1.3) is
t
x = x− u x (t ) dt , (1)
0
t
y = y − uy (t ) dt , (2)
0
t
z =z− uz (t ) dt . (3)
0
It follows, from differentiation of the above equations with respect to time, that
dx dx
= − ux, (4)
dt dt
dy dy
= − uy , (5)
dt dt
dz dz
= − uz , (6)
dt dt
and
d2 x d2 x du x
= − , (7)
dt dt dt
d 2 y d2 y duy
= − , (8)
dt dt dt
d 2 z d2 z duz
= − . (9)
dt dt dt
Let v = (dx/dt, dy/dt, dz/dt) and v = (dx /dt, dy /dt, dz /dt) be the velocities in
the former and latter frames of reference, respectively. Likewise, let a = (d2 x/dt2 ,
d2 y/dt2 , d2 z/dt2 ) and a = (d2 x /dt2 , d2 y /dt2 , d2 z /dt2 ) be the corresponding ac-
celerations. It follows from (4)–(6) that
v = v − u, (10)
1
,2 Newtonian mechanics
which is identical to (1.7), and from (7)–(9) that
du
a = a − , (11)
dt
which is identical to (1.12).
1.2 We have
r × F = τ. (12)
Let us write
F
r = r⊥ + λ , (13)
F
where F · r⊥ = 0. The previous two expressions combine to give
r⊥ × F = τ. (14)
Since λ does not appear in the above equation, we deduce that this is quantity unde-
termined. Moreover,
F × (r⊥ × F) = F 2 r⊥ = F × τ, (15)
where use has been made of (A.1), as well as the fact that F · r⊥ = 0. Hence,
F×τ
r⊥ = , (16)
F2
and
F×τ F
r= +λ , (17)
F2 F
where λ is arbitrary.
1.3 The net force acting on an isolated system is zero, which implies that
FA + FB = 0. (18)
Hence,
FB = −FA . (19)
Likewise, the net torque acting on an isolated system is zero (assuming that the in-
ternal forces are central), which implies that
τA + τB = 0, (20)
where τA and τB are the torques associated with the forces FA and FB , respectively.
Hence,
τB = −τA . (21)
The line of action of FA is
FA × τA FA
rA = + λA . (22)
F A2 FA
, 3 Solutions to exercises
Likewise, the line of action of FB is
FB × τB FB
rB = + λB . (23)
F B2 FB
Thus, it follows from (19) and (21) that
FA × τA FA
rB = 2
− λB . (24)
FA FA
However, (22) and (24) are the equations of the same straight-line, since λA and λB
are arbitrary (so we can set λB = −λA without loss of generality). Hence, FB = −FA ,
and the two forces have the same line of action.
1.4 Let τ1 , τ2 , and τ be the torques associated with the forces F1 , F2 , and F, respectively.
In order for the single force F to be equivalent to the combined action of the two
forces F1 and F2 , the former force must give rise to the same net force and net torque
as the latter combination of forces. In other words,
F1 + F2 = F, (25)
τ1 + τ2 = τ. (26)
If r1 , r2 , and r are the locii of the lines of action of the three forces F1 , F2 , and F,
respectively, then the previous equation implies that
r1 × F1 + r2 × F2 = r × F. (27)
Let us write
r1 = r1 ⊥ + λ1 F1 , (28)
r2 = r2 ⊥ + λ2 F2 , (29)
where r1 ⊥ · F1 = r2 ⊥ · F2 = 0, and λ1 , λ2 are arbitrary. It follows from (27) that
r1 ⊥ × F1 + r2 ⊥ × F2 = r × F. (30)
Taking the scalar product with F, and making use of (25), we obtain
(r2 ⊥ − r1 ⊥ ) · F1 × F2 . (31)
According to the above equation, there are two possibilities: 1) F1 is parallel (or
anti-parallel) to F2 , in which case F is parallel (or anti-parallel) to F1 and F2 , but the
direction of r2 ⊥ − r1 ⊥ is arbitrary; or 2) F2 is not parallel (or anti-parallel) to F2 , in
which case r2 ⊥ − r1 ⊥ is constrained to be coplanar with F1 and F2 . In the first case,
the forces F1 , F2 , and F are all parallel (or anti-parallel) to one another, but their lines
of action do not necessarily coincide. In the second case, we can write
r2 ⊥ = r1 ⊥ + α F 1 + β F 2 , (32)
where α and β are fixed constants. Thus, the lines of action of F1 and F2 take the
forms
r1 = r1 ⊥ + λ1 F1 , (33)
r2 = r1 ⊥ + α F1 + (β + λ2 ) F2 . (34)
Solutions to Exercises
RICHARD FITZPATRICK
University of Texas at Austin
,
,1 Newtonian mechanics
Solutions to exercises
1.1 Consider the motion of a point object whose position vector is r = (x, y, z) in one
frame of reference, and r = (x , y , z ) in another. Let the origin of the latter frame
be moving at the instantaneous velocity u(t) = [u x (t), uy (t), uz (t)] with respect to that
of the former, and let both origins coincide at t = 0. The appropriate generalization
of the Galilean coordinate transform (1.1)–(1.3) is
t
x = x− u x (t ) dt , (1)
0
t
y = y − uy (t ) dt , (2)
0
t
z =z− uz (t ) dt . (3)
0
It follows, from differentiation of the above equations with respect to time, that
dx dx
= − ux, (4)
dt dt
dy dy
= − uy , (5)
dt dt
dz dz
= − uz , (6)
dt dt
and
d2 x d2 x du x
= − , (7)
dt dt dt
d 2 y d2 y duy
= − , (8)
dt dt dt
d 2 z d2 z duz
= − . (9)
dt dt dt
Let v = (dx/dt, dy/dt, dz/dt) and v = (dx /dt, dy /dt, dz /dt) be the velocities in
the former and latter frames of reference, respectively. Likewise, let a = (d2 x/dt2 ,
d2 y/dt2 , d2 z/dt2 ) and a = (d2 x /dt2 , d2 y /dt2 , d2 z /dt2 ) be the corresponding ac-
celerations. It follows from (4)–(6) that
v = v − u, (10)
1
,2 Newtonian mechanics
which is identical to (1.7), and from (7)–(9) that
du
a = a − , (11)
dt
which is identical to (1.12).
1.2 We have
r × F = τ. (12)
Let us write
F
r = r⊥ + λ , (13)
F
where F · r⊥ = 0. The previous two expressions combine to give
r⊥ × F = τ. (14)
Since λ does not appear in the above equation, we deduce that this is quantity unde-
termined. Moreover,
F × (r⊥ × F) = F 2 r⊥ = F × τ, (15)
where use has been made of (A.1), as well as the fact that F · r⊥ = 0. Hence,
F×τ
r⊥ = , (16)
F2
and
F×τ F
r= +λ , (17)
F2 F
where λ is arbitrary.
1.3 The net force acting on an isolated system is zero, which implies that
FA + FB = 0. (18)
Hence,
FB = −FA . (19)
Likewise, the net torque acting on an isolated system is zero (assuming that the in-
ternal forces are central), which implies that
τA + τB = 0, (20)
where τA and τB are the torques associated with the forces FA and FB , respectively.
Hence,
τB = −τA . (21)
The line of action of FA is
FA × τA FA
rA = + λA . (22)
F A2 FA
, 3 Solutions to exercises
Likewise, the line of action of FB is
FB × τB FB
rB = + λB . (23)
F B2 FB
Thus, it follows from (19) and (21) that
FA × τA FA
rB = 2
− λB . (24)
FA FA
However, (22) and (24) are the equations of the same straight-line, since λA and λB
are arbitrary (so we can set λB = −λA without loss of generality). Hence, FB = −FA ,
and the two forces have the same line of action.
1.4 Let τ1 , τ2 , and τ be the torques associated with the forces F1 , F2 , and F, respectively.
In order for the single force F to be equivalent to the combined action of the two
forces F1 and F2 , the former force must give rise to the same net force and net torque
as the latter combination of forces. In other words,
F1 + F2 = F, (25)
τ1 + τ2 = τ. (26)
If r1 , r2 , and r are the locii of the lines of action of the three forces F1 , F2 , and F,
respectively, then the previous equation implies that
r1 × F1 + r2 × F2 = r × F. (27)
Let us write
r1 = r1 ⊥ + λ1 F1 , (28)
r2 = r2 ⊥ + λ2 F2 , (29)
where r1 ⊥ · F1 = r2 ⊥ · F2 = 0, and λ1 , λ2 are arbitrary. It follows from (27) that
r1 ⊥ × F1 + r2 ⊥ × F2 = r × F. (30)
Taking the scalar product with F, and making use of (25), we obtain
(r2 ⊥ − r1 ⊥ ) · F1 × F2 . (31)
According to the above equation, there are two possibilities: 1) F1 is parallel (or
anti-parallel) to F2 , in which case F is parallel (or anti-parallel) to F1 and F2 , but the
direction of r2 ⊥ − r1 ⊥ is arbitrary; or 2) F2 is not parallel (or anti-parallel) to F2 , in
which case r2 ⊥ − r1 ⊥ is constrained to be coplanar with F1 and F2 . In the first case,
the forces F1 , F2 , and F are all parallel (or anti-parallel) to one another, but their lines
of action do not necessarily coincide. In the second case, we can write
r2 ⊥ = r1 ⊥ + α F 1 + β F 2 , (32)
where α and β are fixed constants. Thus, the lines of action of F1 and F2 take the
forms
r1 = r1 ⊥ + λ1 F1 , (33)
r2 = r1 ⊥ + α F1 + (β + λ2 ) F2 . (34)
