MATH10232: SOLUTION SHEET IX
1. Working with forces
(a) The resultant force is simply the vector sum of the three forces
F = F 1 + F 2 + F 3,
⇒
F = ai + 7j − 2k + bj + 5k + i − 7j + ck
⇒
F = (a + 1) i + b j + (3 + c) k.
(b) If the particle moves with a constant velocity then Newton’s first
law states that the resultant force must be zero, so
F = (a + 1) i + b j + (3 + c) k = 0.
Resolving the vector equation into three scalar components gives
(a + 1) = 0,
b = 0,
(3 + c) = 0,
so a = −1, b = 0 and c = −3.
2. Bodies at rest
(a) Newton’s second law states that
m a = F = −W j.
From the lecture notes, or first principles, unit vectors tangential
and normal to the plane are given by
t̂ = cos α i + sin α j
and n̂ = − sin α i + cos α j.
Thus, resolving Newton’s second law into normal and tangential
components gives
Tangential: m a·t̂ = −W j·t̂ = −W j· (cos α i + sin α j) ,
⇒
m a·t̂ = −W sin α.
Normal: m a·n̂ = −W j·n̂ = −W j· (− sin α i + cos α j) ,
⇒
m a·n̂ = −W cos α.
1
(b) If the particle is at rest then the acceleration is zero, so from the
equations above
0 = −W sin α and 0 = −W cos α,
which cannot both be true for general α and W > 0. Therefore,
there must be another force acting on the particle so that the total
force in the tangential and normal directions is zero.
(c) Introducing normal and tangential components of the other force
we have
0 = T − W sin α and 0 = N − W cos α,
where T > 0 and N > 0, provided that α < π/2. Thus, T acts
“up” the plane in the opposite direction to tangential component
of the downward force and resists downward motion. It is called
a friction force. N acts normally away from the plane and is a
reaction force that ensures that the particle does not penetrate
the plane. N can also be called a force of constraint.
3. Gravity: A satellite in orbit
r̂
h
m
R
ME
A satellite of mass m kg is in orbit around at a distance h = 720 km
from the surface of the Earth, which has mean radius R = 6370 km.
The origin of the coordinate system is chosen to be the centre of the
Earth.
2
(a) Newton’s law of gravitation states that
F =−
GmME
r̂.
(R + h)2
We are given that G ≈ 7 × 10−11 m3 kg−1 s−2 and the mass of the
Earth ME ≈ 6 × 1024 kg, so
F =−
=−
7 × 10−11 × 6 × 1024 m [m3 kg−1 s−2 kgkg]
r̂,
((6370 + 720) × 103 )2 [m2 ]
42 × 1013
42 × 1013
mr̂
=
−
mr̂ [kg m s−2 ]
70902 × 106
7.092 × 1012
420
mr̂ ≈ −8.4mr̂ [kg m s−2 ]
=−
7.092
(b) Newton’s second law states that
mr̈ = F ≈ −8.4mr̂,
⇒ r̈ = −8.4r̂
[m s−2 ].
(c) For uniform circular motion
r̈ = −ω 2 r = −ω 2 r r̂,
where r is the radius of the circle. Using the result from part (b)
gives
−8.4r̂ = −ω 2 (R + h)r̂,
8.4
⇒ ω2 =
≈ 1 × 10−6
[s−2 ]
(6370 + 720) × 103
⇒
ω ≈ 0.001
[s−1 ].
(d) The satellite covers ω = 0.001 radians per second, so the time
taken for a complete orbit is
T =
2π
≈ 6283 s;
ω
dividing by 60 we have that the time taken for a complete orbit
is approximately 105 minutes.
3
4. Electromagnetism: A charged particle in a magnetic field
(a) The Lorentz force is given by
F = q (E + v × B) ,
and in the present example q = e, E = 0 and B = B0 k, so
F = e v × B0 k = e B0 v × k.
Hence, from Newton’s second law
ma = F
⇒
mv̇ = e B0 v × k.
(1)
(b) Writing equation (1) into component form gives



  


u̇
u
0
v
m  v̇  = e B0  v  ×  0  = e B0  −u  ;
ẇ
w
1
0
and the three components are
u̇ = Ωv,
v̇ = −Ωu,
ẇ = 0,
where Ω = eB0 /m.
(c) Differentiating the x-component equation with respect to time
gives
ü = Ωv̇,
and using the y-component equation to replace v̇ gives
ü = Ω(−Ωu)
⇒
ü + Ω2 u = 0.
This is the harmonic equation and the general solution is
u = A sin(Ωt) + B cos(Ωt),
where A and B are real constants. Using the x-component equation gives
v = u̇/Ω = A cos(Ωt) − B sin(Ωt).
The z-component is completely decoupled and integrating once
gives
w = V, a constant.
4
(d) Assuming that B = 0 and A = −RΩ then
u = ẋ = −RΩ sin(Ωt),
v = ẏ = −RΩ cos(Ωt),
w = ż = V,
so integrating once, we have
x = R cos(Ωt) + a,
y = −R sin(Ωt) + b,
w = V t + c,
where a, b and c are constants of integration. At t = 0, x = R,
y = 0 and z = 0, which implies that a = b = c = 0 and so
x = R cos(Ωt),
y = −R sin(Ωt),
w = V t.
(e) The particle follows a helical path with axis parallel to the direction of the magnetic field. In the x-y plane the particle moves
clockwise about a circle of radius R and it moves in the z-direction
(parallel to the magnetic field) with constant speed V .
A plot, see Figure 1 of the motion for R = Ω = V = 1 may be
generated in MATLAB with the commands
>>
>>
>>
>>
>>
T = [0:0.001:10];
x = cos(T);
y = -sin(T);
z = T;
plot3(x,y,z);
z
10
8
6
4
2
0
1
y
0.5
1
0.5
0
0
−0.5
x
−0.5
−1
−1
Figure 1: Motion of a charged particle in a uniform magnetic field in the
z-direction.
5
5. Computer Exercises
(a) The equation governing the motion is
mr̈ = −
mGM
r̂,
r2
(2)
and writing the acceleration in plane polar coordinates gives
mGM
m r̈ − r θ̇2 r̂ + 2ṙθ̇ + r θ̈ θ̂ = − 2 r̂.
r
Resolving into components in the r̂- and θ̂-directions gives the
two coupled equations
r̈ − r θ̇2 = −
GM
r2
and 2ṙθ̇ + r θ̈ = 0.
Letting r1 = r, r2 = ṙ, θ1 = θ and θ2 = θ̇ yields the coupled
system of first-order ODEs
ṙ1
ṙ2
θ̇1
θ̇2
=
=
=
=
r2 ,
r1 θ22 − GM/r12 ,
θ2 ,
−2r2 θ2 /r1 .
(3a)
(3b)
(3c)
(3d)
(b) Writing the system of ODEs (3a) with GM = 1 as a MATLAB
M-file, sat.m, gives
function [rdot,thetadot]=sat(r,theta)
rdot(1) = r(2);
rdot(2) = r(1)*theta(2)*theta(2) - 1/(r(1)*r(1));
thetadot(1) = theta(2);
thetadot(2) = - 2*r(2)*theta(2)/r(1);
i. Suitable commands to solve for the specified initial conditions
are
>> h = 0.001;
>> t = [0:h:2];
>> r(1,1) = 1; r(1,2) = 0;
>> theta(1,1) = 0; theta(1,2) = 0;
6
>> for n = 1:2000
[rdot,thetadot] = sat(r(n,:),theta(n,:));
r(n+1,:) = r(n,:) + h*rdot;
theta(n+1,:) = theta(n,:) + h*thetadot;
end
>> plot(t,r(:,1));
>> plot(t,theta(:,1));
The resulting plots of the radius and angle of the satellite
with time are shown in Figure 2 (with the y-range chosen to
be positive). The angle does not change and the satellite is
r
θ
1
1
0.8
0.6
0.8
0.4
0.2
0.6
0
−0.2
0.4
−0.4
−0.6
0.2
−0.8
0
0
0.2
0.4
0.6
0.8
1
−1
t
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
t
2
Figure 2: Motion of a satellite with initial conditions r = 1, ṙ = 0, θ = 0,
θ̇ = 0.
attracted towards the planet until it crashes (r = 0 at t ≈ 1.1).
If you wish to put a termination criterion into to stop the
time-stepping, the Euler loop would be
>> for n = 1:2000
[rdot,thetadot] = sat(r(n,:),theta(n,:));
r(n+1,:) = r(n,:) + h*rdot;
if(r(n+1,1) < 0)
error(’Satellite has hit the planet at t = %g’, h*n);
end
theta(n+1,:) = theta(n,:) + h*thetadot;
end
??? Satellite has hit the planet at t = 1.114
>>
7
The fact that the satellite hits the origin in finite time can be proved by analytically
solving the equation
1
r̈ = − 2 .
r
Multiplying both sides by ṙ and integrating gives
1
1 2
ṙ = + C.
2
r
At time t = 0, r = 1 and ṙ = 0, so C = −1 and
1
1 2
ṙ = − 1
2
r
⇒
ṙ = −
p
2/r − 2,
where the negative branch is chosen because the satellite moves towards the planet.
Separating and integrating gives
Z
⇒
At t = 0, r = 1, so
Thus,
√
√
r
dt = −t + K.
2 − 2r
p
1 −1 √
sin ( r) − r(1 − r) = −t + K.
2
π
1
K = √ sin−1 (1) = √ .
2
2 2
p
π
1 −1 √
−t + √ =
sin ( r) − r(1 − r) ,
2
2 2
and the satellite hits the planet when r = 0,
π
−tcrash + √ = 0
2 2
⇒
π
tcrash = √ ≈ 1.11.
2 2
ii. In this case the Euler loop is the same, but the initial conditions must be changed
>> h = 0.001;
>> t = [0:h:2];
>> r(1,1) = 1; r(1,2) = 0;
>> theta(1,1) = 0; theta(1,2) = 1;
>> for n = 1:2000
[rdot,thetadot] = sat(r(n,:),theta(n,:));
r(n+1,:) = r(n,:) + h*rdot;
theta(n+1,:) = theta(n,:) + h*thetadot;
end
>> plot(t,r(:,1));
>> plot(t,theta(:,1));
8
r
2
θ
1.8
2
1.8
1.6
1.6
1.4
1.4
1.2
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
t
2
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
t
Figure 3: Motion of a satellite with initial conditions r = 1, ṙ = 0, θ = 0,
θ̇ = 1.
The resulting plots of the radius and angle of the satellite with
time are shown in Figure 3.
The radius remains constant at r = 1 and the angle increases
linearly with time. In other words, the satellite orbits the
planet.
iii. In this case, we need to integrate for a longer time to establish
the nature of the motion.
>> h = 0.001;
>> t = [0:h:100];
>> r(1,1) = 1; r(1,2) = 0.1;
>> theta(1,1) = 0; theta(1,2) = 1;
>> %Set dummy final values to allocate memory
>> r(100001,2) = 1; theta(100001,2) = 1;
>> for n = 1:100000
[rdot,thetadot] = sat(r(n,:),theta(n,:));
r(n+1,:) = r(n,:) + h*rdot;
theta(n+1,:) = theta(n,:) + h*thetadot;
end
>> plot(t,r(:,1));
>> plot(t,theta(:,1));
The resulting plots of the radius and angle of the satellite with
time are shown in Figure 4.
The satellite performs oscillations in radius as it orbits the
9
2
1.15
100
r
θ
1.1
90
80
70
1.05
60
50
1
40
30
0.95
20
10
0.9
0
10
20
30
40
50
60
70
80
90
t
100
0
0
10
20
30
40
50
60
70
80
90
t
100
Figure 4: Motion of a satellite with initial conditions r = 1, ṙ = 0.1, θ = 0,
θ̇ = 1.
planet, but the radius does not appear to reach zero, so the
satellite does not crash. In this sense, the orbit can be said to
be stable. The orbit can be seen more clearly by plotting it
in Cartesian coordinates against the reference circular orbit,
see Figure 5
>> th = [0:0.01:2*pi];
>> X = cos(th); Y = sin(th);
>> x = r(:,1)*cos(theta(:,1));
>> y = r(:,1)*sin(theta(:,1));
>> plot(x,y);
>> hold on;
>> plot(X,Y,’r’);
are shown in Figure 4.
The orbit is slightly elliptical, rather than being circular and
appears to be displaced from the “ideal” circular orbit.
iv. In this last case, we must use a smaller timestep to avoid
errors. Initially, I solved this problem with h = 0.001 and
the satellite appeared to crash into the planet, which can be
proved to be impossible1 .
>> h = 0.0001;
>> t = [0:h:50];
>> r(1,1) = 1; r(1,2) = 0;
1
Thanks to Chris Johnson for reminding me of this important result.
10
y
1
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1.5
−1
−0.5
0
0.5
1
1.5
x
Figure 5: Motion of a satellite with initial conditions r = 1, ṙ = 0.1, θ = 0,
θ̇ = 1 shown in blue with a reference circular orbit of radius 1 shown in red.
>> theta(1,1) = 0; theta(1,2) = 0.8;
>> %Set dummy final values to allocate memory
>> r(500001,2) = 1; theta(500001,2) = 1;
>> for n = 1:500000
[rdot,thetadot] = sat(r(n,:),theta(n,:));
r(n+1,:) = r(n,:) + h*rdot;
>> plot(t,r(:,1));
>> x = r(:,1).*cos(theta(:,1));
>> y = r(:,1).*sin(theta(:,1));
>> plot(x,y);
The resulting figures are shown in Figure 6. The satellite
should establish a stable elliptical orbit shown in red. If you
look closely you will see that the amplitude of the oscillations
is slowly changing, which is due to a build up of numerical
error in Euler’s method. This is why using a small timestep
is essential. The numerical errors mean that the system loses
angular momentum, which should be conserved (see below),
and the satellite will eventually hit the plant in our simulation
... but this effect is due only to numerical errors. In other
words, to simulate the motion really accurately we should
use a numerical scheme the explicitly conserves all conserved
quantities.
The fact that the satellite can never crash into the (point mass) planet can be proved
11
r
1
0.8
x
0.6
0.9
0.4
0.8
0.2
0.7
0
−0.2
0.6
−0.4
0.5
−0.6
0.4
0
5
10
15
20
25
30
35
40
45
t
−0.8
−0.5
50
0
y
0.5
Figure 6: Motion of a satellite with initial conditions r = 1, ṙ = 0, θ = 0,
θ̇ = 0.8. The analytic solution is shown in red
by the following analysis. The equations of motion are
r̈ − r θ̇ 2 = −
1
r2
and
2ṙ θ̇ + r θ̈ = 0.
We can integrate the second equation analytically after multiplication by r
d 2 r θ̇
dt
2r ṙ θ̇ + r 2 θ̈ =
⇒
r 2 θ̇ = h,
a constant.
The constant h is called the angular momentum.
We can actually convert the remaining equation into a linear second-order ODE with
constant coefficients by letting u = 1/r, so that
1 dr
1 ṙ
ṙ
du
=− 2
=− 2 =−
dθ
r dθ
r θ̇
h
⇒
ṙ = −h
du
;
dθ
and
d2 u
1 r̈
1 d
1 r̈
r̈
=−
=−
(ṙ) = −
=− 2 2
dθ 2
h dθ
h θ̇
h h/r 2
h u
⇒ r̈ = −h2 u2
d2 u
.
dθ 2
The r̂-component of the momentum equation can now be rewritten in terms of u
r̈ − r θ̇ 2 = −
1
r2
⇒
−h2 u2
d2 u
− h2 u3 = −u2 ,
dθ 2
which is a linear ODE for u(θ),
d2 u
1
+u = 2.
dθ 2
h
We now have an analytic solution for the path of the satellite:
u=
12
1
+ A cos θ + B sin θ.
h2
1
Without loss of generality, we can choose B = 0 and then
u=
1
1
= 2 + A cos θ
r
h
⇒
r=
h2
.
1 + Ah2 cos θ
Hence, r can only be zero if h2 = 0, or if (1 + Ah2 cos θ) → ∞, but because A and
cos θ are bounded this latter case is not possible. Thus, r < 0 if h > 0 and the only
case in which the satellite can hit the origin is when h = 0, in other words when the
satellite is not orbiting, case (i).
In the present example h = 1 × 0.82 = 0.64, so
r=
0.64
,
1 + 0.64A cos θ
but r = 1 at θ = 0 and
1=
0.64
1 + 0.64A
⇒
1 + 0.64A = 0.64
Thus,
r=
0.64
,
1 − 0.36 cos θ
which is shown as the red ellipse in Figure 6.
13
⇒ A = 1 − 1/0.64 = −0.5625.