Orbital Motion
1
Newton's Law of Universal Gravitation
Consider any two masses
m1
and
m2
which are separated by a distance
r.
Newton's law of universal
gravitation states that the masses each feel an attractive force toward the other, and this force has a
magnitude given by
m1 m2
r2
Fg = G
where
G
(1)
is the universal gravitational constant, and has a value of
6.67429 × 10−11 N · m2 /kg 2 .
The force
is directed along the line connecting the masses, and is always attractive. In vector form, we can express the
force exerted
on m2 by m1 as
2
F~ 12 = −G mr1 m
r̂12
(2)
2
where r̂12 is a unit vector pointing from m1 to m2 . The negative sign indicates the attractive nature of the
force; see Figure 1.
Figure 1: Gravitational force between two masses.
Note that by Newton's Third Law, the force exerted
opposite in direction to
2
F~ 21 .
on m2 by
~ 12 )
m1 (F
is equal in magnitude and
Earth's Gravity
So far we have considered the gravitational force between pairs of point-like masses. We can also consider
gravitation between larger objects. It can be shown that for a uniform, spherical distribution of mass, we
get the same results whether we treat it as a sphere of mass, or as a point particle located at the sphere's
center (that point would have the same mass as the sphere).
Consider again the gravitational force between two masses.
Let one of the two masses be the Earth,
ME = 5.97 × 1024 kg . We will treat the Earth as a point mass located at the Earth's
center. If the second mass m is on the Earth's surface, then the distance between that mass and the center
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of the Earth is just the Earth's radius, RE = 6.37 × 10 m. Using these values in Eq. (1) yields
which has a mass
Fg = G
ME m
2 .
RE
1
(3)
The force is directed toward the Earth's center.
If the mass
m
is above the Earth's surface, then we need to adjust the distance between
Earth's center. Let
m
h
be the amount by which
and the center of the Earth is
(RE + h).
m
ME m
.
(RE + h)2
We can learn more by combining Eq. (4) with Newton's second law,
Fg = G
the
m
a=g
F = ma:
(5)
since we are talking about the acceleration of
g=G
g
(4)
ME m
= mg
(RE + h)2
from both sides to obtain
which gives the value of
and the
Using this in Eq. (1) gives
Fg = G
where we have used
m
is above the Earth's surface. Then the distance between
m
due to gravity. We can cancel
ME
(RE + h)2
(6)
as a function of height above the Earth's surface. Notice that
g
decreases with
increasing height above the Earth's surface.
3
Conservation of Angular Momentum: Planar Orbits and Kepler's
Second Law
Consider a planet orbiting the Sun, as shown in Figure 2.
The following argument holds as long as the
orbiting mass is much smaller than the central mass, and so the results are also applicable to the Moon
orbiting the Earth, for example. The only force in this situation is the Sun's gravitational pull on the planet,
which is keeping it in orbit. (Since the Sun is much more massive than the planet, we neglect the planet's
gravitational pull on the Sun.)
Figure 2: Gravitational force between a planet orbiting the Sun.
First we note that the torque
is dened as
In this case,
~τ
on the planet due to the Sun's pull is zero. To see this, recall that torque
~.
~τ = ~r × F
~r is
a vector from the Sun to the planet, and
that these two vectors are antiparallel (that is, 180
◦
(7)
F~ is the force on the planet due to the sun. Notice
apart), and so their cross product (the torque) is zero.
The fact that the torque is zero is very important, due to the following equation:
X
~τ =
~
dL
.
dt
A zero torque therefore means that the planet's angular momentum
means it is constant in both direction and magnitude.
2
(8)
L~ is constant. L~ is a vector, so this
~ Planar Orbits
3.1 Constant direction of L:
~ = ~r ×~p, where ~p is the planet's linear momentum. We know that because of the cross product
Recall that L
~ will always be perpendicular to both ~r and ~p. Another way of saying this is that L
~ will
in this equation, L
~
always be perpendicular to a plane that contains ~
r and ~p. The fact that the direction of L is constant
means that this plane never moves. As a result, ~
r and ~p may rotate as the planet moves around the Sun,
but they are conned to a plane. So we see that the conservation of angular momentum is the reason that
the planets orbit in planes.
~ Kepler's Second Law
3.2 Constant magnitude of L:
~ being constant. Consider Figure 3.
We now consider the implications of the magnitude of L
Figure 3: Gravitational force between a planet orbiting the Sun.
~r
As the planet orbits the Sun, the radial vector
rotates. In a small time interval
amount and the line of the vector sweeps out a small area
dA,
dt, ~r
moves a small
shown as the shaded region in Figure 3. This
area is half the area of the dashed parallelogram shown in the gure. A useful property of the cross product
is that its magnitude is equal to the area of a parallelogram formed by the two vectors in the cross product.
In this case, the parallelogram is bound by the vectors
~r
and
d~r.
So,
dA
is equal to half of the magnitude of
the cross product of these two vectors:
dA =
Now using
~v = d~r/dt
and
~p = Mp~v
dA =
Now we recall that
(where
Mp
1
|~r × d~r|.
2
(9)
is the mass of the planet), we get
~p
1
1
1
|~r × d~r| = |~r × ~vdt| = |~r ×
|dt
2
2
2
Mp
(10)
L~ = ~r × ~p, which allows us to write
dA =
~
L dt
1 L
|
dt| =
2 Mp
2Mp
(11)
or, nally,
dA
L
=
dt
2Mp
which is a constant.
(12)
This says that the planet's rate of sweeping out area is constant, which is Kepler's
Second Law of planetary motion.
4
Computation: Simulating the Moon's Orbit Around the Earth
Your task is to use a spreadsheet to simulate the orbit of the Moon around the Earth. You will use Newton's
Law of Universal Gravitation, given in Eq. (1). Since we know that the Moon will orbit in a plane (due
to conservation of angular momentum, as discussed above), you can dene a two-dimensional coordinate
system in that plane, and not have to worry about a third dimension of motion. See Figure 4.
3
y
m
r
Fg
θ
x
ME
Figure 4: Coordinate system for Moon orbiting Earth.
Starting with Newton's Law of Universal Gravitation,
|F~g | = G
we can extract the
x
and
y
F = ma
(13)
components as
Fgx = −Fg cos θ
(14)
Fgy = −Fg sin θ,
(15)
where the negative signs are due to the direction of
Law
m1 m2
r2
F~g
in Figure 4. Combining these with Newton's Second
yields
X
Fx = −Fgx = max = −G
mME
cos θ
r2
(16)
X
Fy = −Fgy = may = −G
mME
sin θ
r2
(17)
ax = −G
ME
cos θ
r2
(18)
ME
sin θ
r2
y = r sin θ to get
ay = −G
In these equations we can use
Now using
r 2 = x2 + y 2
x = r cos θ
and
(19)
ax = −G
ME
x
r3
(20)
ay = −G
ME
y
r3
(21)
we can write, nally,
ax = −G
ay = −G
ME
(x2
3
x
(22)
3
y
(23)
+ y2 ) 2
ME
(x2
4
+ y2 ) 2
These acceleration equations are the main equations in the simulation, as they capture the physics of the
Earth always pulling the Moon toward it. We wish to calculate the position of the Moon for each time step
of the simulation, and so from the Euler method we have
where
∆t
x(t + ∆t) ≈ x(t) + vx (t)∆t
(24)
y(t + ∆t) ≈ y(t) + vy (t)∆t
(25)
is the time step of the simulation. We now have equations for the position as a function of time,
but those equations depend on the velocity. Again using the Euler method, we can dene equations that
give the velocity as a function of the accelerations dened above:
vx (t + ∆t) ≈ vx (t) + ax (t)∆t
(26)
vy (t + ∆t) ≈ vy (t) + ay (t)∆t
(27)
4.1 The Euler-Cromer method
In the Euler-Cromer method, we make a small change: we use
v(t + ∆t) instead
of
v(t) in
Eqs. (24-25). The
new set of equations is then
x(t + ∆t) ≈ x(t) + vx (t + ∆t)∆t
(28)
y(t + ∆t) ≈ y(t) + vy (t + ∆t)∆t
(29)
Notice that with this change, it is necessary to calculate
and
5
y(t + ∆t).
vx (t+∆t) and vy (t+∆t) before calculating x(t+∆t)
With the Euler method, the order did not matter.
Exercises
5.1 Euler Method
Recall that in the lab when we built the gravitational model of the moon orbiting the Earth, we employed the
modied Euler method (also known as the Euler-Cromer method above) to complete the model. To gain an
appreciation of why it is important to use this modied algorithm, reprogram your Earth-moon model using
the unmodied Euler method, instead of the Euler-Cromer method. What happens to your model (Hint:
something disastrous!)? Provide plots of the moon's trajectory to demonstrate this disastrous behavior. Can
you get rid of the articial behavior by making
∆t
smaller? Finally, try to make an intelligent comment
about why the Euler-Cromer algorithm works, and the unmodied Euler algorithm does not.
5.2 Moon's Gravitational Inuence
Assume the moon is located at its average distance from the Earth at 3.83×10
8
m. Use the spreadsheet (use
separate columns than the ones used for the Earth-moon model) to nd the distance from the Earth, along
the line connecting the centers of the Earth and moon, where the total gravitational eld
gtotal
due to both
the moon and Earth is zero. Remember that the magnitude of the gravitational eld is dened as the force
per unit mass (i.e the gravitational eld at a distance
r
from the center of the Earth is
g = GME /r2 ).
5.3 Trans Lunar Injection
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Assume that your Saturn V rocket is in a parking orbit at an altitude of 2×10
m with a tangential speed
of 7800 m/s. This distance and speed were typical of the Apollo missions wherein the Earth was orbited 2-3
times before the nal burn of the remaining fuel in stage 3. How long does it take to orbit the Earth (i.e.
what is the orbital period) for this parking orbit?
5
Further, assume that you have a sucient amount of fuel left to increase your rocket's speed as needed.
Use your gravitational model, and the result from exercise 5.2, to determine what speed must be given to
the rocket in the nal burn for stage 3 such that the payload will travel a sucient distance, via Trans Lunar
Injection, to come under the predominant inuence of the gravitational pull of the moon. How does this
speed compare to the minimum escape velocity for the Earth? How long does it take to get to this point
where the payload is predominantly aected by the moon's gravity?
5.4 Kepler's Third Law
Reprogram the gravitational model to simulate planets orbiting the Sun. With the information in the table
below, use the gravitational model to verify Kepler's 3rd Law. All of the planets except Mercury and Pluto
(Pluto was actually recently demoted from planetary status) move in very nearly circular orbits. Thus, the
values in the table can be considered to be radii of circular orbits for all except Mercury and Pluto, which
case the are their semi-major axes. Hint: This verication can best be accomplished by making another plot
involving
T
and
r.
Planet
radius (AU)
Mercury
0.39
Venus
0.72
Earth
1.00
Mars
1.52
Jupiter
5.20
Saturn
9.54
Uranus
19.19
Neptune
30.06
Pluto
39.53
Table 1: Planetary radii. 1 AU (astronomic unit) is equal to the average distance from the earth to the Sun.
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