Download Two-Body Problem: Motion of Two Massive Objects Relative to an Inertial Frame and more Study notes Mechanical Engineering in PDF only on Docsity!
The Two-Body Problem
Abstract In my short essay on Kepler’s laws of planetary motion and Newton’s law of universal gravitation, the trajectory of one massive object near another was shown to be a conic section. However, the frame of reference was a coordinate system centered at one of the masses (typically the heavier). Unfortunately, since even a heavier object will exhibit some acceleration toward the lighter one, such a frame cannot be regarded as “inertial,” meaning that Newton’s laws are not entirely valid (without the introduction of “fictitious forces”). The present treatment of eliminates this objection; however, relativistic effects are still not considered.
Two-Body Problem Relative to Inertial Frame
center of mass → R → →
mass m 1
mass m 2
x 2 x 1
Figure 1: Two-Body Diagram
We consider the motions of two mas- sive objects—which we’ll call planets—relative to an inertial frame of reference, subject only to the mu- tual forces of gravitation. For con- venience, we’ll refer to the bodies as planets. Thus, we have a planet of mass m 1 whose position vector is −→ x 1 as well as a planet of mass m 2 with position vector −→ x 2. In terms of −→ x 1 and −→ x 2 , the center of mass is located with position vector −→ R =
M
m 1
−→ x 1 +m 2
−→ x 2
where M = m 1 + m 2.
The force applied to each of the planets has, by Newton’s Law of Universal Gravitation, magnitude
F = Gm 1 m 2 r^2
where r = | −→ x 1 − −→ x 2 | is the distance between the planets, and where G is the universal gravitational constant, whose measured value in SI units is approx- imately G = 6. 67 × 10 −^11 N · m^2 /kg^2. Since these forces are opposite each other, Newton’s Second Law immediately implies that
(m 1 + m 2 )
d^2 dt^2
−→ R = m 1
d^2 dt^2
−→ x 1 +m 2
d^2 dt^2
−→ x 2 =
−→
This already says that the center of mass of the two-body system exhibits no acceleration relative to the inertial frame of reference. Therefore the moving center of gravity can itself be taken as an inertial frame of reference.
Equations Relative to Center of Mass
→
m 2 m 1 R
center-of-mass coordinate system
mass m 2
mass m 1
→ R
Figure 2: Center-of-mass frame
Since a non-rotating coordinate system having origin at the center of mass of our two-body system is an inertial frame, we shall ignore the movement of the center of mass and concentrate only on the movements of the plan- ets relative to their common center of mass. Given this, we’ll re-define the
position vector
−→ R to be that point- ing from the center of mass to Planet 1 (having mass m 1 ). Knowing how Planet 1 moves about the center of mass will dictate Planet 2’s motion as well, since Planet 2’s position vector
is just −
m 2 m 1
−→ R. This configuration is depicted in Figure 2. The task, then, is
to compute how the position vector
−→ R of Planet 1 evolves with time. Rather than letting r be the distance between the two planets, we let r be the length of
the position vector
−→ R. With this understanding, the distance between the two
planets is simply
m 1 + m 2 m 2
r. Using Newton’s law of universal gravitation,
and denoting by
−→ r the unit vector in the direction of
−→ R , we immediately obtain
m 1 d^2 dt^2
−→ R = − Gm 1 m 2 ( m 1 +m 2 m 2
r^2
−→ r.
This gives the acceleration of Planet 1 in this frame of reference:
−→ a = d^2 dt^2
−→ R = − Gm^32 (m 1 + m 2 )^2 r^2
−→ r.
Now comes the math! Let’s gather together all the constants into a single
The acceleration is the time derivative of Equation (2): −→ a =
d dt
−→ v =
d dt
dr dt
−→ r +r
dθ dt
−→ s
d^2 r dt^2
−→ r +
dr dt
dθ dt
−→ s +r
d^2 θ dt^2
−→ s −r
dθ dt
−→ r
d^2 r dt^2 − r
dθ dt
−→ r +
dr dt
dθ dt
d^2 θ dt^2
−→ s
However, recalling from Equation (1) that −→ a = −
K
r^2
−→ r , we extract two equations
of interest: d^2 r dt^2
− r
dθ dt
K
r^2
and
2
dr dt
dθ dt
d^2 θ dt^2
We can immediate extract useful information from Equation (4), namely that d dt
r^2
dθ dt
= 2r
dr dt
dθ dt
d^2 θ dt^2
This means that the quantity L = r^2
dθ dt
is a constant.
Next, define the variable u =
L^2
Kr
and that dθ dt
L
r^2
K^2 u^2 L^3
We have, using the Chain Rule, that
dr dt
d dθ
L^2
Ku
dθ dt
L^2
Ku^2
dθ dt
du dθ
K
L
du dθ
Differentiate again and obtain
d^2 r dt^2
d dθ
dr dt
dθ dt
K
L
d^2 u dθ^2
dθ dt
K^3 u^2 L^4
d^2 u dθ^2
Substitute into Equation (3) and get
K^3 u^2 L^4
d^2 u dθ^2
L^2
Ku
K^2 u^2 L^3
K^3 u^2 L^4
Dividing by the common factor of −
K^3 u^2 L^4
results in the inhomogeneous second-
order differential equation: d^2 u dθ^2
The general solution of this has the form
u = u(θ) = 1 + e cos(θ − θ 0 ),
where e and θ 0 are constants of integration which can be determined from the initial conditions. In turn, we now know that the radial distance of Planet 1 from the center of mass is given, in terms of θ, by
r =
L^2
Ku
L^2
K(1 + e cos(θ − θ 0 ))
where the constants K and L are
K =
Gm^32 (m 1 + m 2 )^2
, and L = r^2
dθ dt
The Center-of-Mass Inertial Frame and the Eccentricity
The two-body problem will have in all four initial conditions: the two initial positions of the planets and the two initial velocities. Correspondingly, there are four constants of integration which occur: the initial position and velocity of the center of mass relative to the original inertial frame, and the constants e and θ 0 appearing in Equation (5). The motion of the center of mass is easily determined. If −→ x 1 (0) and −→ x 2 (0) are the initial positions of Planets 1 and 2, respectively, then the initial position of the center of mass is
−→ R (0) =
m 1 + m 2
(m 1 −→ x 1 (0) + m 2 −→ x 2 (0)).
where u 0 is the magnitude of the vector −→ u 1 (0). Therefore,
e sin θ 0 = −
Lu 0 cos α K
Squaring and adding Equations (7) and (8) leads to
e^2 =
L^2
Kr 0
L^2 u^20 cos^2 α K^2
Equation (9) can be written purely in terms of the initial data once we realize that the constant L can be expressed as
L = r^2
dθ dt
= r 0 u 0 sin α,
giving rise to
e^2 =
r 0 u^20 sin^2 α K
r^20 u^40 cos^2 α K^2
Since K is the constant
K = Gm^21 m 2 (m 1 + m 2 )^2
we see that the eccentricity of the orbit of Planet 1 (and hence of Planet 2) about the moving center of mass can be expressed purely in terms of the initial data r 0 , u 0 , the angle α, the masses m 1 and m 2 and the gravitational constant G.
One Planet’s Orbit Relative to the Other
In the above, we have shown that each member planet of a two-body system will orbit about the system’s center of mass in either an ellipse, a parabola, or a hyperbola, depending on the eccentricity. A question that remains is whether one planet’s orbit about the other will likewise be a conic section. If we can show this, then Kepler’s First Law will be validated, namely that the planets in the solar system orbit about the sun in elliptical orbits.
r= L
2 K(1+ecos(θ−θ 0 ))
center of mass
m 2
d=distance of Planet 1 m^1 from Planet 2 = m 1 +m 2
( m 2 )r
r=distance of Planet 1 from center of mass
r
moving coordinate system at Planet 1
x
y
θ θ
g (x ) = 1-^ x^24 f^ (x^ )^ =^ -^ 1-^ x 42
Figure 5: Coordinate System at Planet 1
The demonstration is actually eas- ier than one might think! Figure 5 depicts both the center-of-mass coor- dinate system, with planet 1’s orbit about the center of mass, as well as the moving coordinate system centered at Planet 1. We are to show that, in po- lar coordinates based at Planet 1, the orbit of Planet 2 is also a conic section with the same eccentricity. If r is the polar distance from the center of mass to Planet 1, then in terms of the radial angle θ, we have already shown that
r =
L^2
K(1 + e cos(θ − θ 0 ))
Since the distance d between Planets 1 and 2 is given by
d =
(m 1 + m 2 )r m 2
it follows that the polar distance s from Planet 1 to Planet 2 must be (note the minus sign!):
s = −
L^2 (m 1 + m 2 ) Km 2 (1 + e cos(θ − θ 0 ))
which also describes a conic section with the same eccentricity e.
