Download Areal Velocity and Orbital Period

6.0 Introduction to Orbital Mechanics
We will now discuss the problem of two bodies orbiting each other in space; for example, the sun
and a planet. The only force between the bodies is given by Newton’s law of gravitation:
𝐹12 =
𝐺𝑚1 𝑚2
𝑟2
where
𝐺 = 6.67 × 10−11
m3
kg ∙ s2
is the universal gravitation constant, r is the distance between the two bodies and m1 and m2 are the
masses of the two bodies.
y
x
Y
m2
r2
X
r1
m1
Let us set up our coordinate system as shown above, with the moving x axis aligned with the line
between the bodies. The vectors r1 and r2 are drawn between the center of mass of the two bodies
and the bodies. Thus
𝑟 = |𝐫2 − 𝐫1 |
Note that there are no external forces on the system, so that
∑𝐹 = (𝑚1 + 𝑚2 )𝐫̈𝑐 = 0
or
𝐫̈𝑐 = 0
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i.e. the center of mass is motionless. We can use the center of mass as the origin of the fixed
coordinate system, since it does not move.
Let us now use the five-term acceleration equation to find the acceleration of the second mass, m2.
𝐚2 = 𝐫̈0 + 𝐬̈ 𝑟2 + 𝛂 × 𝐬2 + 2𝛚 × 𝐬̇ 𝑟2 + 𝛚 × (𝛚 × 𝐬2 )
In this case we have
𝐫̈0 = 𝐫̈𝑐 = 0
𝐬̈ 𝑟2 = 𝑟̈2 𝑖̂
𝛚 = 𝜃̇𝑘̂
𝛂 = 𝜃̈ 𝑘̂
Performing the cross products gives
𝑟2
0
𝛂 × 𝐬2 = {0} × { 0 } = 𝜃̈𝑟2 𝑗̂
0
𝜃̈
0
𝑟̇2
2𝛚 × 𝐬̇ 𝑟2 = 2 {0} × { 0 } = 2𝜃̇𝑟̇2 𝑗̂
0
𝜃̇
𝑟2
0
0
0
𝛚 × (𝛚 × 𝐬2 ) = 𝛚 × ({0} × { 0 }) = {0} × {𝜃̇ 𝑟2 } = −𝜃̇ 2 𝑟2 𝑖̂
0
𝜃̇
𝜃̇
0
Thus, the total acceleration of body two is
−𝜃̇ 2 𝑟2 + 𝑟̈2
𝐚2 = { 𝜃̈𝑟2 + 2𝜃̇ 𝑟2̇ }
0
Similarly, the acceleration for body one can be written
𝜃̇ 2 𝑟1 − 𝑟̈1
𝐚1 = {−𝜃̈𝑟1 − 2𝜃̇𝑟1̇ }
0
The only force acting on each mass is the gravitational attraction, which exists on a line between the
bodies (in the local x direction).
2
𝐅1 =
𝐺𝑚1 𝑚2
𝑟2
𝑖̂
𝐅2 = −
𝐺𝑚1 𝑚2
𝑟2
𝑖̂
We can combine these with the accelerations in the x direction to get
𝐺𝑚1 𝑚2
= (𝜃̇ 2 𝑟1 − 𝑟̈1 )𝑚1
𝑟2
𝐺𝑚1 𝑚2
= (𝜃̇ 2 𝑟2 − 𝑟̈2 )𝑚2
2
𝑟
Since there are no forces in the y direction we can write
𝜃̈𝑟1 + 2𝜃̇ 𝑟̇1 = 0
𝜃̈𝑟2 + 2𝜃̇ 𝑟̇2 = 0
We now have four equations to solve. We can simplify them slightly if we note that
𝑟 = 𝑟1 + 𝑟2
We can also use the definition of the center of mass to help eliminate r1 and r2 from the equations.
𝐫𝑐 = 0 =
−𝑟1 𝑚1 + 𝑟2 𝑚2
𝑖̂
𝑚1 + 𝑚2
Solving for r2 gives
𝑟2 =
𝑚1
𝑟
𝑚2 1
Substituting this into the preceding equation gives
𝑟1 = 𝑚
𝑚2
1 +𝑚2
𝑟
𝑟2 = 𝑚
𝑚1
1 +𝑚2
𝑟
Now place these relations into the first force equation above
𝐺𝑚1 𝑚2 𝑚1 𝑚2 𝜃̇ 2
𝑚1 𝑚2
=
𝑟−
𝑟̈
2
𝑟
𝑚1 + 𝑚2
𝑚1 + 𝑚2
𝐺(𝑚1 + 𝑚2 )
= 𝜃̇ 2 𝑟 − 𝑟̈
𝑟2
Now define
𝜇 = 𝐺(𝑚1 + 𝑚2 )
The orbital differential equations are then
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𝜇
= 𝜃̇ 2 𝑟 − 𝑟̈
2
𝑟
Adding together the y direction equations gives
0 = 𝑟𝜃̈ + 2𝑟̇ 𝜃̇
We will use Simulink to solve these equations, subject to initial conditions given by the planets
themselves.
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F
Sun
a
a(1 - )
b
a
Ellipses and Planetary Orbits
Johannes Kepler, after analyzing orbital data taken by Tycho Brahe, found that planetary orbits
follow ellipses (1609, 1619) with the Sun as one focus. This he formulated into his first law of
planetary motion.
Ellipses are similar to circles, except that instead of a radius, they have two axes
a = semimajor axis
b = semiminor axis
ε = eccentricity
where
𝑏 = 𝑎√1 − 𝜀 2
The eccentricity can vary between 0 (a circle) and 1 (a line). The smaller the eccentricity, the more
closely the ellipse resembles a circle. The area of the ellipse is
𝐴 = 𝜋𝑎𝑏
but there is no known formula for the perimeter of an ellipse! The minimum distance between the
sun an a planet is given by
𝑟𝑝 = 𝑎(1 − 𝜀)
perihelion
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and the maximum distance is
aphelion
𝑟𝑎 = 𝑎(1 + 𝜀)
The mean distance is then
𝑟𝑚 =
𝑟𝑎 + 𝑟𝑝 𝑎(1 − 𝜀) + 𝑎(1 + 𝜀)
=
=𝑎
2
2
We can also use the perihelion and aphelion to find the eccentricity
𝜀=
𝑟𝑎 − 𝑟𝑝
𝑟𝑎 + 𝑟𝑝
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Sun
A
A
= slow
= fast
Areal Velocity and Orbital Period
Kepler also noted that the area swept out by the planet per unit time is constant throughout its orbit.
We can calculate an expression of areal velocity using geometry
d
r
r d /2
As the angle dθ gets smaller, the swept area approaches that of a triangle.
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1
𝑑𝜃
𝑟 2 𝑑𝜃
𝑑𝐴 = 2 ∙ ( 𝑏ℎ) = 2 ∙ ( ∙ 𝑟 ∙ 𝑟 ) =
2
2
2
2
Kepler’s second law states that
𝑑𝐴 𝑟 2
= 𝜃̇ = constant
𝑑𝑡
2
The orbital period can be calculated by dividing the total area of the ellipse by the rate at which area
is swept
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𝑃=
𝐴 𝜋𝑎𝑏 2𝜋𝑎𝑏
= 2 = 2
𝐴̇
𝑟 𝜃̇
𝑟 𝜃̇
2
This is, of course, a planet’s year. We usually know the planet’s year from observation, and we
would like to solve for its distance from the sun and velocity (r, 𝜃̇) at any point in its orbit.
Typical Planet Problem
Known:
mass of planet and sun, perihelion, aphelion, period
Find:
orbit
Governing Differential Equations:
𝜇
𝑟2
̈
̇
𝑟𝜃 + 2𝑟̇ 𝜃 = 0
𝜃̇ 2 𝑟 − 𝑟̈ =
Subject to initial conditions:
𝑟0 , 𝑟̇0 , 𝜃0 , 𝜃̇0
Let us take r0 to be the perihelion distance (closest point to the sun). We will (arbitrarily) assign this
point as θ0 = 0. Since the perihelion is the minimum distance, we have
𝑟̇0 = 0
Finally, we can use the orbital period relation so solve for the initial angular velocity.
𝑃=
2𝜋𝑎𝑏
𝑟02 𝜃̇0
Use Simulink to solve for the orbits!
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