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Gravity (Chapter 11)
Kepler’s Laws
Example of a successful phenomenology at the cradle of science (astronomical observations by Brahe).
- All planets move on elliptical orbits with the sun at one focus. (Ellipse: see mathworld.)
- A line joining any planet to the sun sweeps out equal areas in equal times (figure 11-4 of Tipler-Mosca).
- The square of the period of any planet is proportional to the cube of the semimajor axes of its orbit T 2 = C r^2.
Astronomical unit (AU): The mean earth-sun distance
1 AU = 1. 50 × 1011 m = 93. 0 × 106 mi.
Newton’s Law of Gravity
Force of attraction between each pair of point particles (figure 11-6 of Tipler-Mosca)
F^ ~ 12 = −G m^1 m^2 r^212
ˆr 12 = −
G m 1 m 2 r 123
~r 12.
Where G is the gravitational constant
G = 6. 67 × 10 −^11 N · m^2 /kg.
Free-fall acceleration for objects near the surface of the earth:
g =
G ME
R^2 E
= 9.81 m/s^2.
Measurement of G: Torsion balance (first Cavendish 1798).
Derivation of Kepler’s Laws
Kepler’s first law: With a bit more involved mathematics than we have presently at our disposal, one can show that the only closed solutions to Newton’s two body force are elliptical orbits (intermediate mechanics for physicists).
Kepler’s second law: (Figure 11-8 of Tipler-Mosca.)
The area swept out by the radius vector ~r in the time dt is
dA =
|~r × ~v dt| =
2 m
|~r × m ~v| dt =
L
2 m
dt
where L = |~r × ~p| is the magnitude of the orbital angular momentum of the planet about the sun. Since the force on the planet is along the line from the planet to the center of the ellipse, there is no torque acting on the planet, and L is conserved. Therefore, dA dt
L
2 m
= constant.
Kepler’s third law:
We will show this for the special orbit of a circle. Then
F = G
m 1 m 2 r^2
= m 2 a = m 2
v^2 r
and
v^2 =
2 π r T
G m 1 r follows, which can be written as Kepler’s third law:
T 2 =
4 π^2 G m 1
r^3.
Escape Speed
Initial kinetic energy needed to escape from the earth’ surface to r = ∞:
0 = Ki + Ui =
m v e^2 − G
ME m RE
where ve is called escape speed. Solving for ve:
ve =
2 G ME
RE
2 g RE.
ve =
2 (9.81 m/s^2 ) (6. 37 × 106 m) = 11.2 km/s.
If E = Ki + Ui < 0 the system is bound. If E = Ki + Ui ≥ 0 the system is unbound.
Gravitational field (of the earth):
~g =
F~
m
G ME
r^2
