Kepler’s Laws
1. The orbits of the planets are ellipses, with the sun at one focus of the
ellipse.
2. The line joining the planet to the sun sweeps out equal areas in equal
times as the plates travels around the ellipse.
3. The ratio of the squares of the revolutionary periods (P) of two planets
is equal to the ratio of the cubes of their semimajor axes (a)
P21 / P22 = a31 / a32
Note: If P is measured in fraction of earth years, and a is measured in
a.u., then
P2 = a3 (See Table 2.1)
Period is the amount of time it takes for the planet to complete one orbit
Kepler’s laws
The period of the orbit of Mercury is 0.241 years. The eccentricity of the
orbit is 0.206. What is the distance to Mercury’s aphelion? What is the
distance to Mercury’s perihelion?
According to the Kepler’s Second Law, P2 = a3 . As a result,
a = P2/3 = (0.241)2/3 = 0.387 au = 0.579 x 1011 m = 5.79 x 1010 m
Then,
Distance to aphelion = a (1 - e) = 5.79 x 1010 m (1 - 0.206) = 4.6 x 1010 m
Distance to perihelion = a (1 + e) = 5.79 x 1010 m (1 + 0.206) = 6.98 x 1010 m
Kepler’s Laws
The period of the orbit of Neptune is 163.7 years. The eccentricity of the
orbit is 0.009. What is the distance to Neptune’s aphelion? What is the
distance to Neptune’s perihelion?
According to the Kepler’s Second Law, P2 = a3 . As a result,
a = P2/3 = (163.7)2/3 = 30.7 au = 45.93 x 1011 m = 4.593 x 1012 m
Then,
Distance to aphelion = a (1 - e) = 4.593 x 1012 m (1 - 0.009) = 4.55 x 1012 m
Distance to perihelion = a (1 + e) = 4.593 x 1012 m (1 + 0.009) = 4.63 x 1012 m
The orbit of Neptune is not a perfect circle, but it’s low eccentricity indicates
that the orbit is close to a circle. The distance to aphelion and perihelion
differ by only (approximately) 1.7%.
The “New” Astronomy
Isaac Newton
Mass is the generator of
gravity.
m1
m2
The force of gravity exists
between any objects that
possess mass.
The Universal Law of Gravity
F = G {m1 m2 / r2 }
G = 6.67 x 10-11 Nm2/Kg2
r
r is measured center to center
The Universal Law of Gravity
What is the force of gravity exerted by the sun on Neptune at aphelion?
What is the force of gravity exerted by the sun on Neptune at perihelion?
According to the Universal Law of Gravity, at aphelion
F = G (MSun MNeptune ) / rSun to Neptune 2
= 6.67 x 10-11 N m2 / kg2 (2 x 1030 kg ) (1 x 1026 kg) / (4.55 x 1012 m)2 = 0.644 x
1021 N = 6.44 x 1020 N.
At Perihelion,
F = G (MSun MNeptune ) / rSun to Neptune 2
= 6.67 x 10-11 N m2 / kg2 (2 x 1030 kg ) (1 x 1026 kg) / (4.63 x 1012 m)2 = 0.622 x
1021 N = 6.22 x 1020 N.
The Universal Law of Gravity
The mass of the Hubble Space Telescope (HST) is 11110 kg. It orbits the earth at a
distance for 600 km from the surface of the eath. What is the acceleration of gravity
of the HST?
Potential answer: 9.8 m/sec2 . This answer would be wrong because the HST is not
“near the surface of the earth.”
Use
a = G MEarth / (REarth + H)2
= 6.67 x 10-11 N m2 / kg2 (6 x 1024 kg) / (6378000 m + 600000 m)2 = 8.21 m/sec2 .
Note this acceleration is less than 9.8 m/sec2 because the HST is not near the surface of
the earth.
The Universal Law of Gravity
What was the weight of the HST at launch? What is the weight of the HST in orbit?
Assume the mass of the HST is 11110 kg.
At launch, that is, on the surface of the earth,
W = mg = (11110 kg) gSurface of Earth = (11110 kg) (9.8 m/sec2 ) = 1.09 x 105 N
In orbit,
W = mg = (11110 kg) gat Orbital Location = (11110 kg) (8.21 m/sec2 ) = 0.912 x 105 N
The HST’s weight in orbit is about 84% of it’s weight at launch.
The Universal Law of Gravity
What is the force of gravity on a satellite that is one Earth radius
above the surface of the Earth? The weight of the satellite on Earth is
10,000 N?
At one Earth radius above the surface of the Earth, h = RE. Therefore
g = G Mp / (Rp + h)2
g = G Mp / (Rp + RE )2
g = G Mp / (2 Rp )2
g = (¼) G Mp / ( Rp )2 = (¼) 9.8 m/sec2
Therefore W = m (1/4) g = ¼ m g = ¼ (10,000 N) = 2,500 N
Circular Motion
Using the data provided for the HST in the previous examples, what is the orbital
speed of the HST? Assume a circular orbit.
a = v2 / (REarth + H)
Therefore,
v = (a (REarth + H))1/2 = ((8.21 m/sec2) (6.9 x 106 m))1/2 = 7.53 x 103 m/sec = 7.53 km/sec.
Note: This is about 17,000 mph.