Gravitational potential energy
r2
 
U  U 2  U 1  W    Fdr    Fdr
r2
r1
r1
r2
Mm
GMm GMm
U   G 2 dr  

r2
r1
r
r1
Mm
U (r )  G
r
Near the Earth’s surface: r = R + h; h/R <<1
U (h)  G
Mm
Mm
1
Mm
1  h / R   G Mm  G Mmh
 G
 G
Rh
R 1 h / R
R
R
R2
M
U (h)  G 2 mh  const  mgh  const
R
Escape speed
vi
Ki  Ui  K f  U f
R
2
mvi2
Mm mv f
Mm
G

G
2
ri
2
rf
ri  R
rf  
vf  0
vi  ?
mvi2
Mm
G
0
2
R
vi 

vi 

2 6.67  10 11 N  m 2 / kg 2 5.79  10 24 kg
6.38  10 6 m
2GM
R

vi  1.12  10 4 m / s  11.2 km / s
Question
A spherical asteroid has a mass of
2.0  10 24 kg and a radius of
6.67  10 5 m.
The escape speed from the
asteroid’s surface is __ km/s.
1. 20
2. 80
3. 160
4. 320

r
M
Gravitational field
m
 GMm
 rˆ 
F
2
r

 GMm
 rˆ 
F /m  g 
2
r
Satellites Orbits
Trajectories of a projectile launched
from point A in the direction AB
with different speeds.
Open orbits (6 and 7)
11.2 km/s
Circular orbit (4)
7.9 km/s
Elliptical orbits (1-5)
Circular Orbits
mE m mv2
G 2 
r
r
Velocity:
Period:
GmE
v
r
2r
2r 3 / 2
T

v
(GmE )1/ 2
Example: We want to place 2000-kg satellite into a circular orbit 400 km
above the earth’s surface. For the earth: RE = 6380 km, mE = 5.971024kg
GmE
(6.67 1011 N  m2 / kg 2 )(5.97 1024 kg)
v

 7664m / s
6
6
r
(6.38 10  0.4 10 )m
2r 2 (6.78 106 m)
T

 5556 s  92.6 min
v
7664m / s
Elliptical Orbits
Copernicus in 1543 proposed that the sun was the center of
the Solar System with the planets moving in circular orbits.
In 1619 Kepler showed that planets followed elliptical orbits
using huge amount of high quality data gathered by Tycho
Brave by naked eye astronomy.
Kepler characterized planetary orbits using “Kepler’s Three
Laws”.
In 1683 Newton showed that Kepler’s 3 laws follow from
Newton’s law of gravity.
Kepler’s 1st Law: Geometry of Ellipse
Each planet moves in an
elliptical orbit, with the sun
at one focus of the ellipse.
•Ellipse: foci S and S’.
SP + PS’ = constant
•The longest dimension is the
major axis with half-length a.
•Half-length a is called the
semi-major axis.
•e = eccentricity (0≤e<1)
•If e = 0, the orbit is a circle:
S = S’ at center of circle
Kepler’s 2nd Law: Equal Areas in Equal Time
An imaginary line drawn from the Sun to the planet sweeps out
equal areas in equal periods of time unit of time.
Kepler’s 3rd Law: Period Proportional to 3/2 Power
The period of the planets are proportional to the 3/2 powers of the
major axis length of their orbits:
2
 T1   a1 
    
 T2   a2 
3
2a 3 / 2
T
GM
a is the semi-major axis, M is the sun’s mass.