Lecture 29
Orbits. Kepler’s laws.
A. Closer to the smaller star
B. Closer to the larger star
C.
It depends on the mass of the spaceship.
ACT: Gravitational forces
Two stars of masses
M
and 2
M
are separated by a distance
d
. A spaceship travels between the two stars. The net
gravitational force on the spaceship is zero
when the
spaceship is:
d
M
2
M
1
2
1
2
;
In terms of magnitude:
net
net
F
F
F
F
F
F
=
+
= −
+
G
G
G
m
F
2
F
1
M
1
M
2
F
2
>
F
1
(M
2
>
M
1
, and distance to
M
2
is shorter)
M
1
M
2
m
F
2
F
1
F
2
=
F
1
(M
2
>
M
1
, but distance to
M
2
is larger)
F
2
d
M
1
M
2
m
F
1
x
d  x
= −
+
=
1
2
0
net
F
F
F
The actual calculation:
=
1
2
Mm
F
G
x
(
)
=
−
2
2
2
Mm
F
G
d
x
(
)
−
+
=
−
2
2
2
0
Mm
Mm
G
G
x
d
x
−
−
+
=
2
2
2
0
x
dx
d
(
)
(
)
=
−
=
<
= −
+
2
1
0.41
2
2
1
d
x
d
d
x
d
(
)
=
−
2
2
1
2
x
d
x
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Circular orbits
A communications satellite orbits at 400 km above the
surface of the Earth. What is its period?
The only force acting on the satellite is the
gravitational attraction of Earth (“neglecting”
the attraction of other bodies, like the Sun)
=
=
2
S
E
S
2
g
M m
v
F
G
m
r
r
F
g
→
=
E
M
v
G
r
π
=
2
But we also know that
r
v
T
π
⇒
=
E
2
M
r
G
r
T
=
+
E
with
r
R
h
r
= 6.38 × 10
6
+
400 × 10
3
=
6.78 × 10
6
m
M
E
= 5.98 × 10
24
kg
G = 6.67 × 10
11
N m
2
kg
2
π
=
=
=
2
3
E
4
5545 s
1.54 h
r
T
GM
Important formulas
for circular orbits
Example: Geosynchronous orbit
A geosynchronous satellite orbits above the Equator
with
T
= 24 h. How high above the Earth’s surface
does such a satellite have to orbit?
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 Fall '06
 Johnson
 Physics, Force, Mass, Orbits, Celestial mechanics

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