Lecture 19
Satellites
Quick Review
• Newton’s Law of gravity
Gm1m2
F1on 2 (r )  F2 on1 (r ) 
2
r
2
11 N m
G  6.67 *10
2
kg
Circular Orbits
• M is mass of parent and m the mass of the
satellite. M>>m
GMm
m v2
FM on m (r )  2  mac 
r
r
GM
v
r
2

4

T2  
 GM
 3
r

• Kepler’s Law
Geostationary Satellite
• At the equator, how far above the earth’s
surface should a satellite orbit so it stays
overhead?
• Rearth = 6.37 x 106 m
• G=6.67 x 1011 N•m2/kg2
Mearth = 5.98 x 1024 kg
A cyclist goes around a level, circular
track at constant speed. Since the
cyclist’s speed is constant, her
50%
50%
acceleration is constant
se
Fa
l
Tr
ue
1. True
2. False
QUIZ
• A motor cycle is in a loop-the-loop. At the top
of the loop…
• A) Draw a free body diagram
• B) What is the minimum speed the motor
cycle must have if the radius
of the loop is 15. m?
Homework due Friday
• Problems: 6: 24, 25, 27, 29, 33, 44, 45, 51, 58,
59
Problem 6:27
• What is the ratio between the sun’s
gravitational force on you and the earth’s
gravitational force on you?
• Rearth = 6.37 x 106 m
• G=6.67 x 1011 N•m2/kg2
Mearth = 5.98 x 1024 kg
Rearth-sun =1.5 x 1011 m
Msun = 1.99x1030 kg
Problem 6:27 cont
• Take ratio
Fsun
Fearth
GM sun m
2
R 2 earth  sun
M sun Rearth



2
GM earth m M earth Rearth  sun
2
Rearth
2
1.99*10  6.37 *10 
4
 6.0*10
24 
11 
5.98*10  1.5*10 
30
6
6:33
The space shuttle is in an orbit 250 miles high.
What are its orbital period (in minutes) and orbital
speed?
Rearth = 6.37 x 106 m
G=6.67 x 10-11 N•m2/kg2
Mearth = 5.98 x 1024 kg
1 mi = 1.609 km
2

4

2
T 
 GM
 3
r

6:33
• 250 mi *1.609km/mi=326.5 km
• r= 6.37 x 106 m+3.265 x 105 m =6.70 x 106
• T=5.46 x 103 s = 90.9 minutes
6:44
• A 5g coin is placed 15cm from the center of a
turntable. µs = 0.80, µk = 0.50. As turntable
starts from rest and speeds up to 60 rpm,
does coin slide off?
mv 2
f s   s ,required mg 
r
60rpm
f 
 1 Hz
60 s / m
 s ,required
  2 f
.15*  2 *1
v
r



 .604
rg
rg
9.8
2
2
2
So coin does not slip
2
6:45 Conical Pendulum Find T, ω
1.0m
T
θ
r=0.20m
m  r 
FNET , x  T cos   
r
FNET , y   w  T sin   0
 .2 
0
  cos    78.46
1
1
W=mg
2
6:51 Ultracentrifuge
R=0.10 m 70,000 rpm Δm= 10 mg = 0.000010 kg
What is net force?
r
a
6:58 Martian geostationary satellite
What are its speed and altitude?
dayMars:=24.8 hours
6:59 g on Mars, Maximum walking speed?
g MARS
G * M MARS

2
RMARS
vMAX gr
r  .70m