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Extra notes for circular motion:Circular motion : v keeps changing, maybe both speed anddirection are changing At least v direction is changingdirection are changing. At least v direction is changing.Hence a
≠
Acceleration NEEDED to stay on circular orbit:
2
/
i
ti
t
th
t
l
di
di
ti
a
cp
=v
2
/r , pointing to the center along radius direction.
Total force NEEDED in the radius direction to stayand maintain circular motion:F
needed
= ma
cp
=mv
2
/r , pointing to center
- When total force in r direction pointing to center equals tomv
2
/r
it has what it needs to stay in orbit
Σ
F = mv
2
/r
mv /r , it has what it needs to stay in orbit.
Σ
F
r
= mv /r ,
it stays in circular orbit. Neither flies away, nor falls into thecenter.Thi
i
h
l
l t d f
V l
iti
di
t
This is how you solve related forces. Velocities or radius, etc.1. Find ALL REAL Forces and force components,2. ADD ALL in radius direction
Σ
F
r , toward center.
1
r ,
- Make
Σ
F
r
At that position equals to mv
2
/r at that position
- Solve Equation
Σ
F
r
=mv2/r
Chapter 12 Gravity
p
y
Where does mg come from?Why is g=9 81m/s
(^2)
on earth surface?
Why is g=9.81m/s
(^2)
on earth surface?
Sample problems:Sample problems: •
What speed is needed to launch a object to fly around earth (close to earth surface)? •
Where are the synchronous satellite (for TV) l
t d?
located? •
How to find the period of a planet orbiting aroundHow
to find the period of a planet orbiting around
the Sun?
12-1 Newton’s Law of Universal Gravitation
The gravitational force is always attractive, andpoints along the line connecting the two massespoints along the line connecting the two massescenters:r is the distance between mass centers of m
1
and m
2
2
1
2
2
1
m
r M
G
r
m
M
G
F
g
r
r
The two forces shown are an action-reaction pair.
4
12-1 Newton’s Law of Universal GravitationG=6.67 x 10
Nm
2
/kg
2
, the universal gravitational
constantconstant G
is a very small number; this means that the force
of gravity is negligible unless there is a very largeof gravity is negligible unless there is a very largemass involved (such as the Earth).
1
2
1
M G m M G F
When m
1
or m
2
increases F
g
increases.
2
2
1
2
2
1
m
r M
G
r
m
M
G
F
g
1
2
g
When r increases, F
g
decreases.
F
g
is proportional to 1/r
2
Att
ti
! N
t 1
Attention: 1 over r square! Not 1 over r.When r is doubled, F
g
will reduce to 1/4 of initial F
g
5
-2 Gravitational Attraction between Earth
and an object on it
The center of the Earth is one Earth radius awayThe center of the Earth is one Earth radius awayfrom object m, so this is the distance we use:
M
Earth
Earth
Earth
g
mg
R
mM
G
F
2
M R
G
g
Earth
Earth
Earth
2
m
R
kg
M
Earth
Earth
6
24
Therefore,
m
R
Earth
kg
N
s
m
g
Earth
2
When m is on other planets M
p
, R
p
are different.
g
planet
will be different from 9.8 m/s
2
7
Example 1: Circular orbit around earthTo let an object fly around the earth close to earthsurface, you need to launch it with what velocity?In the radius direction, Force it has = what’s neededF
l
di
/
2
F
net
along radius = mg = m = g
so v
2
= g r
so
v =
r
v
/
2
v
2
r
g
.
g , so v
= g.r
so,
v =
Plug in numbers:
r
r
g
.
g
Close to earth surface. g=9.81 m/s
2
Earth radius r = 6370km = 6.37 x 10
6
m
V
3
il /
V= 7.9 x 10
3
(m/s) ~ 5 mile/s
Time to circle the world :T= 2
π
r / v = 2
π
6.4 x 10
6
/ 7.9 x
3
= 5.1 x 10
3
(s) = 85 (min)
12-2 Gravitational Attraction of Spherical
Bodies
Once the altitude becomes comparable to theradius of the Earth, the decrease in theacceleration of gravity is much larger:
F
i
ti
l t
2
F
g
is proportional to 1/r
2
Example 2, What is the distance h betweensynchronous satellite (for TV) and the earth surface? Synchronous satelliterotates around earth
t
t th
i
d
center at the same periodas earth does, 24 hours.
To keep circular orbit In the radius directionTo keep circular orbit, In the radius direction,The net Force the satellite has = what’s needed
M
2
h
R
r
=
r
v
m
r
M
m
G
satellite
satellite
Earth
satellite
2
2
=
s
T
v
r
s
hr
h
R
r
Earth
86400
/
2
86400
24
π
=
=
=
M
T
r
v
s
T
v
r
/
2
86400
/
2
π
π
=
=
=
2 v
r
M
G
Earth
=
2
)
2 (
T
r
r
M
G
Earth
π
=
3
2
2
4
r
T
GM
Earth
π
=
r=4.2E7 m, h=3.6E7m
Wow, all synchronous satellites are 36000km above earth surface.
11
Summary of Chapter 12
Force of gravity between two point masses:
G
is the universal gravitational constant:
In calculating gravitational forces,
spherically symmetric bodies can be replacedby point masses.
Summary of Chapter 12
Acceleration of gravity on earth :
Combine gravity force with circular motion for orbits.
Gravitational force = GmM/r
2
= mv
2
/r
What speed is needed to launch a object to fly around earth (close to earth surface)?Know mg and r
find v
Know mg and r, find v. •
Where are the synchronous satellite (for TV) located? Know T, Earth mass, G, find r ; (know v=
π
r/T)
Know T, Earth mass, G, find r ; (know v 2
π
r/T)
How to find the period of a planet orbiting around the Sun?Know Sun mass, G, r ; find v and T (know T=
π
r/v)
