Download Extra notes for circular motion: Circular motion : v keeps changing

Extra notes for circular motion:
Circular motion : v keeps changing, maybe both speed and
direction are changing
changing. At least v direction is changing
changing.
Hence a≠0.
Acceleration NEEDED to stay on circular orbit:
acp=v2/r
/ , pointing
i ti to
t the
th center
t along
l
radius
di direction.
di ti
Total force NEEDED in the radius direction to stay
and maintain circular motion:
Fneeded= macp=mv2/r , pointing to center
1. When total force in r direction pointing to center equals to
mv2/r , it has what it needs to stay in orbit.
orbit ΣFr = mv2/r ,
it stays in circular orbit. Neither flies away, nor falls into the
center.
Thi is
This
i how
h
you solve
l related
l t d forces.
f
V l iti or radius,
Velocities
di
etc.
t
1. Find ALL REAL Forces and force components,
2. ADD ALL in radius direction ΣFr, toward center.
3. Make ΣFr At that position equals to mv2/r at that position
1
4. Solve Equation ΣFr =mv2/r
Chapter
p
12 Gravity
y
Where does mg come from?
Why is g=9.81m/s
g=9 81m/s2 on earth surface?
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
located?
t d?
How to find the period of a planet orbiting around
•How
the Sun?
2
12-1 Newton’s Law of Universal Gravitation
The force accelerating an apple downward is
the same force that keeps the Moon in its orbit.
Hence, Universal Gravitation.
3
12-1 Newton’s Law of Universal Gravitation
The gravitational force is always attractive, and
points along the line connecting the two masses
centers:
r is the distance between mass centers of m1 and m2
M 1m 2
M1
= G 2 ⋅ m2
Fg = G
2
r
r
4
The two forces shown are an action-reaction pair.
12-1 Newton’s Law of Universal Gravitation
G=6.67 x 10-11 Nm2/kg2, the universal gravitational
constant
G is a very small number; this means that the force
of gravity is negligible unless there is a very large
mass involved (such as the Earth).
M 1m 2
M1
Fg = G
= G 2 ⋅ m2
2
r
r
When m1 or m2 increases Fg increases.
When r increases, Fg decreases.
Fg is proportional to 1/r2
Att ti
Attention:
1 over r square!! N
Nott 1 over r.
5
When r is doubled, Fg will reduce to 1/4 of initial Fg.
12-2 Gravitational Attraction of Spherical
Bodies
Gravitational force between a point mass and a
sphere: the force is the same as if all the mass
of the sphere were concentrated at its center.
6
12-2 Gravitational Attraction between Earth
and an object on it
The center of the Earth is one Earth radius away
from object m, so this is the distance we use:
mM
M Earth
Fg = G
= mg Earth
2
R Earth
g Earth
M Earth
=G
2
R Earth
M Earth = 5 .97 * 10 kg
24
Therefore,
R Earth = 6 .37 * 10 6 m
g Earth = 9 .81 m / s = 9 .81 N / kg
2
When m is on other planets Mp , Rp are different.7
g planet will be different from 9.8 m/s2
Example 1: Circular orbit around earth
To let an object fly around the earth close to earth
surface, you need to launch it with what velocity?
In the radius direction, Force it has = what’s needed
2
v
Fnet along
l
radius
di = mg = m / r
2
v
= g , so v2 = g.r
gr
r
so v =
so,
g .r
g in numbers:
Plug
Close to earth surface. g=9.81 m/s2
Earth radius r = 6370km = 6.37 x 106m
V 7
V=
7.9
9 x 103(m/s)
( / ) ~ 5 mile/s
il /
Time to circle the world :
T= 2 π r / v = 2 π 6.4 x 106 / 7.9 x103
= 5.1 x 103 (s) = 85 (min)
8
12-2 Gravitational Attraction of Spherical
Bodies
The acceleration of gravity decreases with altitude:
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12-2 Gravitational Attraction of Spherical
Bodies
Once the altitude becomes comparable to the
radius of the Earth, the decrease in the
acceleration of gravity is much larger:
Fg is
i proportional
ti
l to
t 1/r
1/ 2
10
Example 2, What is the distance h between
synchronous satellite (for TV) and the earth surface?
Synchronous satellite
rotates around earth
center
t att th
the same period
i d
as earth does, 24 hours.
To keep circular orbit
orbit, In the radius direction
direction,
The net Force the satellite has = what’s needed
r = R Earth + h
24 hr = 86400 s
2π r / v = T = 86400 s
v = 2π r / T
GM
m satellite v 2 satellite
m satellite M Earth
G
=
2
r
r
M Earth
G
= v2
r
2
2 3
T
=
4
π
r
Earth
M Earth
( 2π r )
G
=
r
T2
2
r=4.2E7 m, h=3.6E7m
11
Wow, all synchronous satellites are 36000km above earth surface.
Example 3, How to find the period of a planet orbiting
around the Sun? Assuming the orbit is a circle.
Again, to keep circular orbit, in the radius direction,
The net Force the planet has = what’s needed
m Planet M SUN
m Planet v 2 planet
=
G 2
r SuntoPlaen t
rSuntoPlaen t
G
M SUN
r
= v 2 planet
SuntoPlaen t
v = 2πr / T
30
Mass of the sun =1.98892 × 10
T = 2πr / r
kilograms
You don’t need to memorize any results above.
You only need to learn and solve ΣFr = Fg = mv2/r
Use the correct r, correct mass and correct v. 12
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 replaced
by point masses.
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Summary of Chapter 12
•
Acceleration of gravity on earth :
Combine gravity force with circular motion for orbits.
Gravitational force = GmM/r2 = mv2/r
•What speed is needed to launch a object to fly
around earth (close to earth surface)?
Know mg and r,
r find v
v.
•Where are the synchronous satellite (for TV) located?
Know T, Earth mass, G, find r ; (know v=2πr/T)
v 2πr/T)
•How to find the period of a planet orbiting around the
Sun?
Know Sun mass, G, r ; find v and T (know T=2πr/v)
14