Download AP Physics – Gravity and Circular Motion

AP Physics – Gravity and Circular Motion
Newton’s theory is very simple. Gravity is a force of attraction between any two objects that have
mass. Two objects sitting on a desktop attract each other with a force that we call gravity. They
don’t go flying together because gravity is a very weak force and is only significant when one or the
other of the masses is enormous – planet size. This is why we aren’t attracted to objects that we
pass on our daily wanderings.
The size of the force of gravity is given by this equation:
mm
F = G 12 2
r
Newton’s law of gravity
G is the universal gravitational constant.
G = 6.67 x 10!11
N " m2
kg 2
m1 is the mass of one of the bodies and m2 is the mass of the other body.
r is the distance between the two bodies.
The value of G is the same everywhere throughout the universe.
On the old AP Physics Test the equation is written as:
FG = !
G m1m2
r2
Newton’s law of gravity is an inverse-square Law. This means that the force of gravity gets
smaller or larger by the square of the distance. The force is directly proportional to the masses, so
if the mass of one of the objects doubles, the force of gravity would double. But if the distance
doubled, the force of gravity would decrease by a factor of four. That’s because it decreases by the
square of the distance. Inverse-square laws are very common in physics. We’ll see more of them in
our explorations.
To give the particulars of the theory:
Universal Law of Gravitation
1.
2.
3.
4.
5.
6.
Gravitational force is a field force between two particles -- in all mediums.
Force varies as the inverse square of the distance
Force is proportional to mass of objects.
The gravity force acts from the center of the two objects.
The gravitational force is always attractive.
The gravitational force cannot be shielded or canceled.
Solving gravity problems is quite simple. Let’s do one.
•
A girl, Brandy (42.5 kg), sits 1.50 m from a boy (63.0 kg), George. What is the force of gravity
between them? (This will tell us how attracted they are to each other.)
F =G
2 # 42.5 kg 63.0 kg
"
(
)(
)
!11 N $ m
F = % 6.67 x 10
&
2
2
%
kg &(
(1.50 m )
'
m1m2
r2
F = 7940 x 10!11 N
7.94 x 10!8 N
=
You can see that this is a tiny force, one so small that Brandy will never
notice its presence. George must generate some other attractive force if
there is to be a relationship between the two of them.
One useful application of Newton’s law of gravity was to weigh Earth – this allowed physicists to
make an accurate determination of the earth’s mass. Let’s do that.
•
Find the mass of the earth. re = 6.38 x 106 m. We will use a 10.0 kg mass in our solution (not
that it matters), the other mass will be that of the earth.
F =G
m1mE
r2
m1g = G
This is the force of gravity using Newton’s law. But we know that the
force of gravity must also equal F = mg, so we set them equal to one
another:
m1mEarth
r2
The mass of the object cancels out:
mE =
m
s
2
(
6.38 x 106 m
2
)
2
"
m #
!11 kg $ m
$
% 6.67 x 10
& (10.0 kg )
s2
kg 2 (
'
mE = 59.8 x 1023 kg
=
mE
r2
gr 2
mE =
G
Solve for the mass of the earth
9.8
g =G
5.98 x 1024 kg
= 59.8 x 1023 kg
=
5.98 x 1024 kg
Force and Circular Motion - In order for an object to undergo circular motion, a force
must act. Picture an object that has some velocity. What will happen to it if no forces act on it?
Well, according to the first law, it will continue to move with a constant velocity. It will follow a
straight-line path.
To make it change direction a force must act on it. In order to make it changes direction constantly,
a force must act on it constantly.
What is the direction of the force needed to do this? Well, when you spin something in a circle,
what do you have to do? You just pull it towards the center as you go around and around.
The object gets accelerated towards the center. We call this the centripetal acceleration. The
equation for the centripetal acceleration is:
v2
ac =
r
ac is the centripetal acceleration, v is the linear or tangential speed, and r is the radius
of the circular path.
This equation will be provided to you for the AP Physics Test.
The force that brings about this acceleration is called the centripetal force. Its direction is also
towards the center of the circular path. Centripetal means "center seeking". The centripetal force
changes the direction of the object’s velocity vector. Without it, there would be no circular path.
The centripetal force is merely a convenient name for the net force that is towards the center. It is
always caused by something – it could be caused by the force of gravity, the reaction force between
the control surfaces of an airplane with the air, &tc.
When you rotate a ball around your head in a circle, the centripetal force is supplied by the tension
in the string.
What is the source of the centripetal force that causes a racecar to travel in a circular path on the
racetrack?
The force is brought about by the tires pushing on the racetrack. The friction between the
road and the tire is very important, so race tires are designed to maximize friction.
What is the source of the centripetal force required to make the earth revolve around the sun?
This is where the apple falling on Newton story fits in. Before Newton no one could explain the
orbits of the planets and moons. Newton, the story goes, was relaxing under an apple tree
pondering the problem of the moon’s orbit. He knew that there had to be a force acting on the
moon to accelerating it towards the earth, but had no idea what was the source of the force. Then he
saw an apple fall and the simple solution struck him like the old thunderbolt. Just as the earth’s
gravity reached out and made the apple fall, so it reached out and made the moon fall. Thus, the
force that keeps the planets and moon following their orbital paths is gravity.
The AP Test equation sheet will not give you the equation for centripetal force. It does give you the
equation for centripetal acceleration. It also gives you the equation for the second law. Using these
two equations you can easily derive the formula for centripetal force. Here’s how to do it:
v
r
so plug in the value of the centripetal acceleration: aC = 2
F = ma
! v2 "
F = ma = m # $
% r &
mv 2
FC =
r
That’s all there is to it.
• A car is traveling at a constant speed and makes a turn with a radius of 50.0 m. Its speed is15.0
m/s. Find the minimum coefficient of friction needed to
keep the car traveling along the path.
n
Let’s look at the FBD:
The frictional force must equal the centripetal force.
The centripetal force is given by:
mv
FC =
r
2
Fc
We know that this must equal the frictional force. We also
know that the frictional force is:
fs = µs N
Assume the road is flat, so
n = mg
Set the two equal to each other and solve for the coefficient of friction:
mv 2
µ s mg =
r
v2
µs =
gr
2
m"
1
!
µ s = #15.0 $
s & !
m "
%
# 9.8 2 $ 50.0 m
s &
%
=
0.459
f
mg
•
A child twirls a yo yo. If angle of the cord with the vertical is 30.0°, find ac.
0
Look at the forces in the y direction:
∑ Fy = 0
T cos! " mg = 0 T =
mg
cos!
The horizontal component of T is the centripetal force.
FC = T sin !
FC =
so
Plug into equation for T:
mg
sin !
cos!
FC = mg tan !
maC = mg tan !
m
aC = 9.8 2 tan 30.0o
s
T
We know that:
FC = maC
0
aC = g tan !
aC =
m
5.66 2
s
mg
Centripetal Force and Gravity: You may have seen a silly demonstration involving a
bucket of water that was spinning in a vertical circle. The water stayed in the bucket and did not
fall out.
So what was the deal? Does spinning something
in a vertical circle somehow cancel out gravity?
Well, no, gravity is a force that cannot be stopped
or canceled. It is always there, anytime you have
the appropriate masses.
The water does fall, it falls but the bucket falls
with it and catches it.
This only works if the bucket is moving fast enough to catch the water. If the bucket is too slow,
then the water will fall out of it.
The minimum linear speed for this is called the critical velocity.
Critical velocity ≡ minimum velocity for an object to travel in vertical circle
and maintain its circular path against the force of gravity.
The same thing is needed for satellites in orbit around the earth or planets in orbit around the sun.
They too must travel at the critical velocity.
The critical velocity formula is not provided on the AP Test, but it is very simple to figure out.
You just set the centripetal force equal to the weight of the object that is in circular motion. If the
two forces are equal, then the object won’t be able to “fall out” of the bucket.
mv 2
FC =
F = mg
and
Set them equal to each other:
r
mv 2
mg =
r
v 2 = gr
v = gr
v = gr
So here is the critical velocity
Orbital Equations:
Let us assume that the orbit of a satellite about the earth (or any other
massive body) is a circle. Most orbits are not actually circles but are instead ellipses. This was
discovered by Johanes Keppler in the 1600’s. But let’s keep it simple and look at a circular orbit.
In order to have a circular path, a centripetal force is required. This is supplied by the force of
gravity between the two bodies.
So we can set the centripetal force equal to Newton’s law of gravity:
mm
F = G 12 2
r
FC =
gravity
G
Set them equal to one another:
m1 m2
Notice how the mass of the object canceled out.
v2 =
Gm1
r
v=
m2v 2
r
r2
centripetal force
m2v 2
=
r
Gm1
r
This gives us an equation for the orbital velocity:
v=
Gm
r
The mass, m, in the equation is the mass of the body being orbited. If we are talking about a planet
orbiting the sun, then the mass we would use would be that of the sun. The mass of the satellite
cancels out, so its mass is not involved in the orbital velocity equation at all.
The equation for the orbital velocity will not be given you on the AP Physics Test. So be prepared
to derive it if you need it.
Period of satellite: This is another simple derivation job.
The period of a satellite is T, the
time to make one orbit. What would be the period of the earth around the sun?
Let’s develop the equation for the period of a satellite. We’ll use the equation for distance and
solve it for the time:
v=
x
t
t=
x
v
d, the distance traveled is the circumference of the orbit. We know that it would be:
x = 2! r
So we can plug that in to the equation we solved for time:
t=
2! r
v
but v is also given by the equation we just derived for the orbital velocity:
Gm
r
v=
If we plug the orbital velocity into our working equation, i.e., put them together, we get:
t=
Square both sides:
2! r
Gm
r
4! 2r 2
t =
Gm
r
2
Clean up everything up nice and neatlike using our potent algebra skills:
2 3
2 4! r
2
2 2 r
t = 4! r
t=
t =
Gm
4! 2r 3
Gm
t = 2!
Gm
r3
Gm
And we end up with an equation for the period of a satellite. Again the mass in the thing is the mass
of the body being orbited:
t = 2!
r3
Gm