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PH 121 8.3-5

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Richard Datwyler

on 9 June 2015

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Transcript of PH 121 8.3-5

Orbits
Your book has a clever illustration
that describes how projectile motion
can turn into an orbit.
Qualitatively this helps us realize that
when an object is in orbit, it is in free fall
This idea is a bit of an odd one, but it is true.
When we throw a ball, or rock, we can visualize
that it is in free fall. If we ignore drag, the only
force acting on it is gravity.
Back to my picture, the orbital projectile is one
that is in freefall, but one that never gets closer
to the surface, thus its trajectory is closed. This is
by definition an orbit, specificaly a circular orbit.
Recalling our centripital accelertion:
If the only force acting on an orbital problem is
gravity, then we can calculate the velocity of an
object orbiting the earth.




for example if the earth had no atmosphere, and an
object was literally skimming along the surface then
this speed would be:

7900 m/s ~ 16,000 mph
Now this is obviously way to fast, and in truth the
satalite doing this would burn up by friction in our
atmosphere.
Rather, consider an orbit that is about 200 miles up
(this represents about 5%) of the radius of the earth
and is a fairly good approximation of low earth orbit
for a satalite or even the suttle.

We can solve not only for its velocity but also the
period of orbit.
At this height, the orbital time for a satalite would be
~87 minutes. (it is actually a bit more, because gravity is
not 9.8 here. But this is close enough.
We will not dive any deeper into orbital information
in this chapter, save to make a disclaimer about gravity.

We couldn't use the previous equation to solve for the
orbital period of the moon, this is because Gravity gets
smaller the further away you get.

If we had the period of the moons orbit would be 11 hours.

Not true.

Rather we could solve for g, in the equation, for we
know the period of the moons orbit.
0.00272 m/s
Fictitious forces
Is there really such a thing as a fictitious force?
You recall my discription of playing corners in a car.

Even simplier, think of just slamming on your brakes
you feel a force pushing you towards the dash.

But yet if you think about it, there isn't a push, least ways
not on your back.
Both the force 'acting' on you as you come to a stop
and the centrifugal force are fictious.
They are simply a result of Newton's 1st law and the
fact that we are not in a inertial reference frame.
This is because we are acclerating.
We've already talked about this.
I bring it up as we are diving into circular dynamics.

Specifically we bring it up because the earth is spinning
thus we have a centripital acceleration.

Can we then use Newton's laws on earth?
The answer is, sure, close enough.
Amongst friends.
But realize there is a bit of an adjustment
that would need to be made.
To give an example of this consider the weight of an object
as it moves around on the equator. Specificaly if it was sitting
on a spring scale.
Solving for that spring is what we would call mg, or rather Fg.
Fg is what we have always calculated as =mg
Here this g_earth is 9.83 m/s^2. and the last term is ~0.033 m/s^2
Giving the value of g that we use as 9.80.

Although this is a nice result, the point is, the rotational effects of the fact
we are spinning doesn't change our calculations much, and we'll go on
pretending we are in an inertial reference frame.
Amazingly now, the author spends a
section on why water stays in a bucket
This can be answered almost immediately.
It has been the whole purpose of this chapter

When objects move in circular path there must
be a force pointing to the center of the circular
path.
In addition, our fictious forces, or rather our
understanding of Newton's laws help us see
that the water would keep on a straight trajectory
but the bucket gets in the way.
However, we can gleen something from this section.
That of a critical speed.

Illustrate the roller coaster ride.

Then solve for the water in bucket.
Because we can, and have the time.
Lets look at section 8.7
We finish this chapter off with a section
on non-uniform circular motion.
This simply means that we now have an
acceleration in the tangential direction as well
and likewise there would have to be a force in this
direction.
Orbits are extended projectile motion

Only force acting on orbiting object is gravity, thus in
free Fall motion

Orbital speed and period is


Note g can change
For this chapter stay close to earth and
g = 9.8

Really just
Newton’s 1st law
application

Happen because not in
inertial reference frame

Examples of water in bucket and roller coasters

Effect on gravity
g=9.83 - inertial effect (.033) = g = 9.80
Small enough we will
neglect rotational effects


Kinematic equations
of motion in rotational reference frame




Dynamics
in Nonuniform Circular Motion





Find total Force/acceleration using Pythagorean theorem

Orbital velocity depends upon:
1. Mass of satellite 2. Mass of planet
3. Acceleration due to gravity
4. Radius 5. Rotation of planet

A. 1,3,5
B. 1,2,3,4
C. 2,3,4
D. 1,2,4,5
E. 3,4
F. 1,2,5
Water stays in a bucket as you swing it around your head because:

A. It is an application of Newton's 3rd law
B. The centrifugal force acts on it
C. Its inertia is keeps it there
D. The bottom of the bucket pulls it up
E. Vanderwall attraction
F. It is frozen and stuck in bucket
Non uniform circular motion problems are:

A. Solvable using circular kinematic equations
B. Can't be solved unless transformed into linear terms
C. Are just simply too hard to solve, give up.
#60 Father stands on conical 20 degree hill and spins his 20 kg child on a 5.0 kg cart, with 2 meter rope. Friction is negligible and speed is 14 rpm, what is tension
Example
20
0
20
0
2 m
radius = r = 2 cos(20) = 1.88
20
20
0
0
G
N

F
F
T
A 500 g ball moves in a vertical circle on a 102 cm string. If the speed at the top is 4.0 m/s and the bottom is 7.5 m/s
a. What is the gravitational force acting on the ball?
b. What is the tension in the string when the ball is at the top?
c. What is the tension in the string when the ball is at the bottom?

A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed. What is the ratio of the normal force to the gravitational force?
Communications satellites are placed in circular orbits where they stay directly over a fixed point on the equator as the earth rotates. The altitude of a geosynchronous orbit is 3.58 x 10^7 m.
a. What is the period of a satellite in this orbit?
b. Find the value of g at this altitude.
c. What is the weight of a 2000 kg satellite in this orbit?

32.5 N
2.94 N
4.9 N
3:1
24 hr
zero
.223 m/s^2
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