1. PHYS 1443 – Section 003 Lecture #10
Wednesday,Oct. 1, 2003
Dr. Jaehoon Yu
Exam review
Newton’s Law of Gravitation
Kepler’s Laws
No homework today!!
Homework #5 deadline extended till noon, next Wednesday, Oct. 10!!
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

2. Exam Results Class average: 53 Top score: 90.5 Remember that
There are three exams of which two bests are used for final grade
Each exam accounts for 25%  50% total
Other factors
Lab: 20%
Homework: 15%
Quiz: 15%
Extra credits: 10%
Final grades will be assigned on a sliding scale
Exam can be easily made up by other credits
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

3. Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

4. Exam problems 14 – 18
A 5 kg block is placed on a frictionless inclined plane of angle 30o and pushed up the plane with a horizontal force of magnitude 30 N.
14. Draw a free-body diagram for this motion. +x +y n
F= -Mg
q=30o
F=30N
q=30o M a Fg n +x +y
F=30N
Free-body
Diagram
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

5. 16. What are the direction and magnitude of block’s acceleration?
x comp.
thus
y comp.
Thus the block’s acceleration is up the incline. The magnitude is
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

6. 17. What is the magnitude of force exerted on the plane by the block?
Since this force is the same magnitude as the normal force but opposite direction, one can obtain the magnitude from y-component of the net force
18. What is the smallest horizontal force needed to move the block up the plane?
In order for the block to be pushed up the plane, the component of the horizontal force up along the plane must overcome the component of the gravitational force downward.
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

7. Newton’s Law of Universal Gravitation
People have been very curious about the stars in the sky, making observations for a long time. But the data people collected have not been explained until Newton has discovered the law of gravitation.
Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
With G
How would you write this principle mathematically?
G is the universal gravitational constant, and its value is
Unit?
This constant is not given by the theory but must be measured by experiment.
This form of forces is known as an inverse-square law, because the magnitude of the force is inversely proportional to the square of the distances between the objects.
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

8. Free Fall Acceleration & Gravitational Force
Weight of an object with mass m is mg. Using the force exerting on a particle of mass m on the surface of the Earth, one can get
What would the gravitational acceleration be if the object is at an altitude h above the surface of the Earth?
What do these tell us about the gravitational acceleration?
The gravitational acceleration is independent of the mass of the object
The gravitational acceleration decreases as the altitude increases
If the distance from the surface of the Earth gets infinitely large, the weight of the object approaches 0.
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

9. Example for Gravitation
Using the fact that g=9.80m/s2 at the Earth’s surface, find the average density of the Earth.
Since the gravitational acceleration is
So the mass of the Earth is
Therefore the density of the Earth is
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

10. Kepler’s Laws & EllipseF1 F2 b c a
Ellipses have two different axis, major (long) and minor (short) axis, and two focal points, F1 & F2
a is the length of a semi-major axis
b is the length of a semi-minor axis
Kepler lived in Germany and discovered the law’s governing planets’ movement some 70 years before Newton, by analyzing data.
All planets move in elliptical orbits with the Sun at one focal point.
The radius vector drawn from the Sun to a planet sweeps out equal area in equal time intervals. (Angular momentum conservation)
The square of the orbital period of any planet is proportional to the cube of the semi-major axis of the elliptical orbit.
Newton’s laws explain the cause of the above laws. Kepler’s third law is the direct consequence of law of gravitation being inverse square law.
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

11. The Law of Gravity and the Motion of Planets
Newton assumed that the law of gravitation applies the same whether it is on the Moon or the apple on the surface of the Earth.
The interacting bodies are assumed to be point like particles.
Newton predicted that the ratio of the Moon’s acceleration aM to the apple’s acceleration g would be RE
Moon
Apple g aM v
Therefore the centripetal acceleration of the Moon, aM, is
Newton also calculated the Moon’s orbital acceleration aM from the knowledge of its distance from the Earth and its orbital period, T=27.32 days=2.36x106s
This means that the Moon’s distance is about 60 times that of the Earth’s radius, its acceleration is reduced by the square of the ratio. This proves that the inverse square law is valid.
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

12. Kepler’s Third Law
It is crucial to show that Keper’s third law can be predicted from the inverse square law for circular orbits. Ms v r
Since the gravitational force exerted by the Sun is radially directed toward the Sun to keep the planet circle, we can apply Newton’s second law
Since the orbital speed, v, of the planet with period T is
The above can be written
Solving for T one can obtain
and
This is Kepler’s third law. It’s also valid for ellipse for r being the length of the semi-major axis. The constant Ks is independent of mass of the planet.
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

13. Example of Kepler’s Third Law
Calculate the mass of the Sun using the fact that the period of the Earth’s orbit around the Sun is 3.16x107s, and its distance from the Sun is 1.496x1011m.
Using Kepler’s third law.
The mass of the Sun, Ms, is
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

14. Kepler’s Second Law and Angular Momentum Conservation
Consider a planet of mass Mp moving around the Sun in an elliptical orbit. S B A D C r dr
Since the gravitational force acting on the planet is always toward radial direction, it is a central force
Therefore the torque acting on the planet by this force is always 0.
Since torque is the time rate change of angular momentum L, the angular momentum is constant.
Because the gravitational force exerted on a planet by the Sun results in no torque, the angular momentum L of the planet is constant.
Since the area swept by the motion of the planet is
This is Keper’s second law which states that the radius vector from the Sun to a planet sweeps our equal areas in equal time intervals.
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

15. Motion in Accelerated Frames
Newton’s laws are valid only when observations are made in an inertial frame of reference. What happens in a non-inertial frame?
Fictitious forces are needed to apply Newton’s second law in an accelerated frame.
This force does not exist when the observations are made in an inertial reference frame.
Let’s consider a free ball inside a box under uniform circular motion.
What does this mean and why is this true?
How does this motion look like in an inertial frame (or frame outside a box)? v
We see that the box has a radial force exerted on it but none on the ball directly Fr r
How does this motion look like in the box?
The ball is tumbled over to the wall of the box and feels that it is getting force that pushes it toward the wall.
Why?
According to Newton’s first law, the ball wants to continue on its original movement but since the box is turning, the ball feels like it is being pushed toward the wall relative to everything else in the box.
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu

16. Example of Motion in Accelerated Frames
A ball of mass m is hung by a cord to the ceiling of a boxcar that is moving with an acceleration a. What do the inertial observer at rest and the non-inertial observer traveling inside the car conclude? How do they differ?
This is how the ball looks like no matter which frame you are in. ac q
How do the free-body diagrams look for two frames? m
How do the motions interpreted in these two frames? Any differences?
For an inertial frame observer, the forces being exerted on the ball are only T and Fg. The acceleration of the ball is the same as that of the box car and is provided by the x component of the tension force.
Fg=mg m q T
Inertial Frame
In the non-inertial frame observer, the forces being exerted on the ball are T, Fg, and Ffic.
For some reason the ball is under a force, Ffic, that provides acceleration to the ball.
Non-Inertial Frame
Fg=mg m q T
Ffic
While the mathematical expression of the acceleration of the ball is identical to that of inertial frame observer’s, the cause of the force is dramatically different.
Wednesday, Oct. 1, 2003
PHYS , Fall 2003
Dr. Jaehoon Yu