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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

PowerPoint® Lectures forUniversity Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 12

Gravitation


Goals for Chapter 12

• To study Newton’s Law of Gravitation

• To consider gravitational force, weight, and gravitational energy

• To compare and understand the orbits of satellites and celestial objects

• To explore the existence and nature of black holes (beyond science fiction)


Homework

Read Ch 12

Try them in that order (increasing difficulty)

3, 11, 7, 15, 21, 23, 27, 31, 35, 43, 45


Introduction

• Looking at the picture of Saturn, we see a very organized ring around the planet. Why do the particles arrange themselves in such orderly fashion?

• From Copernicus and Galileo to Hubble and NASA, centuries of scientists have struggled to characterize gravitation and celestial motion.


Newton’s Law of GravitationThe gravitational force is always attractive and depends on both the masses of the bodies involved and their separations. Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distances between them.

Equation 12.1:

See F

igur

es

12.1

and

12.2

.

221

r

mGmFg =


G and g?

Both values use the letter “gee”, but only one is a universal constant.“g” is the acceleration due to gravity, which differs from place to place and planet to planet.“G” is a constant related to the gravitational force between two objects in the universe, anywhere.Eq 12.1 deals with distance from center of mass.Consider a situation where we drill a hole through the earth to its “chewy” center.

How would the force of gravity change as we went from the surface toward the center of the earth?

What would be the force of gravity at the center of the earth?


Henry Cavendish determines G

Gravitational forces were relative until 1798 when Henry Cavendish made the sensitive measurement to determine a numerical value for the constant G.Cavendish calculated a value very close to the accepted value of6.67310 x 10-11 Nm2/kg2 or for us: G = 6.67 x 10-11 Nm2/kg2


Calculate gravitational variables• The mass m1 of one of the small spheres of a Cavendish balance is

0.0100 kg, the mass m2 of one of the large spheres is 0.500 kg, and the center-to-center distance between each large sphere and the nearer small one is 0.0500 m.

• Find the gravitational force Fg on each sphere due to the nearest other sphere.

• Suppose one large sphere and one small sphere are detached from the apparatus in the previous example and placed 0.0500 m from each other at a point in space far removed from all other bodies.

• What is the magnitude of the acceleration of each, relative to an inertial system?


Calculate gravitational variables• Many stars in the sky are actually systems of two or more stars that

remain together due to their mutual gravitational attraction. Below is a trinary star system at an instant when the stars are at the vertices of a 45o right triangle. We assume that the stars are spherical so that we can replace each star by a point mass at its center.

• Find the magnitude and direction of the total gravitational force exerted on the small star by the two other large ones.


Homework

On page 468:

3, 11, 7


Read 447 carefully.


12.2 WeightThe weight of a body is the total gravitational force exerted on the body by all other bodies in the universe.Most other gravitational forces are neglected when close to a large body.The weight close the surface of the earth is:

We can use this equation and w = mg to calculate the mass of the earth, like on page 442.

2E

Eg R

mGmFw ==


Weight (skip Weight Watchers, just climb upward)• Gravity (and hence, weight) decreases as altitude rises.


If the earth was uniform . . .

How would the graph in Figure 12.7 look if we extended the plot to the center of the earth?The density of the earth, rho, can be calculated by dividing the earth’s mass by its volume:

By contrast, the density of water is 1.00 g/cm3

Rocks on the surface have a density of 3-5 g/cm3

So the earth cannot be uniform.

24

21 3 3 33

5.97 104 1.09 103

5500 5.5E

E

m x kg kg g

x m m cmRπρ = = = =


If it were uniform:


Gravitational force changes densities below sea level• Just as it’s interesting to remember

that all gravitational forces are calculated from the center of the planet, it’s interesting to follow the density as one proceeds from crust to mantle to core.


With this information on the density it would look like:


Gravity on Mars

You’re involved in the design of a mission carrying humans to the surface of the planet Mars, which has a radius rM = 3.40 X 106 m and a mass mM = 6.42 X 1023 kg. The earth weight of the Mars Lander is 39,200 N. Calculate its weight Fg and the acceleration gM due to the gravity of Mars:

6.0 X 106 m above the surface of Mars.

At the surface of Mars.


Gravitational Potential Energy

Close to the surface U = mgy was sufficient, when Fg = mg. We have to use another relationship, one based on

when r changes enough that the gravitational force is no longer constant.To find this relationship we must follow the same steps we did to find U = mgy, we must integrate Fg.

Because Wgrav= U1 - U2:

221

r

mmGFg =

2 2 2

2 22 1

1 1 1

E E E

r r rm m Gm m Gm mdr

grav r E r rr rr r r

W F dr G dr Gm m= = = = − +∫ ∫ ∫

rmmEGU −=


Gravitational potential energyWhen the astronaut moves away, U becomes less negative. When the astronaut moves toward the earth, U becomes more negative.U is zero when r = infinity.


Jules Verne had no way to know…To escape from the earth, an object must have escape velocity (not a small number). In Jules Verne’s 1865 story three men were shot to the moon in a shell fired from a giant cannon sunk in the earth in Florida.Find the muzzle speed needed to shoot the shell straight up to a height above the earth equal to the earth’s radius.Find the escape speed—that is, the muzzle speed that would allow the shell to escape from the earth completely.Read page 447 carefully.


Homework

On page 469:

15, 21, 23

Read 447 - 456


Satellite motionTrajectories 1-5 are called closed orbits, 6 and 7 are open orbits.For a circular orbit (like 4) a certain velocity must be maintained:

rGmEv =


Circular OrbitsWe can derive an expression relating the orbital radius, r, to the period of the orbit, T. The speed, v, is the distance traveled in one revolution, 2πr, divided by the time, T:

Substitute our previous expression for v:

We can also use the velocity equation to calculate the total energy in a circular orbit (conservation applies)

Trv π2=

E

Gmr

vr

Gm

rrT

E

23

222 πππ ===

( ) ( )rmm

rmm

rGm

rmm

E

EEE

GE

GmGmvUKE

2

212

21

−=

−=−+=+=


Consider satellite orbits

• Several images of things in orbit to consider are shown below.


A satellite orbitSuppose you want to place a 1000-kg weather satellite into a circular, 300 km orbit above the earth’s surface.

What speed, period, and radial acceleration must it have?

How much work has to be done to place this satellite in orbit?

How much additional work would have to be done to make this satellite escape the earth?


Homework

On page 469:

27 and 31


Kepler’s laws for planetary motion

• Each planet moves in an elliptical orbit with the sun at one focus.

• A line connecting the sun to a given planet sweeps out equal areas in equal times.

• The periods of the planets are proportional to the 3/2 powers of the major axis lengths in their orbits.


Kepler, Newton, and ellipse theoryKepler did not know why the planets followed his laws, but Newton figured it out three generations later. Newton found that each of Kepler’s Laws can be derived. Lets do that now.The longest dimension is the major axis, with half-length a, called the semi-major axis. The sum of the distances from S to P and S’ to P is the same for all points on the curve. S and S’ are the foci of the ellipse, the sun is on S and the planet is on P. Nothing is on S’.


Kepler, Newton, and ellipse theory

Newton was able to show that for a body acted on by an attractive force proportional to 1/r2, the only paths could be a circle or an ellipse. (see figure 12.13 on page 448)

The distance of each foci to the center of the ellipse is the product of the semi-major axis a and a dimensionless number e from 0 to 1 called the eccentricity. If e = 0 the ellipse is a circle. The closest approach of the planet to the sun is called the Perihelion, the furthest distance is the Aphelion. (anti � Aphelion)


Kepler’s 2nd Law �� derivedThe figure shows that for a small time interval dt, the line from the sun to the planet turns through an angle dθ. The area of the triangle swept out is ½ base rtimes the height rdθ (the arclength). The rate this area is swept out is called the sector velocity:

This sector value must be the same for similar time intervals for Kepler’s 2nd

law to hold. When the planet is close to the sun, r is small and dθ/dt is large, when the planet is far away from the sun r is large then dθ/dt is small.

dt

dr

dt

dA θ2

2

1=


Kepler’s 2nd Law �� derivedTo see how this relates to Newton’s law we need to express it in terms of the velocity vector v of the planet. The perpendicular component of the velocity to the radial line v┴ = v sinφ. The displacment along the direction of v┴durring time interval dt is rdθ so we also have v┴ = rdθ/dt. Using this we get:

Now: Which is 1/m times the angular momentum:

φsin2

1rv

dt

dA =

vrrv��×=φsin

vmrL��

�

×=


Kepler’s 2nd Law �� derivedSo we can rewrite the sector velocity dA/dt as:

So Kepler’s 2nd law states that the planets angular momentum is constant.Remember that:

The vector r is directed along the straight line path between the planet and the sun, just like the gravitational force Fg. Because they are always in the same direction, the cross product is always zero, so is change in momentum.

1

2 2

dA Lr v

dt m= × =� �

Frdt

Ld �

��

�

×==τ


Kepler’s 3rd law

We already derived this law for the special case when e = 0, an ellipse that is a circle. Newton was able to show that the same relationship holds true for an elliptical orbit when we replace the circular radius r with the semi-major axis a:

SGm

aT

23

2π=


Orbital questions• At what point in an elliptical orbit does a planet have the

greatest speed?

• The asteroid Pallas has an orbital period of 4.62 years and an orbital eccentricity of 0.233. Find the semi-major axis of its orbit.

• Comet Halley moves in an elongated elliptical orbit around the sun. At perihelion, the comet is 8.75 x 107 km from the sun; at aphelion it is 5.26 x 109 km from the sun. Find the semi-major axis, eccentricity, and period of the orbit.


Homework

On page 470:

35

Read 461 - 466


Black Holes

Read on own, but important info:

Derivation of Schwarzschild radius, Rs, is not as simple as book has, but math is too complicated for this level.

2

2

c

GMRs =


A visit to a black hole

• Refer to Example 12.11.


Homework

On page 470:

43 and 45


Homework

Read Ch 12


3, 11, 7, 15, 21, 23, 27, 31, 35, 43, 45

ch 12 notes - fcps 12 notes.pdf · jules verne had no way to know… to escape from the earth, an...

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