Giancoli - Chapter 5 - Section 5.6 - 5.9 - Universal Gravitation

Newton’s Law of Universal GravitationGIANCOLI - CHAPTER 5 – SECTION 5.6 – 5.9

MR. SHARICK

Section 5.6 – Newton’s Law of Universal Gravitation

The geocentric model had now been proven incorrect, and the heliocentric model of the universe was very widely accepted.

Sir Isaac Newton, at the same time he developed the three laws of motion, was also examining the motions of the planets and the moon.

He was also trying to determine what the cause of gravity was (Where did this force come from?)

He determined that gravity must act at a distance, as opposed to most forces which are contact forces.

I wonder what causeth the apple to falleth from the tree to the ground?

Section 5.6 – Deriving Newton’s Universal Law

Newton found that the moons centripetal acceleration was 1/3600 of g.

He also realized that the moon was about 384,000km away from the Earth, which was about 60 times the radius of the earth (6380 km).

Newton proposed that the proportionality between the gravitational force and the distance between the two centers of the two objects was as follows:

𝐹 ∝1

𝑑2


2

2

2 3

2 2

2 3

2 2

2

, 2

(get the total acceleration since sin )

(using r = sin )

(since c = d)

41. ( sin )

42. sin

4 13.

44. = P

P S

a a

da dT

ca

T d

ca

T d

MF

d

3

2 (using )

cF ma

T


3

2

2 3

2

Using Kepler's 3rd law of planetary motion Newton knew that

c was a constant for all planets orbiting the sun. Therefore, to

simplify Equ. # 4 Newton defined a new variable...

45. K

T

c

T

3

2

20

m

, 2

, , 2

which he calculated as 1.33 10

Subsituting Equ # 5 into Equ. # 4 we get

6.

Using his own 3rd Law of motion Newton realized that

7.

s

P

P S

P

S P P S

MF K

d

MF F K

d

HERE IT COMES!!! ANOTHER STROKE OF GENIUS FROM NEWTON

Newton sees symmetry between these forces. If FS,P is proportional to the

mass of planet, then the FP,S must be proportional to the mass of the sun,

and therefore sets

where he called G the Universal Gravitational Constant.

With one last substitution Newton finds…


sK GM

, , 2

s PS P P S

M MF F G

d


Realize Newton still did not know G or MS

But, he did know GMS =

Newton assumed that his law should work for any two objects not just

planets. Newton’s Law of Universal Gravitation for any two objects is

stated as follows…

The direction of the force is along the line joining the two objects and is always an attractive force.

1 2

2

m mF G

d

320

2

m1.33 10

s

Section 5.6 – Cavendish Experiment (1798)

•Used to measure the universal constant “G” performed by British scientist Henry Cavendish

1. Two small masses are placed at the end of a light rod.

2. The force required to rotate the fiber from its untwisted position (a), is measured.

3. The force required to twist the fiber is calibrated as a function of the angle,

twist measuredF

Section 5.6 – Cavendish Experiment (1798)

4. The light beam and scale are used to magnify theta.

5. When the mass M, is brought near to mass m it applies a gravitational force onto the fiber. To check for symmetry it is placed in positions AA and BB.

6. Knowing that the force causing the twist was caused by gravity Cavendish now used The Law of Universal Gravitation.

7. Cavendish was now able to isolate G, and calculated it as… 2Twist Gravity

mMF F G

d

2

2

116.67 10 N m

kgG x

Example Problem – Using Newton’s Universal Law of GravitationAssume the earth is moving in a circular orbit around the sun. Using the following data calculate the speed of the earth in its orbit in miles/hr.

Mean Radius of Orbit = 1.5 x 1011 m

Mass of Sun = 1.99 x 1030 kg

1 mile = 1.61 km

G = 6.67 x 10-112

2

N m

kg

Section 5.6 – Example 5.11 – Can you attract another person gravitationally?A 50 kg person and a 75 kg person are sitting on a bench so that their centers are about 50 cm apart. Estimate the magnitude of the gravitational force each exerts on the other.

Section 5.6 – Example 5.12 – Spacecraft at 2reWhat is the force of gravity acting on a 2000 kg spacecraft when it orbits two Earth radii from the Earth’s center (re = 6380 km)? The mass of the Earth is me=5.98x1024kg.

Section 5.6 - Finding ‘g’ from Newton’s Law of Universal GravitationShow that g = 9.81 m/s^2 using Newton’s Law of Universal Gravitation, mass of the earth me=5.98x1024kg, and the radius of the earth is 6380 km.

Section 5.6 – Example Problem - Gravity from the moon

What is the gravitational force on a 50 kg person due to the mass of the moon? The mass of the moon is 7.36 x 1022 kg and the distance to the moon is 3.82 x 108m.

Section 5.7 – Gravity Near the Earth’s Surface

Which ninja will experience a larger acceleration from the earth?

How does this differ from g=9.81 m/s^2 that we calculate at sea level?

Section 5.7 – Gravity Near the Earth’s SurfaceThe force from gravity a person experiences is his weight, which is:

Newton says that the attractive force between two objects is given by:

Gravity is the attractive force between the Earth and you!

1 2

2

gF mg

Gm mF

r

Section 5.7 – Gravity Near the Earth’s SurfaceSo making object 1 you, and object 2 the earth:

Cavendish was the first person to measure G, and used this to find the mass of the Earth!

How could you use this equation to find ‘g’ on another planet or moon?

2

2

you earth

you

earth

earth

earth

Gm mm g

r

Gmg

r

Section 5.7 – Example 5.14 – Gravity on EverestEstimate the effective value of g on top of Mt. Everest, 8848m (29,028ft) above the Earth’s surface.

What are some things that lower the precision of this calculated value?

Section 5.7 – Problem 5.32What is the distance from the Earth’s center to a point outside the Earth where the gravitational acceleration is 1/10 of its value at the earth’s surface?

Section 5.7 – Losing weightHow would the weight of a 100 kg person change as he increases in altitude? Find his weight at (a) sea level, (b) 5000 meters, (c) 10,000 meters, and (d) 20,000 meters.

Section 5.8 – Satellites and “Weightlessness”

Section 5.8 – Newton’s cannonballWhat causes a satellite to stay in orbit, rather than falling to the Earth?

To answer this, let’s consider a cannon mounted on the top of Mt. Everest.

Section 5.8 – Newton’s cannonballWhat causes a satellite to stay in orbit, rather than falling to the Earth?

To answer this, let’s consider a cannon mounted on the top of Mt. Everest.

Section 5.8 – Newton’s cannonballIf we were to load the cannon with enough gun powder and fire it HORIZONTALLY to the surface of the Earth below, we may hit a town far away.

Section 5.8 – Newton’s cannonballIf we were to load the cannon with more gun powder and fire it HORIZONTALLY to the surface of the Earth below, we may even hit a different country far away.

Section 5.8 – Newton’s cannonballIf we were to load the cannon with more gun powder and fire it HORIZONTALLY to the surface of the Earth below, we may even hit a different country far away.

Section 5.8 – Newton’s cannonballWell, let’s say we were to load the cannon with perfect amount of gun powder and fire it HORIZONTALLY to the surface of the Earth below, so that it just keeps falling for ever as the Earth curves away from its fall. If this cannon ball does not hit the back of the cannon which it was fired from, then this cannon ball will continuously go around the world; falling towards the Earth… FOREVER!!!!!

Section 5.8 – Newton’s cannonballWell, let’s say we were to load the cannon with perfect amount of gun powder and fire it HORIZONTALLY to the surface of the Earth below, so that it just keeps falling for ever as the Earth curves away from its fall. If this cannon ball does not hit the back of the cannon which it was fired from, then this cannon ball will continuously go around the world; falling towards the Earth… FOREVER!!!!!

Section 5.8 – Escape VelocityAnd if it’s fired at a speed that is higher than its escape velocity, the following happens!

Section 5.8 – Calculating Escape Velocity

2

c c

mvF ma

r

RECALL:


1 2

2G

m mF G

r

2

c c

mvF ma

r

RECALL:

And


1 2

2G

m mF G

r

2

c c

mvF ma

r

C GF FRECALL:

And


1 2

2G

m mF G

r

2

c c

mvF ma

r

C GF F

2

2

sat Earth satm v M mG

r r

RECALL:

And


1 2

2G

m mF G

r

2

c c

mvF ma

r

C GF F

2

2


r r

Earthorbiting

Mv G

r

RECALL:

And


1 2

2G

m mF G

r

2

c c

mvF ma

r

C GF F

2

2


r r

Earthorbiting

Mv G

r

Earthescape

Mv G

r

RECALL:

And

Section 5.8 – Example 5.15 –Geosynchronous satelliteA geosynchronous satellite is one that stays above the same point on the equator of the Earth. Such satellites are used for such purposes as cable TV transmission, weather forecasting, and as communication relays. Determine (a) the height above the Earth’s surface such a satellite must orbit and (b) such a satellite’s speed.

Section 5.8 – Orbiting satelliteA 5000.0 kg satellite is moving in a stable circular orbit at altitude of 12,800 km above the earth's surface.

Rearth= 6.38 x 106 m Mearth= 5.98 x 1024 kg

G= 6.67 x 10-11 N*m2/kg2

a. Calculate the orbiting velocity of the satellite.

b. What is its period (time to make one orbit) in hours.

Section 5.8 – “Weightlessness”Objects like satellites in orbit are said to experience apparent “weightlessness”.

Let’s first consider a simpler case, a person standing in an elevator.

Section 5.8 – “Weightlessness”If a = 0, then it is as simple as standing on solid ground.

W = mg

Does this change if the elevator is moving with constant speed?

Section 5.8 – “Weightlessness”If the elevator is accelerating up at a rate of a=1/2 g, then we have

ma = W – mg

W = ma + mg = (1/2)mg + mg = (3/2)mg

Section 5.8 – “Weightlessness”If the elevator cable breaks, you have more problems than just worrying about your weight!

…like, dying when the elevator crashes into the ground.

Section 5.8 – “Weightlessness”But, let’s look at the physics anyway!

The overall acceleration is a = g, so

ma = mg - W

W = mg - ma

W = 0; thus “weightless!”

Section 5.8 – “Weightlessness”

These two situations are nearly the same!

Section 5.8 – Finding the acceleration of a “weightless” astronutThis man is plummeting toward the Earth with an acceleration very close to 9.8 m/s^2.

If the satellite is orbiting 36,000 km above the surface of the Earth, what is the value of “g” this astronut experiences?

Section 5.8 – Problem 5.41A cylindrical spaceship is rotated such that its occupants are to experience simulated gravity of ½ g. Assume the spaceship’s diameter is 32m. Find the time needed for one revolution (period).

Section 5.8 – Problem 5.46A ferris wheel, 24.0m in diameter rotates once every 12.5s. What is

the person’s apparent weight (mass = 80kg) at the top and at the bottom, compared to the person’s weight at rest?

Section 5.9 – Kepler’s Laws1st Law- The orbits of the planets are ellipses, with the sun at one focal point.

Section 5.9 – Kepler’s Laws2nd Law- The line joining the sun and a planet (its radius vector) sweeps over equal areas in equal times.

The time it takes to go from 1 to 2 (moving faster) is the same as it takes to go from 3 to 4 (moving slower)

Section 5.9 – Kepler’s LawsAlso called the Law of Equal Areas

Section 5.9 – Kepler’s Laws3rd - The squares of the periods of the planets’ motions are proportional to the cubes of the semimajor axes of their elliptical paths; that is c3/T2 is the same for all the planets, where T is the time for the planet to complete one orbit about the sun and c is the semimajor axis.

3 3

2 2

A B

A B

c c

T T

Deriving Kepler’s 3rd Law

2

1 1 1

2

1 1

11

1

2

1 1 1

2 2

1 1

Setting forces equal:

We know:

2

Substituting gives:

4

S

S

Gm M m v

r r

rv

T

Gm M m r

r T

2 2

1

3

1

2 2

1 2

3 3

1 2

Rearrange to get:

4

This is the constant the Kepler derived!

Do this for a 2nd planet and then you can write:

S

T

r GM

T T

r r

Section 5.9 – Using Kepler’s 3rd LawMars’ period was first noted by Kepler to be about 687 days (Earth days), which is 1.88 years. Determine the distance of Mars from the Sun using the Earth as a reference (distance of the earth from the sun is 1.50 x 1011m).

Section 5.9 – Using Kepler’s 3rd LawHow long does it take the planet Saturn to make one revolution about the Sun if its orbital radius is 9.5 times larger than the earth?

Section 5.9 – Using Kepler’s 3rd LawDetermine the mass of the Sun given the Earth’s distance from the Sun as r = 1.50 x 1011m.

2 2

1

3

1

4

S

T

r GM

Section 5.9 – Problem 5.56The Sun rotates about the center of the Milky Way Galaxy at a distance of about 30,000 light years from the center (1 light year = 9.5x1015m). If it take about 200 million years to make one rotation, estimate the mass of our galaxy. Assume that the mass is concentrated mostly in a central uniform sphere.

If all the stars had the mass of our Sun (about 2 x 1030kg) how many stars would there be in our galaxy?

Section 5.9 – Kepler’s LawsWhat happens next?

•Galileo used his telescopes to discover the four moons of Jupiter and times their

periods of orbit.

•He found that for all four moons Kepler’s ratio of c3/T2 was constant.

•This convinced most scientists that Kelper’s laws were not an accidental fit, and

Copernicus’ heliocentric universe model must be correct.

•The Catholic Church was not convinced and Galileo was arrested for his

teachings and excommunicated from the church. He has since been re-instated.

-Kepler and Galileo had worked to show how the planets behaved, and how the

motions of the moons of Jupiter and the planets of our sun obeyed the same

rules. It was Newton’s turn to explain what causes planets to obey the same

rules.

Giancoli - Chapter 5 - Section 5.6 - 5.9 - Universal Gravitation

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