Gravity
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Gravity . Newton’s Law of Gravitation Kepler’s Laws of Planetary Motion Gravitational Fields. G m 1 m 2. F G =. r 2. Newton’s Law of Gravitation. r. m 2. m 1.

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Gravity

Gravity

Newton’s Law of Gravitation

Kepler’s Laws of Planetary Motion

Gravitational Fields


Newton s law of gravitation

G m1m2

FG =

r2

Newton’s Law of Gravitation

r

m2

m1

There is a force of gravity between any pair of objects anywhere. The force is proportional to each mass and inversely proportional to the square of the distance between the two objects. Its equation is:

The constant of proportionality is G, the universal gravitation constant. G = 6.67 · 10-11 N·m2 / kg2. Note how the units of G all cancel out except for the Newtons, which is the unit needed on the left side of the equation.


Gravity example

G m1m2

FG =

r2

Gravity Example

How hard do two planets pull on each other if their masses are 1.231026 kg and 5.211022 kg and they 230 million kilometers apart?

(6.67 · 10-11 N·m2 / kg2) (1.23 · 1026 kg) (5.21 · 1022 kg)

=

(230 · 103 · 106 m)2

= 8.08 · 1015 N

This is the force each planet exerts on the other. Note the denominator is has a factor of 103to convert to meters and a factor of 106 to account for the million. It doesn’t matter which way or how fast the planets are moving.


3rd law action reaction

3rd Law: Action-Reaction

In the last example the force on each planet is the same. This is due to to Newton’s third law of motion: the force on Planet 1 due to Planet 2 is just as strong but in the opposite direction as the force on Planet 2 due to Planet 1. The effects of these forces are not the same, however, since the planets have different masses.

For the big planet: a = (8.08 ·1015 N) / (1.23 ·1026 kg) = 6.57 ·10-11 m/s2.

For the little planet: a = (8.08 ·1015 N) / (5.21 · 1022 kg) = 1.55 ·10-7 m/s2.

1.23 ·1026 kg

5.21 · 1022 kg

8.08 ·1015 N

8.08 ·1015 N


Inverse square law

G m1m2

FG =

r2

Inverse Square Law

The law of gravitation is called an inverse square law because the magnitude of the force is inversely proportional to the square of the separation. If the masses are moved twice as far apart, the force of gravity between is cut by a factor of four. Triple the separation and the force is nine times weaker.

What if each mass and the separation were all quadrupled?

answer:

no change in the force


Calculating the gravitational constant

Calculating the Gravitational Constant

In 1798 Sir Henry Cavendish suspended a rod with two small masses (red) from a thin wire. Two larger mass (green) attract the small masses and cause the wire to twist slightly, since each force of attraction produces a torque in the same direction. By varying the masses and measuring the separations and the amount of twist, Cavendish was the first to calculateG.

SinceGis only6.67 · 10-11N·m2 / kg2, the measurements had to be very precise.


Calculating the mass of the earth

G m1m2

FG =

r2

Calculating the mass of the Earth

Knowing G, we can now actually calculate the mass of the Earth. All we do is write the weight of any object in two different ways and equate them. Its weight is the force of gravity between it and the Earth, which is FG in the equation below. ME is the mass of the Earth, RE is the radius of the Earth, andmis the mass of the object. The object’s weight can also be written as mg.

G MEm

=

=mg

RE2

The m’s cancel in the last equation. gcan be measured experimentally; Cavendish determinedG’s value; and REcan be calculated at 6.37 · 106 m (see next slide). MEis the only unknown. Solving forMEwe have:

g RE2

= 5.98 ·1024kg

ME=

G


Calculating the radius of the earth

Calculating the radius of the Earth

This is similar to the way the Greeks approximated Earth’s radius over 2000 years ago:

Red pole blocks incoming solar rays and makes a shadow. is measured.

RE

s

The green pole is on the same meridian as the red one, but it casts no shadow at this latitude. sis the measured distance between the cities in which the poles stand.

Earth

is also the central angle of the arc. s = RE

RE= s/ 6.37 ·106 m


Net force gravity problem

Net Force Gravity Problem

3106kg

3 asteroids are positioned as shown, forming a right triangle. Find the net force on the 2.5 million kg asteroid.

40 m

2.7106kg

2.5106kg

60 m

Steps:1. Find each force of gravity on it and draw in the vectors.2. Find the angle at the lower right.3. One force vector is to the left; break the other one down into components.4. Find the resultant vector: magnitude via Pythagorean theorem; direction via inverse tangent.

answer:

0.212 N at 14.6 above horizontal (N of W)


Falling around the earth

x = vt

Falling Around the Earth

v

y = 0.5gt2{

Newton imagined a cannon ball fired horizontally from a mountain top at a speedv. In a timetit falls a distancey = 0.5gt2while moving horizontally a distancex = vt. If fired fast enough (about 8 km/s), the Earth would curve downward the same amount the cannon ball falls downward. Thus, the projectile would never hit the ground, and it would be in orbit. The moon “falls” around Earth in the exact same way but at a much greater altitude. .

continued on next slide


Necessary launch speed for orbit

Necessary Launch Speed for Orbit

R = Earth’s radiust = small amount of time after launchx = horiz. distance traveled in timety = vertical distance fallen in timet

x2 + R2 = (R + y)2

= R2 + 2Ry + y2 Sincey << R, x2 + R2R2 + 2Ryx2 2Ry v2 t22R(g t2/2)

v2Rg. So,

v  (6.37 ·106 m ·9.8 m/s2)½

v  7900 m/s

(If t is very small, the red segment is nearly vertical.)

x = vt

y = gt2/2

R

R


Early astronomers

Early Astronomers

In the 2nd century AD the Alexandrian astronomer Ptolemy put forth a theory that Earth is stationary and at the center of the universe and that the sun, moon, and planets revolve around it. Though incorrect, it was accepted for centuries.

In the early 1500’s the Polish astronomer Nicolaus Copernicus boldly rejected Ptolemy’s geocentric model for a heliocentric one. His theory put the sun stated that the planets revolve around the sun in circular orbits and that Earth rotates daily on its axis.

In the late 1500’s the Danish astronomer Tycho Brahe made better measurements of the planets and stars than anyone before him. The telescope had yet to be invented. He believed in a Ptolemaic-Coperican hybrid model in which the planets revolve around the sun, which in turn revolves around the Earth.


Early astronomers1

Early Astronomers

Both Galileo and Kepler contributed greatly to work of the English scientist Sir Isaac Newton a generation later.

In the late 1500’s and early 1600’s the Italian scientist Galileo was one of the very few people to advocate the Copernican view, for which the Church eventually had him placed under house arrest. After hearing about the invention of a spyglass in Holland, Galileo made a telescope and discovered four moons of Jupiter, craters on the moon, and the phases of Venus.

The German astronomer Johannes Kepler was a contemporary of Galileo and an assistant to Tycho Brahe. Like Galileo, Kepler believed in the heliocentric system of Copernicus, but using Brahe’s planetary data he deduced that the planets move in ellipses rather than circles. This is the first of three planetary laws that Kepler formulated based on Brahe’s data.


Kepler s laws of planetary motion

Kepler’s Laws of Planetary Motion

Here is a summary of Kepler’s 3 Laws:

1. Planets move around the sun in elliptical paths with the sun at one focus of the ellipse.

2. While orbiting, a planet sweep out equal areas in equal times.

3. The square of a planet’s period (revolution time) is proportional to the cube of its mean distance from the sun: T2 R3

These laws apply to any satellite orbiting a much larger body.


Kepler s first law

Kepler’s First Law

Planets move around the sun in elliptical paths with the sun at one focus of the ellipse.

F1

F2

Sun

P

Planet

An ellipse has two foci, F1 and F2. For any pointPon the ellipse, F1P + F2Pis a constant. The orbits of the planets are nearly circular (F1and F2are close together), but not perfect circles. A circle is a an ellipse with both foci at the same point--the center. Comets have very eccentric (highly elliptical) orbits.


Kepler s second law

Kepler’s Second Law

(proven in advanced physics)

While orbiting, a planet sweep out equal areas in equal times.

A

D

Sun

C

B

The blue shaded sector has the same area as the red shaded sector. Thus, a planet moves fromCtoDin the same amount of time as it moves fromAtoB. This means a planet must move faster when it’s closer to the sun. For planets this affect is small, but for comets it’s quite noticeable, since a comet’s orbit is has much greater eccentricity.


Kepler s third law

mv2

=

R

GMm

R2

42R

=

T2

m[2R/T]2

=

42

R

R3

=

T2

GM

Kepler’s Third Law

The square of a planet’s period is proportional to the cube of its mean distance from the sun: T2R3

Assuming that a planet’s orbit is circular (which is not exactly correct but is a good approximation in most cases), then the mean distance from the sun is a constant--the radius. Fis the force of gravity on the planet. Fis also the centripetal force. If the orbit is circular, the planet’s speed is constant, and v = 2R/T. Therefore,

m

Cancel m’s and simplify:

GM

Planet

F

R2

M

R

Sun

Rearrange:

Since G, M,andare constants,T2R3.


Third law analysis

42

R3

=

T2

GMm

GM

R2

Third Law Analysis

We just derived

  • It also shows that the orbital period depends on the mass of the central body (which for a planet is its star) but not on the mass of the orbiting body. In other words, if Mars had a companion planet the same distance from the sun, it would have the same period as Mars, regardless of its size.

  • This shows that the farther away a planet is from its star, the longer it takes to complete an orbit. Likewise, an artificial satellite circling Earth from a great distance has a greater period than a satellite orbiting closer. There are two reasons for this: 1. The farther away the satellite is, the farther it must travel to complete an orbit; 2. The farther out its orbit is, the slower it moves, as shown:

mv2

G M

=

 v =

R

R


Third law example

Third Law Example

One astronomical unit (AU) is the distance between Earth and the sun (about 93 million miles). Jupiter is 5.2 AU from the sun. How long is a Jovian year?

answer:Kepler’s 3rd Law saysT2 R3, so T2= kR3, wherek is the constant of proportionality. Thus, for Earth and Jupiter we have:

TE2= kRE3andTJ2= kRJ3

k’s value matters not; since both planets are orbiting the same central body (the sun), k is the same in both equations. TE = 1 year, and RJ /RE = 5.2, so dividing equations:

TJ2

RJ3

TJ2 = (5.2)3

=

TJ = 11.9 years

TE2

RE3

continued on next slide


Third law example cont

Third Law Example (cont.)

What is Jupiter’s orbital speed?

answer: Since it’s orbital is approximately circular, and it’s speed is approximately constant:

Jupiter is 5.2 AU from the sun (5.2 times farther than Earth is).

1 year

d

1 day

2 (5.2)(93·106 miles)

=

v =

·

·

t

11.9 years

365 days

24 hours

Jupiter’s period from last slide

 29,000 mph.

This means Jupiter is cruising through the solar system at about 13,000 m/s! Even at this great speed, though, Jupiter is so far away that when we observe it from Earth, we don’t notice it’s motion.

Planets closer to the sun orbit even faster. Mercury, the closest planet, is traveling at about 48,000 m/s!


Third law practice problem

Third Law Practice Problem

Venus is about 0.723 AU from the sun, Mars 1.524 AU. Venus takes 224.7 days to circle the sun. Figure out how long a Martian year is.

answer:

686 days


Uniform gravitational fields

Uniform Gravitational Fields

We live in what is essentially a uniform gravitational field. This means that the force of gravity near the surface of the Earth is pretty much constant in magnitude and direction. The green lines are gravitational field lines. They show the direction of the gravitational force on any object in the region (straight down). In a uniform field, the lines are parallel and evenly spaced. Near Earth’s surface the magnitude of the gravitational field is9.8 N/kg. That is, every kilogram of mass an object has experiences 9.8 N of force. Since a Newton is a kilogram·meter per second squared, 1N/kg = 1m/s2. So, the gravitational field strength is just the acceleration due to gravity,g.

continued on next slide

Earth’s surface


Uniform gravitational fields cont

Uniform Gravitational Fields (cont.)

A 10 kg mass is near the surface of the Earth. Since the field strength is9.8 N/kg, each of the ten kilograms feels a9.8 Nforce, for a total of98 N. So, we can calculate the force of gravity by multiply mass and field strength. This is the same as calculating its weight (W = mg).

10 kg

98 N

Earth’s surface


Nonuniform gravitational fields

Nonuniform Gravitational Fields

Near Earth’s surface the gravitational field is approximately uniform. Far from the surface it looks more like a sea urchin.

  • The field lines

  • are radial, rather than parallel, and point toward center of Earth.

  • get farther apart farther from the surface, meaning the field is weaker there.

  • get closer together closer to the surface, meaning the field is stronger there.

Earth


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