Gravity - Physics - Lecture Slides, Slides of Physics

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Gravity

Newton’s Law of Gravitation

Kepler’s Laws of Planetary

Motion

Gravitational Fields

Newton’s Law of Gravitation

m 1

r m 2

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:

F G =

G m 1 m 2 r^2

The constant of proportionality is G, the universal gravitation constant. G = 6.67 · 10-11^ N·m^2 / kg^2. 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.

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/s^2.

For the little planet: a = (8.08 · 1015 N) / (5.21 · 10^22 kg)

= 1.55 · 10 -7^ m/s^2.

· 10^22 kg

1.23 · 1026 kg

8.08 · 1015 N 8.08 · 1015 N

Inverse Square Law

F G =

G m 1 m 2 r^2

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 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, and m is the mass of the object. The object’s weight can also be written as mg.

F G = =

G m 1 m 2 r^2

G M E m R E 2 =^ m^ g

The m’s cancel in the last equation. g can be measured experimentally; Cavendish determined G ’s value; and R E can be calculated at 6.37 · 10^6 m (see next slide). ME is the only unknown. Solving for ME we have: g R E^2 M E = (^) G = 5.98 ·^10 (^24) kg

Calculating the radius of the Earth

R E

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

s

is also the central angle of the arc. s = R E R E = s / 6.37 · 106 m

Earth

Falling Around

the Earth v

Newton imagined a cannon ball fired horizontally from a mountain top at a speed v. In a time t it falls a distance y = 0.5 g t^2 while moving horizontally a distance x = v t. 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..

x = v t

y = 0.5 g t^2 {

continued on next slide

Necessary Launch Speed for Orbit R = Earth’s radius t = small amount of time after launch x = horiz. distance traveled in time t y = vertical distance fallen in time t

R R

y = g t 2 / 2

x = v t

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

x 2 + R 2 = (R + y) 2 = R 2 + 2 R y + y 2

Since y << R, x 2 + R^2 R 2 + 2 R y x 2 2 R y

v 2 t 2 2 R (g t 2 / 2 )

v 2 R g. So,

v (6.37 · 106 m · 9.8 m/s^2 ) ½ v 7900 m/s

Early Astronomers

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.

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

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Kepler’s Laws of Planetary Motion

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:

T^2 R^3

Here is a summary of Kepler’s 3 Laws:

These laws apply to any satellite orbiting a much

larger body.

Kepler’s Second Law

Sun

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

C

D

A

B

The blue shaded sector has the same area as the red shaded sector. Thus, a planet moves from C to D in the same amount of time as it moves from A to B. 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.

(proven in advanced physics)

Kepler’s Third Law

The square of a planet’s period is proportional to the

cube of its mean distance from the sun: T^2 R^3

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. F is the force of gravity on the planet. F is also the centripetal force. If the orbit is circular, the planet’s speed is constant, and v = 2 R / T. Therefore,

Sun

F^ Planet

R

M

m

G M m R^2

m v^2 = R

G M R^2

m [2 R / T ] 2 = R

Cancel m ’s and simplify:

4 2 R

= T 2

Rearrange: G M

T^2

= R^3

Since G , M , and are constants, T^2 R Docsity.com^3.

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 says T^2 R^3 , so T^2 = k R^3 , where k is the constant of proportionality. Thus, for Earth and Jupiter we have:

T E 2 = k R E 3 and T J 2 = k R J 3

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

T J 2

T E 2

R J 3

R E 3

= T J 2 = (5.2) 3 T J = 11.9 years

continued on next slide

1 year 365 days

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:

v = d^2 (5.2)^ (93^ ·^106 miles) t

11.9 years

Jupiter is 5.2 AU from the sun (5. times farther than Earth is).

· (^) ·

1 day 24 hours

29,000 mph. Jupiter’s period from last slide

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!