Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.
Upcoming SlideShare
Loading in …5
×

# Gravity 2

899 views

Published on

• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here
• Be the first to comment

### Gravity 2

1. 1. GravityNewton’s Law of GravitationKepler’s Laws of Planetary Motion Gravitational Fields
2. 2. 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: G m1 m2 FG = r2The constant of proportionality is G, the universal gravitationconstant. G = 6.67 · 10-11 N·m2 / kg2. Note how the units of G allcancel out except for the Newtons, which is the unit needed on theleft side of the equation.
3. 3. Gravity ExampleHow hard do two planets pull on each other if their masses are1.23 × 1026 kg and 5.21 × 1022 kg and they 230 million kilometersapart? G m1 m2 FG = r2 (6.67 · 10-11 N·m2 / kg2) (1.23 · 1026 kg) (5.21 · 1022 kg) = (230 · 103 · 106 m) 2 = 8.08 · 1015 NThis is the force each planet exerts on the other. Note thedenominator is has a factor of 103 to convert to meters and a factor of106 to account for the million. It doesn’t matter which way or howfast the planets are moving.
4. 4. 3rd Law: Action-ReactionIn the last example the force on each planet is the same. This is due toto Newton’s third law of motion: the force on Planet 1 due to Planet 2is just as strong but in the opposite direction as the force on Planet 2due 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. 5.21 8.08 · 1015 N 8.08 · 1015 N 1.23 · 1026 kg · 1022 kg
5. 5. 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. G m1 m2 FG = r2What if each mass and the separation were all quadrupled? answer: no change in the force
6. 6. Calculating the Gravitational ConstantIn 1798 Sir Henry Cavendish suspended a rod with two small masses(red) from a thin wire. Two larger mass (green) attract the smallmasses and cause the wire to twist slightly, since each force ofattraction produces a torque in the same direction. By varying themasses and measuring the separations and the amount of twist,Cavendish was the first to calculate G. Since G is only 6.67 · 10-11 N·m2 / kg2, the measurements had to be very precise.
7. 7. 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. G m1 m2 G ME m FG = r2 = RE 2 = mgThe m’s cancel in the last equation. g can be measured experimentally;Cavendish determined G’s value; and RE can be calculated at6.37 · 106 m (see next slide). ME is the only unknown. Solving forME we have: g RE 2 ME = = 5.98 · 1024 kg G
8. 8. Calculating the radius of the EarthThis is similar to the way the Greeks approximated Earth’s radius over2000 years ago: θ Red pole blocks incoming solar rays and makes a shadow. θ is measured. RE s θ The green pole is on the same meridian as the red one, but it casts Earth no shadow at this latitude. s is the measured distance between the cities in which the poles stand. θ is also the central angle of the arc. s = RE θ RE = s / θ ≈ 6.37 · 106 m
9. 9. Net Force Gravity Problem 3 × 106 kg 3 asteroids are positioned as shown, forming a right triangle. Find the net40 m force on the 2.5 million kg asteroid. 2.7 × 106 kg 2.5 × 106 kg 60 mSteps: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)
10. 10. Falling Around x = vt vthe Earth y = 0.5 g t 2 { 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. . continued on next slide
11. 11. Necessary Launch Speed for OrbitR = Earth’s radius x 2 + R 2 = (R + y) 2t = small amount of time after launchx = horiz. distance traveled in time t = R2 + 2 R y + y 2y = vertical distance fallen in time t Since y << R, (If t is very small, the red segment is nearly vertical.) x 2 + R2 ≈ R2 + 2 R y ⇒ x2 ≈ 2R y x = vt ⇒ v 2 t 2 ≈ 2 R (g t 2 / 2)y = gt 2 / 2 ⇒ v 2 ≈ R g. So, v ≈ (6.37 · 106 m · 9.8 m/s2) ½ R R v ≈ 7900 m/s
12. 12. Early AstronomersIn the 2nd century AD the Alexandrian astronomer Ptolemy put forth atheory that Earth is stationary and at the center of the universe and thatthe sun, moon, and planets revolve around it. Though incorrect, it wasaccepted 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.
13. 13. Early AstronomersBoth 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.
14. 14. 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: T 2 ∝These laws apply to any satellite orbiting a much R3 larger body.
15. 15. 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 PlanetAn ellipse has two foci, F1 and F2. For any point P on the ellipse,F1 P + F2 P is a constant. The orbits of the planets are nearly circular(F1 and F2 are close together), but not perfect circles. A circle is a anellipse with both foci at the same point--the center. Comets have veryeccentric (highly elliptical) orbits.
16. 16. 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 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.
17. 17. 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 3Assuming that a planet’s orbit is circular (which is not exactly correctbut is a good approximation in most cases), then the mean distancefrom the sun is a constant--the radius. F is the force of gravity on theplanet. F is also the centripetal force. If the orbit is circular, theplanet’s speed is constant, and v = 2 π R / T. Therefore, GMm m v2 m [2 π R / T] 2 = = R2 R R m Cancel m’s GM 4 π2 R F Planet and simplify: = R2 T2 M R 4 π2 3 Sun Rearrange: T 2 = R GM Since G, M, and π are constants, T 2 ∝ R 3.
18. 18. Third Law Analysis 4 π2 3 We just derived T 2 = R GM• It also shows that the orbital period depends on the mass of thecentral body (which for a planet is its star) but not on the mass of theorbiting body. In other words, if Mars had a companion planet thesame 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 ittakes to complete an orbit. Likewise, an artificial satellite circlingEarth from a great distance has a greater period than a satellite orbitingcloser. There are two reasons for this: 1. The farther away the satelliteis, the farther it must travel to complete an orbit; 2. The farther out itsorbit is, the slower it moves, as shown: GMm m v2 GM = ⇒ v= R2 R R
19. 19. Third Law ExampleOne astronomical unit (AU) is the distance between Earth and thesun (about 93 million miles). Jupiter is 5.2 AU from the sun. Howlong is a Jovian year?answer: Kepler’s 3rd Law says T 2 ∝ R 3, so T 2 = k R 3, where k isthe constant of proportionality. Thus, for Earth and Jupiter we have: TE 2 = k RE 3 and TJ 2 = k RJ 3k’s value matters not; since both planets are orbiting the samecentral body (the sun), k is the same in both equations. TE = 1 year,andRJ / RE = 5.2, so 3 dividing equations: TJ 2 RJ = ⇒ TJ 2 = (5.2) 3 ⇒ TJ = 11.9 years TE 2 RE 3 continued on next slide
20. 20. Third Law Example (cont.)What is Jupiter’s orbital speed?answer: Since it’s orbital is approximately circular, and it’s speedis approximately constant: Jupiter is 5.2 AU from the sun (5.2 times farther than Earth is). v= d = 2 π (5.2) (93 · 106 miles) 1 year 1 day t · · 11.9 years 365 days 24 hours≈ 29,000 mph. Jupiter’s period from last slideThis means Jupiter is cruising through the solar system at about13,000 m/s ! Even at this great speed, though, Jupiter is so far awaythat when we observe it from Earth, we don’t notice it’s motion.Planets closer to the sun orbit even faster. Mercury, the closestplanet, is traveling at about 48,000 m/s !
21. 21. Third Law Practice Problem Venus is about 0.723 AU from the sun, Mars 1.524 AU. Venus takes 224.7 days to circlethe sun. Figure out how long a Martian year is. answer: 686 days
22. 22. Uniform Gravitational FieldsWe live in what is essentially a uniform gravitational field. This meansthat the force of gravity near the surface of the Earth is pretty muchconstant in magnitude and direction. The green lines are gravitationalfield lines. They show the direction of the gravitational force on anyobject in the region (straight down). In a uniform field, the lines areparallel and evenly spaced. Near Earth’s surface the magnitude of thegravitational field is 9.8 N / kg. That is, every kilogram of mass anobject has experiences 9.8 N of force. Since a Newton is a kilogram ·meter per second squared, 1 N / kg = 1 m/s2. So, the gravitational fieldstrength is just the acceleration due to gravity, g. continued on next slide Earth’s surface
23. 23. Uniform Gravitational Fields (cont.)A 10 kg mass is near the surface of the Earth. Since the fieldstrength is 9.8 N / kg, each of the ten kilograms feels a 9.8 Nforce, for a total of 98 N. So, we can calculate the force ofgravity by multiply mass and field strength. This is thesame as calculating its weight (W = mg). 10 kg 98 N Earth’s surface
24. 24. Nonuniform Gravitational FieldsNear 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 Earth the surface, meaning the field is weaker there. • get closer together closer to the surface, meaning the field is stronger there.