Gravity Newton’s Law of Gravitation Kepler’s Laws of Planetary Motion Gravitational Fields

Newton’s Law of Gravitation m1m1 m2m2 r 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 2r 2 The constant of proportionality is G, the universal gravitation constant. G = 6.67 x 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.

Gravity Example F G = G m 1 m 2 r 2r 2 How hard do two planets pull on each other if their masses are 1.23 x kg and 5.21 x kg and they 230 million kilometers apart? This is the force each planet exerts on the other. Note the denominator is has a factor of 10 3 to convert to meters and a factor of 10 6 to account for the million. It doesn’t matter which way or how fast the planets are moving. (6.67 x N·m 2 / kg 2 ) (1.23 x kg) (5.21 x kg) = (230 x 10 3 x 10 6 m) 2 = 8.08 x N

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 x N) / (1.23 x kg) = 6.57 x m/s 2. For the little planet: a = (8.08 x N) / (5.21 x kg) = 1.55 x m/s x kg 1.23 x kg 8.08 x N

Inverse Square Law F G = G m 1 m 2 r 2r 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 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 calculate G. Since G is only 6.67 x N·m 2 / kg 2, the measurements had to be very precise.

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 F G in the equation below. M E is the mass of the Earth, R E 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 2r 2 G M E m R E 2 = mg 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). M E is the only unknown. Solving for M E we have: g R E 2 G M E = = 5.98 x kg

Calculating the radius of the Earth   RERE Here’s how the Greeks approximated Earth’s radius over 2000 years ago: Red pole blocks incoming solar rays and makes a shadow.  is measured. The green pole is on the same meridian as the red one, but it casts no shadow at this latitude. s is the measured distance between the cities in which the poles stand. s  is also the central angle of the arc. s = R E  R E = s /   6.37 x 10 6 m Earth

Net Force Gravity Problem 3 x 10 6 kg 2.7 x10 6 kg2.5 x 10 6 kg 40 m 60 m 3 asteroids are positioned as shown, forming a right triangle. Find the net force on the 2.5 million kg asteroid. 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: N at 14.6  above horizontal (N of W)

Falling Around the Earth v Newton imagined a cannon ball fired horizontally from a mountain top at a speed v. If fired fast enough 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

Earth Orbits Consider a satellite of mass m, circling close to the earth’s surface (so that the extra radius above the surface of the earth is negligible). In order to maintain this orbit, there is a minimum speed with which the satellite must travel. The force which is required to maintain this orbit is:

This must be equal to the gravitational force Cancelling m on both sides leaves: v 2 =rg Thus v=√(rg) (now r= 6.4x10 6 m and g=9.8Nkg -1 ) v=8x10 3 ms -1 or 8kms -1

What is the period of this orbit? Time= Distance/Speed =2πr/v = (2xπx6.4x10 6 )/ 8000= 83 mins

Geostationary Orbits It is possible to extend the distance the satellite has to travel by increasing its height above the earth. There will be a height at which the period of the satellite will equal the period of rotation of the earth (i.e 24hrs). At this height the satellite will remain at the same point above the earth’s surface. It will be in geostationary orbit.

Height required for Geostationary Orbit: (this is the height above the centre of the earth) Height above the surface of the earth is: radius of the earth= 36000km G=6.67x m 3 /kgs 2 T=24hrs M=6x10 24 kg

Early Astronomers In the 2 nd 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. Copernicus Tycho Brahe

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.

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 First Law Planets move around the sun in elliptical paths with the sun at one focus of the ellipse. An ellipse has two foci, F 1 and F 2. For any point P on the ellipse, F 1 P + F 2 P is a constant. The orbits of the planets become more circular if F 1 and F 2 are close together (A circle is a an ellipse with both foci at the same point--the center.) Comets have very eccentric (highly elliptical) orbits. F1F1 F2F2 Sun Planet P

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 Planet F R M m G M m R 2R 2 m v 2 R = m [2  R / T] 2 R = Cancel m’s and simplify: G M R 2 4 2 R4 2 R T 2 = Rearrange: G MG M T 2 4  2 = R3R3 Since G, M, and  are constants, T 2  R 3.

Third Law Analysis G MG M T 2 4  2 = R3R3 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: G M m R 2R 2 m v 2 R = v = G M R

Third Law Example One astronomical unit (AU) is the distance between Earth and the sun (about 150 million km). Jupiter is 5.2 AU from the sun. How long is a Jovian year? answer: Kepler’s 3 rd 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: TJ 2TJ 2 TE 2TE 2 RJ 3RJ 3 RE 3RE 3 = T J 2 = (5.2) 3 T J = 11.9 years continued on next slide

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: t v = d = Jupiter is 5.2 AU from the sun (5.2 times farther than Earth is). 1 year 365 days 2  (5.2) (150 x 10 6 km) 11.9 years · · 1 day 24 hours  46,400 km/h. 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 !

Third Law Practice Problem Venus is about AU from the sun, Mars AU. Venus takes days to circle the sun. Figure out how long a Martian year is. answer: 686 days

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 is 9.8 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, 1 N / kg = 1 m/s 2. So, the gravitational field strength is just the acceleration due to gravity, g. Earth’s surface continued on next slide

Uniform Gravitational Fields (cont.) Earth’s surface 10 kg 98 N A 10 kg mass is near the surface of the Earth. Since the field strength is 9.8 N / kg, each of the ten kilograms feels a 9.8 N force, for a total of 98 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).

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

Escape Velocity If a body thrown skywards it will accelerate due to the gravitational field of the earth. As this has a value of -9.8m/s 2 the body normally will eventually stop and fall back towards the earth. Obviously if the body is given sufficient initial velocity it will be possible for it to escape the earth’s gravitational pull. It would then travel into outer space. We call this value the ESCAPE VELOCITY.

Escape Velocity We know that as the body moves upwards the kinetic energy it loses is gained as potential energy (neglecting air friction). Ek ->Ep The work done in moving from the surface of the earth, re, to the furthest extent of the earths gravitational field, ∞, is: At this point (∞), the potential is zero

The kinetic energy required to produce the initial velocity must be equal to the work required, thus:

Escape velocity in terms of ‘g’ But Thus Therefore

Energies of Orbiting Bodies A body orbiting a planet has both Potential Energy (Ep) due to position and Kinetic Energy (Ek) due to its speed of orbit. Consider a satellite orbiting at a height r above the centre of a planet. Its potential energy is :

Energies of Orbiting Bodies The Kinetic Energy of the satellite is : The total energy is the sum of its potential and kinetic energies.

Energies of Orbiting Bodies Now from centripetal force equations: therefore rearranging we get

Energies of Orbiting Bodies thus

Graphs of Kinetic, Potential and Total Energies of a mass of 1kg in a circular orbit around the earth; Ep Ek Etot