Chapter 5 Part 2

Chapter 5 Part 2Newtons Law of Universal Gravitation, Satellites, and WeightlessnessNewtons ideas about gravityNewton knew that a force exerted on an object causes an acceleration.Most forces occurred because of contact with an object.The idea of a force without contact met resistance.Newton theorized the force that holds all things to the Earth, is the same force that acts to hold the moon in its nearly circular path.

Force at a distanceNewton compared the force Earth exerts on objects at its surface to the force Earth exerts on the moon.On Earths surface, acceleration due to gravity, g = 9.80m/s2.The centripetal acceleration of the moon is aR=v2/R. In terms of g, that is about 1/3600g.The moon is 60 times farther from Earths center than objects are from the surface.Newton concluded Fg on any object decreases with distance from Earths center by 1/d2.What about mass?Newton reasoned, according to his 3rd law of motion, that the force of gravity on an object must also be directly proportional to the masses of BOTH objects.

mB is the mass of a body, mE is the mass of Earth, and r is the distance between their centers.

Force vs distance appliedIn analyzing gravity, Newton found, by examining the orbits of planets around the sun, the force required to hold different planets in orbit around the Sun seemed to diminish as the inverse square of their distance from the Sun.

Universal GravitationIn examining the planets, Newton stepped further analyzing gravity.Newton theorized the gravitational force that attracts an apple to the Earth and the moon to the Earth, is the same force that acts between the Sun and other planets to keep them in their orbits.If gravity acts here, why not between ALL things?Finalizing the Universal law of GravitationEvery particle in the Universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. This force acts along the line joining the two particles.Universal Gravitational constant, G, (very small number) was experimentally discovered to calculate the EXACT force of attraction between 2 objects. G = 6.67 x 10-11 N-m2/kg2 (a constant like pi)Example 1A 50 kg person and a 75 kg person are sitting on a bench so that their centers are 50 cm apart. Estimate the magnitude of the gravitational force each exerts on the other.

Example 1 SolutionUsing the equation for force,

F = 1.0 x 10-2 NWhich is unnoticeably small unless delicate instruments are used.

Example 2What is the force of gravity acting on a 2000kg spacecraft when it orbits two Earth radii from the Earths center (a distance rE = 6380 km above Earths surface)? The mass of the Earth is mE = 5.98 x 1024kg.

Example 2 SolutionInstead of plugging in all the numbers for our equation, we could take a simpler approach.The spacecraft is twice as far from the Earths center as when at the surface of the Earth. Therefore, since the force of gravity decreases as the square of the distance (and 2 = ), the force of gravity on it will be only its weight at the Earths surface.FG = mg = (2000 kg)(9.80m/s2) = 4900 NGeophysical applications of Gravity near Earths surfaceApplying the Universal law between Earth and objects at Earths surface, the FG becomes the objects weight.Thus we rewrite the formula:

Cancelling mass we get:Acceleration of gravity at Earths surface depends on mE and rE. Once G was known, the Earths mass was determined.

ApplicationsWhen dealing with objects at Earths surface, we calculate weight by mg. If we want to consider objects far from Earths surface, we can calculate the acceleration due to gravity there by including their mass and distance from Earths surface.EX: Estimate the effective value of g on the top of Mt. Everest, 8848m above the Earths surface. What is the accel due to gravity of objects that freely fall at this altitude?

Example SolutionCalling the acceleration due to gravity there, g, we replace rE with r = 6380 km + 8.8km = 6389 km = 6.389 x 106 m:

=9.77m/s2Which is a reduction of about 0.3%.

Satellites and WeightlessnessArtificial satellites are put into Earth orbit by high accelerations to give them a high tangential speed, v. If tangential speed is too fast, the satellite escapes Earth orbit. If too slow, it accelerates downward to Earth due to Earths gravitational pull.What keeps a satellite up is a combination between its tangential speed and Earths aR.Satellites moving in an approx circle have an acceleration of aR = v2/r. This force is caused by the force of gravity acting on it.Newtons laws on SatellitesFor an orbiting satellite, the only force present is the force due to gravity.Using Newtons Second law: FR=maR, we find

Note that r is the sum of Earths radius and the satellites height above Earth: r = rE + h.

Geosynchronous SatellitesA geosynchronous satellite is one that stays above the same point on the equator of the Earth. Such are used for purposes like cable tv, weather forecasting, and communication relays. Determine (a) the height above Earths surface a satellite must orbit and (b) such a satellites speed.

See page 130 for worked out solution.Weightlessness Standing on a scale in a stationary elevator, mg ON the scale is equal to the support force, w, BY the scale.If the elevator accelerates upward, the net force on you = ma. So we have w mg = ma.Solving for w = mg + ma and you would weigh more than normal. In a freely falling elevator, the scale falls at the same rate as you and cannot push up (support) your weight, so it would read zero. (Apparent weightlessness).Weightlessness in satellitesSince satellites orbiting Earth are essentially falling around Earth, a passenger experiences the same apparent weightlessness that you would find in a freely falling elevator. Keplers Laws (1571-1630)Before Isaac Newton, a German astronomer named Johannes Kepler spent his lifetime studying planets and their motion. He developed 3 laws of planetary motion:1st Law: The path of each planet about the Sun is an ellipse with the Sun at one focus.2nd Law: Each planet moves so that an imaginary line drawn from the Sun sweeps out equal areas in equal periods of time.Keplers Third Law3rd Law: The ratio of the squares of the periods of any two planets (time for one revolution around the sun) is equal to the ratio of the cubes of their mean distances from the Sun.If T1 and T2 are the periods for any two planets and r1 and r2 are their average distances from the Sun, then

If we rewrite this, then r3/T2 should be the same for each planet.

Your turn to PracticePlease do Chapter 5 Review pg 141 #s 25-30, 39, and 42. Bonus # 53.