Gravitation
Kepler's law and Newton's Synthesis The nighttime sky with its myriad stars and shinning planets has always fascinated people on Earth. Towards the end of the XVI century the astronomer Tycho Brahe studied the motions of the planets and made accurate observations Using Brahe's data Johannes Kepler worked out a detailed description of the motion of the planets about the Sun: three empirical findings that we now refer to as Kepler's laws of planetary motion Kepler s laws provided the basis for Newton's discovery of the law of gravitation
Kepler s Laws and Newton's Synthesis Kepler s laws describe planetary motion. 1.The orbit of each planet is an ellipse with the Sun at one focus.
Kepler s Laws and Newton's Synthesis (cont d) 2. An imaginary line drawn from each planet to the Sun sweeps out equal areas in equal times.
Kepler s Laws and Newton's Synthesis (cont d) The ratio of the square of a planet s orbital period is proportional to the cube of its mean distance from the Sun. Planetary data applied to Kepler s Third Law
Jupiter's orbit Jupiter's mean orbital radius is 5.2 AU? What is the period of Jupiter's orbit around the Sun? τ =11.9 yr J
Kepler's law and Newton's Synthesis (cont d) The periods of the planets as a function of the mean orbital radii The square of the periods of the planets versus the cubes of the mean orbital radii
Newton s Law of Universal Gravitation If the force of gravity is being exerted on objects on Earth, what is the origin of that force? Newton s realization was that the force must come from the Earth. He further realized that this force must be what keeps the Moon in its orbit.
Newton s Law of Universal Gravitation (cont d) The gravitational force on you is one-half of a Third-Law pair: the Earth exerts a downward force on you and you exert an upward force on the Earth When there is such disparity in m s the reaction force undetectable but for bodies more equal in mass it can be significant
Newton s Law of Universal Gravitation (cont d) The gravitational force must be proportional to both masses By observing planetary orbits Newton also concluded that the gravitational force must decrease as the inverse of the square of the distance between the masses In its final form the Law of Universal Gravitation reads: F = G 12 m m 1 2 r² 12 ^ r 12
Newton s Law of Universal Gravitation (cont d) The magnitude of the gravitational constant G can be measured in the laboratory This is the Cavendish experiment G = 6.67 x 10-11 N m²/kg²
Every few hundred years most of the planets line up on the same side of the Sun. Calculate the total force on the Earth due to Venus, Jupiter, and Saturn, assuming all four planets are in a line. The masses are M V = 0.815 M M J = 318 M M = 95.1 M S their mean d from Sun are 108, 150, 778, and 1430 million km, respectively. What fraction of the Sun's force on the Earth is this. F = 9.56 x 10 N Earth/planets 22 F = 3.52 10 N 17 Earth/Sun (Not to scale)
Gravity Near the Earth s Surface: Geophysical Applications Now we can relate the gravitational constant to the local acceleration of gravity. We know that, on the surface of the Earth mg = G m m Solving for g gives m E Knowing g and the radius of the Earth, the mass of the Earth can be calculated: r² E g = G r² 6 gr² 24 m = E (9.80 m/s²)(6.38 x 10 m)² = = 5.98 x 10 kg E -11 G 6.67 x 10 N m²/ kg² E E
Gravity Near the Earth s Surface; Geophysical Applications (cont d) The acceleration due to gravity varies over the Earth s surface due to altitude, local geology, and the shape of the Earth, which is not quite spherical. Acceleration Due to Gravity at Various Locations on Earth Location Elevation (m) g (m/s²) New York 0 9.803 San Francisco 0 9.800 Denver 1650 9.796 Pikes Peak 4300 9.789 Sydney, Australia 0 9.798 Equator 0 9.780 North Pole 0 9.832
Falling to Earth What is the acceleration of an object at the altitude of the space shuttle's orbit, above 400 km above the Earth surface. a= 8.70 m/s²
Given that the acceleration of gravity at the surface of Mars is 0.38 of what it is on Earth, and that Mars radius is 3400 km, Pablo, Uniqua and Austin are trying to determine the mass of Mars. Can you help them? 23 M = 6.4 x 10 kg
Kepler s Laws from Newton's Law Kepler s laws can be derived from Newton s laws. Irregularities in planetary motion led to the discovery of Neptune, and irregularities in stellar motion have led to the discovery of many planets outside our Solar System.
Derivation of Kepler Second Law In a time dt the planet moves a distance vdt and the radius vector r sweeps out the area shaded in the figure This is half the area of the paralelogram form by vectors r and v dt r x mv da = ½ r x v dt = 2m da L = dt 2m L is the angular momentum of the planet around the Sun. Because the force of the planet is along the line from the planet to the Sun, it exerts no torque about the Sun L is constant The rate at which area is swept out is the same for all parts of the orbit dt
Derivation of Kepler Second Law (cont d) L = constant rv sin Φ = constant At aphelion and perihelion Φ = 90º r v = r v a a p p
Newton s Law of Gravitation implies Kepler s Third Law for circular orbit Planet Earth moves with speed v in an approximate circular orbit of radius r about the Sun The gravitational force on the Earth by the Sun provides the centripetal acceleration v²/r GM M ʘ r² = M v² = GM ʘ r Because the Earth moves a distance 2π r in time τ its speed is related v² r to the period by 2π r v = τ Substituting 4π² 3 τ² = r GM ʘ
Gravitational Potential Energy The general definition of the potential energy is du = -F dl where F is a conservative force acting on a particle and dl is a general displacement of the particle du = -F dl = -(-F r) dl = g ˆ GM m r² dr U = GM m r ²dr= GM m r + U 0
Gravitational Potential Energy (cont d) We can set U = 0 and then U = 0 at r 0 U = - GM m r Gravitational potential of the Earth starting at the surface of the planet U = - GM m R and increasing with rising r
Satellites Satellites are routinely put into orbit around the Earth. The tangential speed must be high enough so that the satellite does not return to Earth, but not so high that it escapes Earth s gravity altogether.
Satellites (cont d) The satellite is kept in orbit by its speed it is continually falling, but the Earth curves from underneath it.
Escape speed K + U = K + U f f i i The speed near the Earth s surface corresponding to zero total energy is called the escape speed v e 0 = ½ mv² - e v =( e 2GM R GM m R ½ ) =11.2km/s
Escape speed (cont d) The kinetic energy of an object at a distance r from the center of the Earth is K = E U (r) When E < 0 (E ) When E > 0 (E 2 ) K = 0 at r = r 1 max the object can escape the Earth
Gravitational field The gravitational field at point P is determined by placing a point particle of mass m and calculating the force on it due to all other particle g = lim F g m 0 The gravitational field at a field point due to masses of a collection of point particles is the vector sum of the fields due to individual masses m g = g i i The locations of these point are called source point To find the gravitational field of a continuous object we finf the field dg due to a small element of volume with mass dm and integrate over the entire mass distribution of the object g = dg The gravitational field of the Earth at a distance r R points towards the Earth and has a magnitude g (r) = GM r²
Gravitational field of two point particles Two point particles, each of mass M, are fixed in position on the y axis at y = +a and y = -a. Find the gravitational field at all points in the x axis as a function of x. g = - 2GMx (x²+a²) 3/2 ˆ i
A gravity map of Earth The Gravity Recovery and Climate Experiment (GRACE) is the first mission in NASA's Earth System Science Pathfinder project, which uses satelliteborne instrumentation to aid research on global climate change. Twin satellites launched in March 2002 are making detailed measurements of Earth gravitational field. They are in identical orbits with one satellite directly in front of the other by about 220 km. The distance between the satellites is continuously monitored with micrometer accuracy using onboard microwave telemetry equipment. How does the distance between the two satellites changes as the satellites approach a region of increased mass?
A gravity map of Earth (cont d) As the twin satellites approach a region where there is an excess of mass the increased gravitational field strength due to the excess mass pulls them forward (toward the excess of mass). The pull on the leading satellite is greater than the pull on the trailing satellite because the leading satellite is closer to the excess mass. Consequently the leading satellite is gaining speed more rapidly than is the trailing satellite. This results in an increase in separation between the satellites
A gravity map of Earth (cont'd) Gravity anomaly maps show how much the Earth's actual gravity field differs from the gravity field of a uniform, featureless Earth surface. The anomalies highlight variations in the strength of the gravitational force over the surface of the Earth The gal (sometimes called galileo) is defined as 1 centimeter per second squared 1 gal = 1 cm/s²
The gravitational slingshot effect The figure shows the planet Saturn moving in the negative x direction at its orbital speed (with respect to the Sun) of 9.6 km/. The mass of Saturn is 5.69 x 10 kg. A spacecraft with mass 825 kg approaches Saturn. When far from Saturn, it moves in the plus x direction at 10.4 km/s. The gravitational attraction of Saturn (a conservative force) acting on the spacecraft causes it to swing around the planet (orbit shown as dashed line) and head off in the opposite direction. Estimate the final speed of the spacecraft after it is far enough away to be considered free of Saturn's gravitational pull. 26 v = - 29.6 km/s f
Ocean tides The ocean tides have long been of interest to humans The Chinese explained the tides as the brething of the Earth *Around the sun once a year Galileo tried unsuccessfully to explain the tides effect of Earth s motion * On its own axis once a day (could not account for the timing of the approximately two high tides each day) Mariners have known for at least 4000 yr that tides are related to Moon s phases Exact relationship hidden behind many complicated factors Newton finally gave an adequate explanation
Ocean tides (cont d) To the moon Ocean tides are caused by the gravitational attraction of the ocean Calculation is complicated To the sun Surface of the Earth is not an inertial system! Earth s rotation Timing of tidal events is related to Revolution of the moon around the Earth If the moon was stationary in space tidal cycle would be 24 hours long However The moon is in motion revolving around the Earth 1 revolution takes about 27 days Adding about 50 minutes to the tidal cycle
Ocean tides (cont d) Second factor controlling tides on Earth s surface is then sun s gravity M ʘR² The height of the average solar tide is moon 0.46 the average lunar tide M moon R² ʘ At certain times during the moon s revolution around the Earth The direction of its gravitational attraction is aligned with the sun s During these times the two tide producing bodies act together Creating the highest and lowest tides of the year Spring tides occur every 14-15 days during full and new moons
Ocean tides (cont d) When gravitational pull of moon and sun are right angles to each other Daily tidal variations on the Earth are at their least Neap tides occur during the first and last quarter of the moon
g of a spherical shell of a solid sphere Consider a uniform spherical shell of mass M and radius R GM g = - ˆ r r > R r² g = 0 r < R To understand this last result consider the shell segments with masses m and m that are proportional 1 to the areas A and A which in turn are proportional m m 1 1 2 m 1 r² 2 to the radii r and r = = = 1 r² 2 2 1 A 1 r² 1 A 2 r² 2 m 2 Because the gravitational force falls off inversely as the square of the distance, the force on m due to the smaller mass m on the left, is exactly 0 1 balanced by that due to the more distant but larger mass m 2 on the right 2
g inside a solid sphere Consider a uniform solid sphere of radius R and mass M As we have seen the field inside a spherical shell is zero The mass of the sphere outside r exerts no force inside r Only the mass M whithin the radius r contributes to the gravitational field at r
g inside a solid sphere (cont d) The mass inside r produces a field equal to that of a point mass M at the center of the sphere For a uniform sphere 4 π r 3 3 π 4 R 3 3 M = M The gravitational field at a distance r < R is then g =- GM GM r GMr =- = - r r² r² R 3 R 3 3
Radially Dependent Density A solid sphere of radius R and mass M is spherically symmetric but not uniform. Its density ρ is proportional to the distance r from the center for r < R. That is, ρ = Cr, for r < R, where C is a constant. (a) Find C (b) Find g for all r > R. (c) Find g at r = R/2. C = M πr 4 GM g = - r (r > R) r² ˆ g = - GM ˆ r (@r = R/2) 4R²