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Planetary Motion & Gravitational Fields Historical Background Our ancestors envisioned the earth at the center of the universe and considered the stars to be fixed on a great revolving crystal sphere. The observation of planetary motions depends on the frame of reference of the viewer. Early Theories Geocentric Theory: Ptolemy~140 AD • Motion of planets from the point of view of the observer on Earth Heliocentric Theory: Copernicus, 1543. • Same motions but from the point of view of the observer on the Sun We use the heliocentric theory due to its simpler mathematics. Tycho Brahe (1546-1601) The Royal Astronomer for King Fredrick II of Denmark, •measured accurately the positions of the planets •Charted planets motion for 20 years •Accurate to 1/60 of a degree •“Greatest Observational Genius”. •35 years before the telescope. Johannes Kepler • Imperial Mathematician for Holy Roman Emperor Rudolf II • In 1600 became Brahe’s assistant. • Stole Brahe’s data after his death in 1601, spent 15 years analyzing it. • Deduced 3 laws of planetary motion • Apply to any system composed of a body revolving about a more massive body • Examples: moons, satellites, comets. Kepler’s 1st Law The Law of Orbits Planets move in elliptical orbits with the Sun at one of the focal points. Copyright ©2007 Pearson Prentice Hall, Inc. Kepler’s dA k dt where k is a constant nd 2 Law The Law of Areas A line from the Sun to a planet sweeps out equal areas in equal lengths of time. Copyright ©2007 Pearson Prentice Hall, Inc. Kepler’s rd 3 Law The Law of Periods (the Harmonic Law) The square of the orbital period of a planet is directly proportional to the cube of the average distance of the planet from the sun. 2 3 T r • The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their mean distances from the sun. T1 2 r1 3 T2 r2 Isaac Newton • Recognized that a force must be acting on the planets; otherwise, their paths would be straight lines. • Force Sun (Kepler’s 2nd Law) • Force decreases with the square of the distance (Kepler’s 3rd Law) Newton Continued… Gravity • Compared the fall of an apple with the fall of the moon. • The moon falls in the sense that it falls away from the straight line it would follow if there were no forces acting on it. • Therefore, the motion of the moon and the apple were the same motion. • Showed, everything in the universe follows a single set of physical laws. • Law of Universal Gravitation. Law of Universal Gravitation (1666) Every particle in the universe attracts every other particle with a force. This force is: • proportional to the product of their masses • inversely proportional to the square of the distance between them. • Acts on a line joining the two particles. Gm1m 2 F 2 -11 2 2 r G = 6.67 x 10 Nm /kg Discovered by Henry Cavendish in 1798 Law of Universal Gravitation What is the force of attraction between a 4.8 kg Physics book and a 2.4 kg Math book if they are 5.0 cm apart? Gm1m2 Fg 2 r G = 6.67 x 10-11 Nm2/kg2 Rule of 1’s Practice Gm1m2 Fg r2 a. If m1 is doubled, how is F changed? b. If neither of the masses were changed but r was doubled, how would F change? c. If r is not changed but both masses are doubled, how would F change? d. If r is halved and both masses doubled, how would F change? Kepler’s Law of Periods Proof Assumptions: • Must conform to equations for circular motion • Newton’s Universal Law of Gravity • Planet rotates in a circular (elliptical) path Fnet mv 2 Fc mac r Newton’s 3rd Law symmetry Fgravity Fnet Fc Recall, 2 T GMm mv 2 Fg 2 r r so 2r v r T Therefore, GMm m4 r 2 2 2 r T Law of Periods T 4 rearranging 3 R GM 2 2 Testing the Inverse Square Law of Gravitation • The acceleration due to gravity at the surface of the earth is 9.8m/s2. • If the inverse square relationship for gravity (Fg~1/r2) is correct , then, at a distance ~60 times further away from the center of the earth, the 9.8 m 2.72 10 acceleration due to gravity should be 60 s • The centripetal acceleration of the moon is given by a vr where the radius of the moon’s orbit is r = 3.84 × 108 m and the time period of the 2r moon’s orbital motion is T = 27.3 days. v 3 2 2 c ac 4 r T2 2 ac 4 2 (3.84 10 8 ) 24h (3600s) (27.3days)(1day ) 1h T 2 a c 2.72 10 3 m s2 2 Acceleration due to Gravity When only gravity is present, Fg = Fnet where Fnet = ma (Newton’s 2nd law) Fnet = ma = So, ma = Note: a = g GMm 2 r Gm1m2 Fg 2 r GM g 2 r Changes in Gravity Acceleration due to Gravity on different Planets Consider an object thrown upward v f vo 2 ad 2 2 at the maximum height vf=0 v o 2 gd 2 If the same object was thrown upward with the same initial speed gd constant So, g 1d 1 g 2 d 2 g1 d 2 g 2 d1 Acceleration due to Gravity on different Planets • Example A person can jump 1.5m on the earth. How high could the person jump on a planet having the twice the mass of the earth and twice the radius of the earth? Gravitational Field Strength (g.f.s) The force acting on a 1 kg mass at a specific point in a gravitational field. F g m Units: N/kg http://www.chrisbence.com/projects/powerplant/img/gravity_zoomed_mass.gif From Newton’s 2nd law F/m = a g.f.s is the acceleration due to gravity (g) GM g 2 r Variation of g With distance from a point mass • g.f.s at a point p is GM g 2 r • Variation of g with distance from the center of a uniform spherical mass of radius, R Variation of g GM g 2 r From • Close to earth the acceleration due to gravity changes slowly. • As the distance from earth increases, the change in the acceleration due to gravity increases. www.physicsclassroom.com/Class/energy/U5L1b.cfm Variation of g On a line joining the centers of two point masses • If m1 > m2 then Note: when g = 0 the gravitational fields cancel each other out Gravitational Potential Energy (GPE) of two point sources G P E W Fgd • Gravitational force is not constant but decreases as the separation of masses increase. • Requires calculus to compute the total R GMm work done. W G Mm W r R G Mm R r 2 dr G Mm GPE R Gravitational Potential (V) Background: The potential at a point in a gravitational field is equal to the work done bringing a 1kg mass from infinity to that point W G PE V m m G Mm V mr Recall, G Mm G PE r GM V r Units: J/kg Gravitational Potential (V) Potential (V) is the GPE per unit mass in a gravitational field. GM V Why the negative sign? r • A body at infinity, has zero gravitational potential. • A body normally falls to its lowest state of potential (energy) as r decreases, the potential decreases. • We can do this by including a negative sign in the above equation. • As r decreases, V decreases (becomes a greater negative quantity). Source: http://www.saburchill.com/physics/chapters/0007.html Equipotential Surfaces Equipotential Surface – points that all have the same potential (V) where GM V r • Formula shows that the points will all be the same distance from the mass or lie on a sphere whose center is that of a spherical mass. Cutnell & Johnson, Wiley Publishing, Physics 5th Ed. Equipotential Surfaces Equipotential Surface – points that all have the same potential (V) • This can be shown two dimensionally where the blue lines represent the equipotential lines Similar to topographical maps http://www.iop.org/activity/education/Projects/Teaching%20Advanced%20Physics/Fields/Images%20400/img_tb_4719.gif Equipotential surfaces due to two equal masses. • Note that the field becomes more spherical further away from the masses • At a far enough distance the field appears to come form one mass with the combined mass of the individual masses. Physics for the IB Diploma 5th Edition (Tsokos) 2008 Equipotential Surfaces Equipotential Surfaces • Equipotential surfaces due to two unequal masses Physics for the IB Diploma 5th Edition (Tsokos) 2008 Field Lines Field Lines – Indicate the direction of gravitational force that would act on a point mass at that location. 1. For a gravitational field, the lines will always point inward as gravity is always an attractive force. 2. Note that the field lines are perpendicular to the equipotential lines. http://www.intuitor.com/student/Planet%27s_g-field.jpg http://www.chrisbence.com/projects/powerplant/img/gravity_zoomed_mass.gif www.chrisbence.com/projects/powerplant/index.htm Field Lines Consider moving a test mass from point a to b as shown where the distance between a and b is r and the difference in potential is ΔV. The work W required is W mV W Fg r m gr Setting these expressions equal to each other V g r A more precise result based on using Calculus results in dV g dr Cutnell & Johnson, Wiley Publishing, Physics 5th Ed. The work done could also be calculate as Field Lines V g r This expression implies that on a graph of the variation of gravitational potential with distance (V vs r), the slope would be equal to the gravitational field strength g This also implies that the equipotential surfaces are always perpendicular to the gravitational field lines. • As shown for a single mass • True for any shape field lines and equipotential surfaces http://www.intuitor.com/student/Planet%27s_g-field.jpg Escape Velocity • The velocity needed to overcome (escape) the gravitational pull of a planet. • As the body is moving away from the planet, it is losing kinetic energy and gaining potential energy. • To completely escape from the gravitational attraction of the planet, the body must be given enough kinetic energy to take it to a position where its potential energy is zero. Escape Velocity • The potential energy possessed by a body of mass m, in a gravitational field is given by GPE = Vm • If the field is due to a planet of mass M and radius R, then the escape velocity can be calculated as follows: ΔKE = ΔGPE • So, 2 GM vesc R • as g = GM/R² v es c 2 gR 1 m v 2 esc G Mm 2 rE Satellites: Orbits & Energy G Mm using Newton’s second law (F=ma) U r G Mm v2 F m 2 r r Solving for v2 where v2/r is the centripetal acceleration. GM v r 2 GMm KE mv So, 2r 1 2 2 U KE 2 Total Mechanical Energy of a satellite is: GMm GMm ME KE U 2r r ME G Mm 2r Satellites: Orbits & Energy ME = -KE This is for a circular orbit. • For a satellite in an elliptical orbit with a semi-major axis a G Mm E 2a where a was substituted for r Energy Graphs • As shown the energies for a satellite are, G Mm KE 2r • Graphing these, http://www.opencourse.info/astronomy/introduction/06.motion_gravity_laws/energy_elliptical.gif G Mm U r G Mm ME 2r Energy Wells • If we rotate the potential energy graph, we can generate a 3D picture known as an energy well. http://www.opencourse.info/astronomy/introduction/06.motion_gravity_laws/ Physics for the IB Diploma 5th Edition (Tsokos) 2008 Energy Changes in a Gravitational Field • A mass placed in a gravitational field experiences a force. – If no other force acts, the total energy will remain constant but energy might be converted from g.p.e. to kinetic energy. • If the mass of the planet is M and the radius of the orbit of the satellite is r, then it can easily be shown that the speed of the satellite, v, is given by – if r decreases, v must increase. Energy Changes in a Gravitational Field • If the satellite’s mass is m, then the kinetic energy, K, of the satellite is K 1 m GM 2 r • the potential energy of the satellite, U, is These equations show: GM U m r – that if r decreases, K increases but U decreases (becomes a bigger negative number) – the decrease in U is greater than the increase in K. • Therefore, to fall from one orbit to a lower orbit, the total energy must decrease. – work must be done to decrease the energy of the satellite if it is to fall to a lower orbit. Energy Changes in a Gravitational Field • The work done, W, is equal to the change in the total energy of the satellite, W = ΔK + ΔP. • This work results in a conversion of energy from gravitational potential energy to internal energy of the satellite (it makes it hot!). • Air resistance reduces the speed of the satellite along its orbit. This allows the satellite to fall towards the planet. As it falls, it gains speed. • So, if a viscous drag (air resistance) acts on a satellite, it will – decrease the radius of the orbit – increase the speed of the satellite in it’s new orbit. Energy Changes in a Gravitational Field • In principle, the satellite could settle in a lower, faster orbit. • In practice it will usually be falling to a region where the drag is greater. It will therefore continue to move towards the planet in a spiral path. Energy Changes in a Gravitational Field • Looking at this with a fbd of the satellite spiraling downward due to air resistance. – Note that the velocity is no longer perpendicular to the weight. – Due to this there is a component of gravity which is now acting in the direction of the velocity providing a net force which is accelerating the satellite. Physics for the IB Diploma 5th Edition (Tsokos) 2008 Physics for the IB Diploma 5th Edition (Tsokos) 2008 Weightlessness Astronauts on the orbiting space station are weightless because... a. there is no gravity in space and they do not weigh anything. b. space is a vacuum and there is no gravity in a vacuum. c. space is a vacuum and there is no air resistance in a vacuum. d. the astronauts are far from Earth's surface at a location where gravitation has a minimal affect. e. None of the above Source: http://www.physicsclassroom.com/Class/circles/u6l4d.cfm Weightlessness Source: Kirk, Tim, Physics for the IB Diploma, Oxford University Press 2007 Astronauts on the orbiting space station are weightless because... The astronaut and the space station are both free-falling together - Apparent Weightlessness • The gravitational force on the astronaut provided the needed centripetal acceleration for the astronaut to stay in orbit. • The space station remains in orbit because of the gravitational force on it. • However, since there is no contact force between the satellite and the astronaut here is an apparent weightlessness Weightlessness Source: Kirk, Tim, Physics for the IB Diploma, Oxford University Press 2007 Weightlessness Astronauts on the orbiting space station are weightless because...with calculations • Summing the forces on the astronaut F F g • For circular motion (Will learn in the next unit) • • • • FN Fnet Fnet v2 m r Mm v2 m M N G m G v2 r2 r r r Mm v2 F G r 2 FN m r So, or Recall for a satellite, v GM r Substituting, N m G M G M 0 r r r With no reaction force (normal) the astronaut feels weightless 2 Source: Physics for the IB Diploma 5th Edition (Tsokos) 2008