Physics 1110 Fall 2004L33 - 1Lecture 33 10 November 2004Announcements• Reading Assignment for Fri, Nov. 12: Knight 13.1Newton’s Law of GravitationNewton’s theoretical solution to explain Kepler’s laws was to proposethat there exists a force between all objects, not just planets, and themathematical form of this force wasFg=Gm1m2r2where this is the force between two masses 1 and 2 and r is thedistance between the objects. This law is called “universal” because itapplies to all objects, not just planets or not just on the surface of theearth. The constant G is determined from experiment and it is verysmall: G = 6.67 x 10-11 N-m2/kg2 . Also we have to be careful when wetalk about the distance between objects. For point masses, theanswer is easy; it’s just the position of the point mass. In the case ofreal extended objects we have to use a calculation of a point in theobject called the “center of gravity”. We’ll learn how to calculate this inthe next chapter. For uniform density objects with a regular shape(e.g., sphere or cube), the center of gravity is at the geometric center.This is not obvious, and Newton spent 20 years inventing calculus sohe could prove that you should use the center of the planets, whichare roughly spherical. For now, we’ll stick to simple objects, so youshould use the distance between the centers of two objects when youcalculate the distance r in the gravity force formula.In order to make this more clear, let’s do a few example calculations.1. Force of gravity between two 1 kg masses, 1 meter apart.Fg=Gm1m2r2=(6.67x1011Nm2/ kg2)(1kg)(1kg)(1m)2= 6.67x1011N.Physics 1110 Fall 2004L33 - 2This is a very tiny force! This is why we are not generally aware ofthese forces in everyday life.2. Force of gravity between Earth and a 1 kg mass at the Earth’ssurface. Note that here we actually use the distance betweenthe surface (object position) and the center of the earth (earthposition) to calculate the distance in the gravity formula. This isthe Earth’s radius of course, as shown below.Now we calculate the force on the 1 kg object:Fg=GMemRe2=(6.67x1011Nm2/ kg2)(6x1024kg)(1kg)(6.4 x106m)2= 9.8 N.This is just what we would get using our old mg formula andnow we can see where it comes from. Furthermore, we cannow calculate “g” for any planet by using the equalitymgp=GMpmRp2gp=GMpRp2.Physics 1110 Fall 2004L33 - 33. Finally let’s consider the gravitational force of the earth on themoon:Fg=GMemmrem2=(6.67x1011Nm2/ kg2)(6x1024kg)(7.4 x1022kg)(3.8x108m)2= 2x1020N.This is an enormous force, but then we do need a force capableof keeping the moon in orbit around the earth! In fact, now we’llturn to just this question.OrbitsTo continue with understanding Kepler’s laws, we need to understandjust what an orbit is. When we studied free fall trajectories using mgas the force pointed down towards the ground, we found trajectoriesthat were parabolas. However, if we throw an object with enoughhorizontal velocity so that it would hit the ground 10 or 15 miles away,we would get the wrong answer! The surface of the earth is reallyspherical, and in that distance the curvature would have two effects:1) the ground would have curved away “lower” and 2) the force wouldhave rotated so that it kept pointing straight towards the center of theearth. Nonetheless, the object would eventually hit the ground, just abit further away, after a longer time. If we continue this process andthrough the object faster and faster horizontally, we can get the objectto move farther and farther around the earth before it hits. Eventuallywe would reach a speed where the object never hit the earth again. Itwould continually fall towards the earth’s surface, but the earth’ssurface would keep curving away, and so the object never is able toreach the surface. In this last case, we say that the object is orbitingthe earth.Let’s now study this quantitatively by considering circular orbits first.We’ll take a specific case of a 100 kg satellite orbiting the earth at analtitude of 1 km. (By the way, we are ignoring all air resistance andmountains, etc, here just to keep things simple.) First let’s calculatethe force of gravity on the satellite:Physics 1110 Fall 2004L33 - 4Fg=GMemsres2=(6.67x1011Nm2/ kg2)(6x1024kg)(100kg)(6.4 x106+1000 m)2= 980 N.Note that this is almost identical to just using msg because thisaltitude is only a small fraction of the Earth’s radius. The figure belowillustrates this.Now we have analyzed circular motion before, and we know that thenet force towards the center must be equal to mv2/r . There is onlyone force here, gravity, and it points towards the center. Going backand using the equations we can solve for v in terms of thegravitational force.GMemsres2=msv2resv2=GMeresv =GMeres.This is an amazing formula. Using Newton’s law of gravity I can nowrelate the speed of a satellite to the mass of the earth and the radiusof the orbit. If I look at planets orbiting the sun, or moons orbiting aplanet, I can figure out the mass of the sun (or planet) by justmeasuring the orbit radius and speed with a telescope! If we plug inthe numbers for our satellite at 1 km altitude, we get v = 7900 m/s(17,700 mph)! For the moon, we get v = 1000 m/s. As the radius getsPhysics 1110 Fall 2004L33 - 5larger, the speeds get smaller. Note that they do not depend on themass of the satellite.PeriodAnother quantity that characterizes an orbit is the period, the time tomake one complete orbit. This is now easy to calculate, just using thecircumference divided by the speed.T =2resv= 2res3GMeIf we plug in our satellite numbers, the period for the satellite wouldbe only 85 minutes, while that of the moon around the earth is about27 days (1 lunar month).If we square the formula for the period we get a new formula:T2= 42res3GMeIn fact , this is Kepler’s 3rd law! A circle is a special case of an ellipse,in which the two foci merge at the center, and the semimajor andsemiminor axes are equal to the radius of the circle. Now we caneven explain and predict the constant of proportionality that Keplerfound. In the case of planets around the sun, it depends on the massof the sun.Next lecture, we’ll turn to elliptical orbits and also look at the potentialenergy function for this new law of
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