Phys 201 1nd Edition Lecture 10Outline of Last Lecture I. Solving for uniform circular motion using period, frequency, angular velocity, and centripetal acceleration.Outline of Current Lecture II. Centrifugal ForceIII. Newton’s law of Universal GravitationA. Gravity as an Attractive ForceB. EquationC. Gravity on Earth IV. Kepler’s lawV. Astronomical UnitsCurrent LectureCentrifugal force:Do not confuse centrifugal force with centripetal force. Centrifugal force is a somewhat abstract concept.Essentially, it is a fake force that only exists in our brains to explain what we observe with forces. For example, we recognize gravity as a force that pulls objects down. However, gravity can actually be felt in, certain situations, pushing on one in a certain direction. Centrifugal force describes what our brain perceives as the “unknown” force caused by the actions of an identified force.Newton’s Law of Universal Gravitation:Newton discovered that the force of gravity is what keeps the planets moving in a circular orbit. Gravity works as an attractive force between two bodies and it is universal. This means that the coefficient of gravitation is constant for any given mass, and the attractive force between two objects is proportional to the masses involved and the distance between the two. The equation for Newton’s universal law of gravitation is;F=(GM1M2)/r2For this equation, M1 represents the mass of the first object and M2 represents the mass of the second object. r represents the distance from the center of the first object, to the center of the second object. G represents gravitation and is always constant. The value of G is 6.67x10-11 and the unit is m3/s2kg.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Because the value of G is extremely small, the force of gravity is only significant when dealing with exceptionally large masses at very long distances. This is why when we’ve previously dealt with gravity (as g) we’ve always used the value 9.81N. This value corresponds to the force of gravity on Earth, based on the mass and radius of Earth. An object that is projected anywhere on the planet will not have a big enough mass or cover a large enough distance, to significantly affect the force of gravity between the object and the Earth. Therefore, we assume that the force of gravity of any object on Earth will be affected by a force of gravity of approximately 9.81N.Because we are so familiar with the force of gravity on Earth, most of the problems we do involving gravitational force are ratio problems. This just means that we are comparing the effects of gravitational force on earth to the effects of gravitational force on other masses, such as planets or stars.Example: An astronaut has a weight of 700N on Earth. If the mass of Mars is .107 of Earths mass and the radius of Mars is .53 of Earths radius, what does the astronaut weigh on Mars?For this problem, let the coefficient e represent Earth and the coefficient m represent Mars. Because we know the force of gravity on earth is 9.81, and that the weight of the astronaut on Earth is the force of gravity on Earth times the mass of the astronaut, all we have to do for this problem is multiply the weight of the astronaut on Earth by the change in mass (.107) over the square of the change in the radius (.53)Fe=mg  Fe=M(MeG/re2)=700N Fm=M(MmG/rm2) = M(.107MeG/.532re2) = (.107/.532)[(MeG/re2)]=(.38)(700N)=267NThe astronaut would have a weight of 267N on Mars.Kepler’s Law:Kepler’s Law uses the relation of the orbit of the Earth to the sun to describe the relationship of the forceof gravity of any object in the solar system. Essentially, it’s a very specific extension of Newton’s law of Universal gravitation. The force of gravity of the orbit of Earth around the sun is equal to the force of gravity on earth times the mass of the sun divided by the square of the distance between the two (Fg=GMeMs/r2). However, the force of gravity on earth is equal to the mass of Earth times the centripetal acceleration of Earth (GMe=Meac). Centripetal acceleration is equal to the square of the Earth’s velocity divided by its radius (ac=V2/r), and the velocity is equal to 2πr divided by the period (2πr/Te). If we plug all of these equivalences into the original comparison between the Earth’s orbit and the sun, we simplify to get the equation for Kepler’s law, which is;Te2/r3=4π2/GMswhere Te represents the period of Earth’s orbit, r represents the distance from the center of the earth to the center of the sun, and Ms represents the mass of the sun. When applying this equation to other gravitational systems, the period corresponds to the orbit of whichever object is rotating around the other, the radius corresponds to the distance between the two objects, and the mass corresponds to themass of whichever object is being orbited around. It works for any gravitational systems orbits, including elliptical orbits.Example: Jupiter has a satellite that orbits 4.22x105km from its center with a period of 42.5hrs. What is the mass of Jupiter?First, we need to convert the period from hours to seconds, and the radius from km to meters. Then all we have to do is plug our known values into Kepler’s law equation and solve for M.T=42.5hrs =42.5(3600)= 153000r=4.22x105km=4.22x108mG=6.67x10-11T2/r3=4π2/GM 1530002/(4.22x108)2=4π2/6.67x10-11M3.11x10-16=4π2/6.67x10-11MM=4π2/(3.11x10-16)(6.67x10-11)M=1.9x1027kgAstronomical Units:Obviously, since we are dealing with really large masses in kilogram units, we’re going to get ridiculously large numbers every time we use one of these equations. To make these numbers smaller and more manageable, we convert the left side of the Kepler’s law equation from Te2/r3 to Years2/AU3. Because our basis for comparison when using Kepler’s law is the Earth’s orbit around the sun, 1 astronomical unit (AU) refers to the radius or distance between the earth and the sun, and we know that the period for an Earth’s orbit around the sun is 1 year. Now we can use years and astronomical units as a ratio, so that ourvalues are smaller and more manageable. Example: A massive planet orbits a star at an estimated distance of 5 AU and an orbital period of 6 years.At what distance will it orbit the star at a period of 1 year?All we have to do for this problem is compare the ratios with the two