UNIT 6 (end of mechanics) Universal Gravitation & SHM.
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1UNIT 6 (end of mechanics) Universal Gravitation & SHM
2where ‘r’ is distance from center to center 2 point masses will attract one another with equal and opposite force given by:where ‘r’ is distance from center to centerG = 6.67x Nm2/kg2Gravity acts as an inverse square law
3What is the net gravitational force on the particle due to the planet? Consider a particle at the center of a spherical planet of uniform density…What is the net gravitational force on the particle due to the planet?What would you weigh at the center of the planet?
4What if the particle was moved halfway to the surface… What is the net gravitational force on the particle due to the planet?Now consider the particle at the center of a hollow planet. What would be the net gravitational force on the particle?
5What if the particle was moved off center What if the particle was moved off center? How would that change the net gravitational force on the particle?
6Analysis of gravity inside a planet of mass M and radius R Analysis of gravity inside a planet of mass M and radius R. Assume uniform density.rRConsider a particle of mass, m, located a distance, r, from the center of the planet.
7Gravity as a function of distance for a planet of mass M and radius R.
8What if the density of a planet is NON-uniform? Consider a spherical planet whose mass density is given by ρ = br (where b = constant in kg/m4) and has radius, a. Calculate the gravitational force on an object of mass, m, a distance r (where r < a) from the center of the planet.
9For the outside of the planet the gravitational force would be
10G-fieldThe region around a massive object in which another object having mass would feel a gravitational force of attraction is called a gravitational field. The gravitational field's strength at a distance r from the center of an object of mass, M, can be calculated with the equation.M
11Gravitational Potential Energy (Ug) We must derive an expression that relates the potential energy of an object to its position from Earth since ‘g’ is NOT constant as you move away from Earth. All previous problems dealt with objects ‘close’ to the earth where changes in ‘g’ were negligible.Consider a particle of mass, m, a very far distance, r1 ,away from the center of the earth (M).r1
14All values of the potential energy are negative, approaching the value of zero as R approaches infinity. Because of this we say that the mass is trapped in an "energy well" - that is, it will have to be given additional energy from an external source if it is to escape gravity's attraction.
15Ug between rod and sphere (non-point mass) A thin, uniform rod has length, L, and mass, M. A small uniform sphere of mass, m, is placed a distance, x, from one end of the rod, along the axis of the rod. Calculate the gravitational potential energy of the rod-sphere system.The potential energy of a system can be thought of as the amount of work done to assemble the system into the particular configuration that its currently in. So, to compute the potential energy of a system, we can imagine assembling the system piece by piece, computing the work necessary at eachstep. The potential energy will be equal to the amount of work done in setting the system up in this way.
17Kepler's 1st Law: The Law of Elliptical Orbits All planets trace out elliptical orbits. When the planet is located at point P, it is at the perihelion position. When the planet is located at point A it is at the aphelion position. The distance PA = RP + RA is called the major axis.Semi-major axis is half of the major axis. Eccentricity is a measure of how "oval" an ellipse is.
18Kepler's 2nd Law: Law of Areas A line from the planet to the sun sweeps out equal areas of space in equal intervals of time where dA/dt = constantAt the perihelion, the planet’s speed is a maximum.At the aphelion, the planet’s speed is a minimum.Angular momentum is conserved since net torque is zero. Gravity lies along the moment arm or position vector, r. vARA = vPRP
19Kepler's 3rd Law: Law of Periods The square of the period of any planet is proportional to the cube of the avg distance from the sun.Since the orbits of the planets in our solar system are EXTREMELY close to being circular in shape (the Earth's eccentricity is ), we can set the centripetal force equal to the force of universal gravitation and,
20Energy Considerations for circular Satellite Motion
21Changing the orbit of a satellite Calculate the work required to move an earth satellite of mass m from a circular orbit of radius 2RE to one of radius 3RE.
22How was the energy distributed in changing the orbit? The initial and final potential energies were:The kinetic energies were:
23Escape VelocityIf an object of mass m is launched vertically upwards with speed vo we can use energy to determine minimum speed to escape planet’s gravitational field. Our parameters are when the object can just reach infinity with zero speed.
24Example: Two satellites, both of mass m, orbit a planet of mass M Example: Two satellites, both of mass m, orbit a planet of mass M. One satellite is elliptical and one is circular. Elliptical satellite has energy E = -GMm/6r.a) Find energy of circular satelliteRrMb) Find speed of elliptical satellite at closest approachc) Derive an equation that would allow you to solve for R but do not solve it.