UNIT 6 (end of mechanics) Universal Gravitation & SHM.
Presentation on theme: "UNIT 6 (end of mechanics) Universal Gravitation & SHM."— Presentation transcript:
UNIT 6 (end of mechanics) Universal Gravitation & SHM
2 point masses will attract one another with equal and opposite force given by: where r is distance from center to center G = 6.67x10 -11 Nm 2 /kg 2 Gravity acts as an inverse square law
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?
What 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?
What if the particle was moved off center? How would that change the net gravitational force on the particle?
Analysis of gravity inside a planet of mass M and radius R. Assume uniform density. r R Consider a particle of mass, m, located a distance, r, from the center of the planet.
Gravity as a function of distance for a planet of mass M and radius R.
What 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/m 4 ) 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.
For the outside of the planet the gravitational force would be
G-field The 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
Gravitational Potential Energy (U g ) 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, r 1,away from the center of the earth (M). r1r1
All 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.
U g 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 each step. The potential energy will be equal to the amount of work done in setting the system up in this way.
Kepler'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 = R P + R A is called the major axis. Semi-major axis is half of the major axis. Eccentricity is a measure of how "oval" an ellipse is.
A line from the planet to the sun sweeps out equal areas of space in equal intervals of time where dA/dt = constant At the perihelion, the planets speed is a maximum. At the aphelion, the planets speed is a minimum. Angular momentum is conserved since net torque is zero. Gravity lies along the moment arm or position vector, r. v A R A = v P R P Kepler's 2 nd Law: Law of Areas
The square of the period of any planet is proportional to the cube of the avg distance from the sun. Kepler's 3 rd Law: Law of Periods Since the orbits of the planets in our solar system are EXTREMELY close to being circular in shape (the Earth's eccentricity is 0.0167), we can set the centripetal force equal to the force of universal gravitation and,
Energy Considerations for circular Satellite Motion
Changing the orbit of a satellite Calculate the work required to move an earth satellite of mass m from a circular orbit of radius 2R E to one of radius 3R E.
How was the energy distributed in changing the orbit? The initial and final potential energies were: The kinetic energies were:
Escape Velocity If an object of mass m is launched vertically upwards with speed v o we can use energy to determine minimum speed to escape planets gravitational field. Our parameters are when the object can just reach infinity with zero speed.
R r 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. M a) Find energy of circular satellite b) Find speed of elliptical satellite at closest approach c) Derive an equation that would allow you to solve for R but do not solve it.