UNIT 6 (end of mechanics) Universal Gravitation & SHM

UNIT 5 (end of mechanics)Universal Gravitation, energy, orbits, &

SHM

2 point masses will attract one another with equal and opposite force given by:

221

r

mGmFg where ‘r’ is distance

from center to center

G = 6.67x10-11 Nm2/kg2

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?

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/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.

For the outside of the planet the gravitational force would be

G-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

Gravitational 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

r1

r2

r

r

F

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.

Ug 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.

dM

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 = 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.

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 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

Kepler's 2nd 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 3rd 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 satelliteCalculate the work required to move an earth satellite of mass m from a circular orbit of radius 2RE to one of radius 3RE.

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 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.

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.

We can relate periodic motion with an object undergoing uniform circular motion. Consider a mass sitting on top of a rotating platform where lights cast a shadow of mass on wall.

Shadow of rotating mass will move back and forth on wall exhibiting periodic motion.

As the mass rotates with a constant angular speed, ω through an angle θ in a time t, we can describe the linear movement across the screen as

Since θ = ωt we can relate the position of the object with respect to time and angular movement as

xxdx

dcos)(sin xx

dx

dsin)(cos

Trig Calculus

Cxxdx cossin Cxxdx sincos

Caxa

dxax cos1

)sin( Caxa

dxax sin1

)cos(

axaaxdx

dcos)(sin axaax

dx

dsin)(cos

Expressions for velocity and acceleration

SIMPLE HARMONIC MOTIONA special case of periodic

When periodic motion possesses a restoring force that is directly proportional to the negative of displacement, this is said to be SHM.

Fs vConsider a mass oscillating on a frictionless surface

Energy considerations for SHM

Example: A 0.500-kg mass is vibrating in a system in which the restoring constant is 100 N/m; the amplitude of vibration is 0.20 m. Find:a. the energy of the systemb. the maximum kinetic energy and maximum velocityc. the PE and KE when x = 0.100 md. the maximum acceleratione. the equation of motion if x = A at t =0

a) E = ½ kA2 = 2.0J

b) Kmax = E = 2.0J

b) vmax = 2.83m/s

c) U = ½ kx2 = 0.5J K = E - U = 1.5J

d) amax =| -ω2 A |= 40m/s2

e) x(t) = 0.20 cos (14.1t)

If a mass is at rest, suspended from a spring, the mass is in static equilibrium. The only forces acting on the mass are a gravitational force down, mg, and the spring force up, kxo, where xo is the stretch of the spring at equilibrium.

Does a vertical spring obey SHM due to gravitational force?

A block is moving undergoing SHM having amplitude A0. At instant block passes thru equilibrium, putty of

mass M is dropped vertically onto block and it sticks. Find new amplitude and period

k

Is a simple pendulum a S.H.O.?

Physical pendulumAny rigid body that is suspended from a fixed point that does not pass through the COM and oscillates about that point.

Pivot

d

A uniform rod of mass M and length L is pivoted about one end and oscillates in a vertical plane. Find the period of oscillation if the amplitude is small.

Torsional PendulumA mass suspended to a fixed support by a thin wire can be made to twist about its axis. This is known as a torsional pendulum. The mass attached to the wire rotates in the horizontal plane where θ is the angle of rotation. It is the twisting of the wire that creates a restoring torque due to the resistance of the wire to the deformation.