Transcript Universal Gravitation

UNIT 6 (end of mechanics)
Universal Gravitation & SHM
2 point masses will attract one another with equal and
opposite force given by:
Gm1m2
Fg 
2
r
Gravity acts as an
inverse square law
where ‘r’ is distance
from center to center
G = 6.67x10-11 Nm2/kg2
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.
Consider a particle of mass, m, located a
distance, r, from the center of the planet.
r
R
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-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 (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 each
step. 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.
Kepler'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 = 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 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 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 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.
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 satellite
M
r
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.
R