Download Newton`s Law of Universal Gravitation Script

Newton’s Law of Universal Gravitation Script
Slide One: Insert Start Button
Slide Two: Circular Motion and Gravitation
Slide Three: Newton’s Law of Universal Gravitation
Slide Four: Every mass in the universe is attracting every other mass in the universe.
However, the force of attraction between two masses is very small and not particularly
significant unless the objects have a very large mass. The two factors that determine the
magnitude of the force are the mass and the distance between the masses. Directly
proportional to the mass means we must multiply the two masses. Inversely proportional
to the square of the distance means we must divide by the distance squared. This
distance is measured from the center of mass of one object to the center of mass of the
other. Because of Newton’s third law, when one object pulls on a mass, the mass must
pull in the opposite direction with an equal magnitude.
Slide Five: Once we know the relationship we can introduce a constant to make the
relationship an equality. The constant G converts a kilogram times a kilogram divided by
a meter squared into the unit of Force, a Newton.
Slide Six: On the surface of a planet, in our case the Earth, the force of gravity on an
object is called the weight. So we can set the equation equal to W or to mg. If that is
done, the mass of one of the objects is the mass of the earth, and the other mass would be
the mass of the object on the earth. The radius would be the distance from the center of
mass of the earth.
Slide Seven: If we solve for the acceleration due to the force of gravity, we simply
divide both sides of the equation by the mass of the object and get a value for g. If you
plug in your variables and solve, you will see that we get the acceleration due to gravity.
Slide Eight: Objects In Orbit
Slide Nine: For an object to orbit around the earth, the force of gravity must equal the
centripetal force needed to follow the orbital path around the earth. We can set m v
squared over r equal to the force of gravity. Note that the mass of the object and the
radius cancel out of the left side of the equation and we are left with just the velocity
squared. Because the mass of the earth is a constant, a change in velocity will change the
radius of the orbital path.
Slide Ten: For an object to orbit a planet then, it must be going a precise speed and enter
the orbit at the right angle. After that, changing orbits is simply a matter of changing the
velocity. Note that when a satellite slows down, it’s orbit decays and it comes closer and
closer to the earth.
Slide Eleven: It is a common mistake to think that the radius of the orbit is measured
from the earth’s surface. Orbits are measured from the center of the pathway, so in this
case from the center of the earth. Add the height above the earth’s surface to the radius
of the earth to get the proper value for the equation. The values listed are the mass of the
earth, the radius of the earth, the mass of the moon, and the mass of the sun. AP will
provide these values for you if you need them for a test question.
Slide Twelve: Other Considerations
Slide Thirteen: There are a few other derived relationships that AP would like you to
know. The gravitational field is just the value for the acceleration due to the force of
gravity at a distance from a uniform sphere. We use the gravitational force equation and
solve for g. This gives us the gravitational field for the mass m sub two at a distance of r
from the center of the objects mass.
Slide Fourteen: In our second case, we are asked what the gravitational field is inside an
object with a uniform density. We solve for the mass of the object in terms of its density.
Slide Fifteen: The volume of a sphere is four thirds pi r cubed. We can solve for the
mass of the uniform sphere in terms of density and the radius.
Slide Sixteen: Combing with the equation for gravitational force, we find that the force
is directly proportional to the density of the object, the mass of the object inside the
uniform sphere, and the distance from the center of the object.
Slide Seventeen: The period of revolution is the time it takes to go around one time in
it’s circular path. The generic time t becomes the period T if we use the distance of one
complete trip.
Slide Eighteen: We use the relationship between centripetal force and the force of
gravity to solve for the velocity.
Slide Nineteen: We combine the equations to find the equation for the period of
revolution. It is dependent only on the radius and the mass of the object being orbited.
Slide Twenty: There are a lot of ways to prove the gravitational potential energy, but
this one seems to cover all the questions that arise from just talking through the proof.
Slide Twenty-One: The general definition of potential energy relates the conservative
force on a particle and the displacement d L. This is the definition of work or the transfer
of energy. A later module will explain this in more detail. Working through the problem
we see that the d U is equal to the force of gravity with respect to r.
Slide Twenty-Two: If we integrate both sides we get a negative value.
Slide Twenty-Three: Defining U naught as zero means we are defining an infinite
distance away as the maximum value of gravitational potential energy. This makes sense
if you realize that to continually move the object away from the earth would mean adding
energy to the object.