Transcript gravitational field. - Plain Local Schools

UCM & Gravity – Gravity
http://http://aplusphysics.com/courses/honors/ucm/gravity.html
Unit #5 UCM & Gravity
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Objectives and Learning Targets
1. Explain the acceleration of an object moving in a circle at constant speed.
2. Define centripetal force and recognize that it is not a special kind of force, but
that it is provided by forces such as tension, gravity, and friction.
3. Solve problems involving calculations of centripetal force.
4. Determine the direction of a centripetal force and centripetal acceleration for an
object moving in a circular path.
5. Calculate the period, frequency, speed and distance traveled for objects moving in
circles at constant speed.
6. Analyze and solve problems involving objects moving in vertical circles.
7. Determine the acceleration due to gravity near the surface of Earth.
8. Utilize Newton’s Law of Universal Gravitation to determine the gravitational force
of attraction between two objects.
9. Explain the difference between mass and weight.
10. Explain weightlessness for objects in orbit.
11. Explain how Kepler’s Laws describe the orbits of planetary objects around the
sun.
Unit #5 UCM & Gravity
Universal Gravitation
• All objects that have mass attract each other with a gravitational force. The
magnitude of that force (Fg) can be calculated using Newton's Law of Universal
Gravitation:
• This law tells us that the force of gravity between two objects is proportional
to each of the masses (m1 and m2) and inversely proportional to the square
of the distance between them (r). The universal gravitational constant (G) is
a "fudge factor," so to speak, included in the equation so that your answers
come out in S.I. units. G is given as
Unit #5 UCM & Gravity
Universal Gravitation
• Let's look at this relationship in a bit more detail. Force is directly
proportional to the masses of the two objects, therefore if either of the
masses were doubled, the gravitational force would also double. In similar
fashion, if the distance between the two objects, r, was doubled, the force of
gravity would be quartered since the distance is squared in the denominator.
This type of relationship is called an inverse square law, which describes many
phenomena in the natural world.
• NOTE: the distance between the masses, r, is actually the distance between
the center of masses of the objects. For large objects, such as the Earth, for
example, you must determine the distance to the center of the Earth, not to
its surface.
Unit #5 UCM & Gravity
Universal Gravitation Tips
• Some hints for problem solving when dealing with Newton's Law
of Universal Gravitation:
1. Substitute values in for variables at the very end of the problem only. The
longer you can keep the formula in terms of variables, the fewer
opportunities for mistakes.
1. Before using your calculator to find an answer, try to estimate the order of
magnitude of the answer. Use this to check your final answer.
1. Once your calculations are complete, make sure your answer makes sense by
comparing your answer to a known or similar quantity. If your answer doesn't
make sense, check your work and verify your calculations.
Unit #5 UCM & Gravity
Sample Problem #1
Newton’s Law of Universal Gravitation
• Question: What is the gravitational force of attraction between two asteroids
in space, each with a mass of 50,000 kg, separated by a distance of 3800 m?
• Answer:
• As you can see, the force of gravity is a relatively weak force, and we would expect a
relatively weak force between relatively small objects. It takes tremendous masses
and small distances in order to develop significant gravitational forces. Let's take a
look at another problem to explore the relationships between gravitational force,
mass, and distance.
Unit #5 UCM & Gravity
Sample Problem #2
Meteor
Unit #5 UCM & Gravity
Sample Problem #2
Meteor
Unit #5 UCM & Gravity
Gravitational Fields
• Gravity is a non-contact, or field, force. Its
effects are observed without the two objects
coming into contact with each other. Exactly how
this happens is a mystery to this day, but
scientists have come up with a mental construct
to help us understand how gravity works.
• Envision an object with a gravitational field, such
as the planet Earth. The closer other masses are
to Earth, the more gravitational force they will
experience. We can characterize this by
calculating the amount of force the Earth will
exert per unit mass at various distances from the
Earth. Obviously, the closer the object is to the
Earth, the larger a force it will experience, and
the farther it is from the Earth, the smaller a
force it will experience.
Unit #5 UCM & Gravity
Gravitational Fields
• Attempting to visualize this, picture the strength
of the gravitational force on a test object
represented by a vector at the position of the
object. The denser the force vectors are, the
stronger the force, the stronger the
"gravitational field." As these field lines become
less and less dense, the gravitational field gets
weaker and weaker.
• To calculate the gravitational field strength at a
given position, we can go back to our definition of
the force of gravity on our test object, better
known as its weight. We've been writing this as
mg since we began our study of dynamics. But,
realizing that this is the force of gravity on an
object, we can also calculate the force of gravity
on test mass using Newton's Law of Universal
Gravitation. Putting these together we find that:
Unit #5 UCM & Gravity
Gravitational Fields
• Realizing that the mass on the left-hand side of the equation, the mass of our
test object, is also one of the masses on the right-hand side of the equation,
we can simplify our expression by dividing out the test mass.
• Therefore, the gravitational field strength, g, is equal to the universal
gravitational constant, G, times the mass of the object, divided by the square
of the distance between the objects.
• But wait, you might say... I thought g was the acceleration due to gravity on
the surface of the Earth! And you would be right. Not only is g the
gravitational field strength, it's also the acceleration due to gravity. The units
even work out... the units of gravitational field strength, N/kg, are equivalent
to the units for acceleration, m/s2!
Unit #5 UCM & Gravity
Gravitational Fields
• Still skeptical? Let's calculate the gravitational field strength on the surface of
the Earth, using the knowledge that the mass of the Earth is approximately
5.98*1024 kg, and the distance from the surface to the center of mass of the
Earth (which varies slightly since the Earth isn't a perfect sphere) is
approximately 6378 km in New York.
• As expected, the gravitational field strength on the surface of the Earth is
the acceleration due to gravity. Let's see if we can't solve some problems
using gravitational field strength.
Unit #5 UCM & Gravity
Sample Problem #3
Astronaut
• Question: Suppose a 100-kg astronaut feels a gravitational force of 700N when placed
in the gravitational field of a planet.
A) What is the gravitational field strength at the location of the astronaut?
B) What is the mass of the planet if the astronaut is 2*106 m from its center?
• Answer:
Unit #5 UCM & Gravity
Orbits
• How do celestial bodies orbit each other?
The moon orbits the Earth. The Earth orbits
the sun. Our solar system is in orbit in the
Milky Way galaxy... but how does it all work?
• To explain orbits, Sir Isaac Newton
developed a "thought experiment" in which
he imagined a cannon placed on top of a
very tall mountain, so tall, in fact, that the
peak of the mountain was above the
atmosphere (this is important because it
allows us to neglect air resistance). If the
cannon then launched a projectile
horizontally, the projectile would follow a
parabolic path to the surface of the Earth.
Unit #5 UCM & Gravity
Orbits
• If the projectile was launched with a higher
speed, however, it would travel farther across
the surface of the Earth before reaching the
ground. If its speed could be increased high
enough, the projectile would fall at the same
rate the Earth's surface curves away. The
projectile would continue falling forever as it
circled the Earth! This circular motion
describes an orbit.
• Put another way, the astronauts in the Space
Shuttle aren't weightless. Far from it, actually,
the Earth's gravity is still acting on them and
pulling them toward the center of the Earth
with a substantial force. We can even calculate
that force. If the Space Shuttle orbits the Earth
at an altitude of 380,000 m, what is the
gravitational field strength due to the Earth?
Unit #5 UCM & Gravity
Sample Problem #5
Space Shuttle Orbit
• Question: If the Space Shuttle orbits the Earth at an altitude of 380 km, what is the
gravitational field strength due to the Earth?
• Answer: Recall that we can
obtain values for G, the mass
of the Earth, and the radius of
the Earth from the reference
table.
• This means that the acceleration due to gravity at the altitude the astronauts are
orbiting the earth is only 11% less than on the surface of the Earth! In actuality, the
Space Shuttle is falling, but it's moving so fast horizontally that by the time it falls, the
Earth has curved away underneath it so that the shuttle remains at the same distance
from the center of the Earth -- it is in orbit. Of course, this takes tremendous speeds...
to maintain an orbit of 380 km, the space shuttle travels approximately 7680 m/s,
more than 23 times the speed of sound at sea level!
Unit #5 UCM & Gravity