Regents Physics - 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
on the front page of your Regents Physics Reference Table as
.
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.
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.
2. 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.
3. 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.
Let's see if we can't apply Newton's Law of Universal Gravitation to a simple
problem...
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.
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.
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:
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!
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.
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:
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.
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 themand 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 380,000m, what is the gravitational field
strength due to the Earth?
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. Oof 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!