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APS 1030 Laboratory
1
Gravity
GRAVITY
“To see the world in a grain
of sand,
And a heaven in a wild flower;
Hold infinity in the palm of
your hand,
And eternity in an hour.”
-W. Blake
SYNOPSIS: By measuring the time taken for an object to fall through a distance, you will
determine the acceleration due to Earth's gravity, and will use that information to calculate the mass
of the Earth. You’ll also compare the Earth’s gravity with that of other objects, including black
holes!
EQUIPMENT: Falling body apparatus, tape measure, calculator.
Introduction
Acceleration is the change in speed (velocity) that a moving body experiences in a given time
interval. For example, suppose that an object starts at rest and accelerates at a rate of 10 meters per
second per second (written m/sec2 ). After the first second, it will be traveling at a speed of 10
m/sec. After the next second, its speed will be 10 m/sec plus 10 m/sec, or 20 m/sec, etc. So, for
constant acceleration of an object starting at rest,
speed = (acceleration) (time).
(1)
Because an accelerating object’s speed is constantly increasing, the distance it covers each second
is accordingly greater. For an object starting at rest and undergoing constant acceleration,
1
distance = 2 (acceleration) (time) 2.
(2)
Thus, an object accelerating at 10 m/sec 2 will travel a distance of 5 meters in the first second of
time; two seconds after starting, the distance covered is 20 meters, implying that the object
travelled 15 meters during the second second.
Every particle in the universe exerts a gravitational force on every other particle. A force gives rise
to an acceleration. The combined effects of all of the particles in the Earth acting upon us produces
an acceleration which we constantly experience; we refer to this acceleration simply as gravity, and
we call the force on our bodies weight.
If we drop an object and time how long it takes to fall a certain distance, we can use equation (2) to
measure the acceleration due to gravity:
acceleration = 2
(distance)
(time)2
(3)
APS 1030 Laboratory
2
Gravity
Part I. Measuring the Acceleration due to Gravity
The falling-body apparatus uses a spring-loaded clamp to hold a steel ball above a landing pad.
When we release the clamp, the ball drops and the timer starts. The timer stops when the ball
strikes the landing pad, and the elapsed time in seconds appears on the readout.
Positioning
Clamp
Green
Pull
Knob
Dropping
Mechanism
Timer
Table
Red
Clamping Knob
Ball
Landing Pad
Floor
Work in a team of three individuals with one dropping apparatus. Measure the acceleration due to
gravity using the small steel ball as described below. After one individual has completed his/her
measurements, the other team members should independently conduct their own trial
measurements (step I.2).
I.1
I.2
Position the dropping mechanism over the edge of the table, and the landing pad (inside
the catch basin) on the floor directly underneath it. Switch the timer on.
•
While holding the small steel ball in the clamp of the dropping mechanism, pull on the
green knob (against the spring tension) so that the ball is captured. Tighten the red
clamping knob to hold the ball in place.
• Measure the drop-distance to the nearest millimeter: Use a metric tape measure to
determine the distance from the bottom of the ball to the top of the landing pad (another
team member may assist by holding one end of the tape, but you must actually read the
distance). Record your distance measurement in meters (1 mm = 0.001 meter).
• Press the reset button on the timer to initialize the display.
• Release the ball by loosening the red knob. Record the elapsed time in seconds as
indicated by the timer.
I.3
Write down each individual's name and his/her values for distance and time in Table 1.
For each individual trial, calculate the gravitational acceleration experienced by the small
ball using equation (3), and record the results in the table.
APS 1030 Laboratory
Table 1:
Experimenter
Names
3
Small Ball
Distance
(meters)
Gravity
Long Distance
Time
(seconds)
Acceleration
(m/sec 2 )
Average
Presumably, each of you obtained similar, but not necessarily identical, values for the drop
distance, the elapsed time, and the calculated acceleration for each experimental trial. If the
measurements were not identical, it does not mean that the distance, time, or acceleration actually
changed between trials, nor does it necessarily imply that one or more of you made a mistake. The
results simply point out that there is no such thing as a perfect measuring instrument or procedure,
and therefore there is no such thing as a perfect measurement!
Since every measurement (distance, time) contains an experimental error (meaning an
uncertainty, not necessarily a mistake), then any information derived from those measurements
(acceleration) also contains an uncertainty. We can never know exactly what the "true" value is;
we can only try to get better and better results by improving our measuring equipment or
technique, and by averaging the results of our measurements to statistically reduce any random
errors that may have occurred. Note: You cannot claim to know the value of the acceleration to
any greater accuracy than the measurement error. It is senseless and misleading to list its value to
place-values smaller than the uncertainty. For example, if your calculator comes up with a figure
of 9.3487851 m/sec2 , while your individual trials ranged from 9.342 to 9.361 m/sec2 , you should
list your result as 9.35 m/sec2 , with an uncertainty of about ±0.01 m/sec2 . It may not look as
impressive, but it's honest!
I.4
Suggest at least one way in which the measuring equipment or your measurement
technique might be improved to produce a smaller experimental error.
I.5
Calculate an improved measurement for the gravitational acceleration by averaging all
three of your team's individual values, and enter the results in Table 1.
Aristotle claimed that heavier objects fall faster than ligher objects (that is, heavy objects
experienced a greater gravitational acceleration). Nearly two thousand years later, Galileo was the
first to dispute this belief, claiming that all objects fall at the same rate (and asserted that air friction
was responsible for the seemingly slow drop of light objects). According to lore, Galileo put his
hypothesis to the test by dropping balls of different weights off the leaning tower of Pisa.
I.6
Test Galileo's hypothesis. Repeat procedure I.2 using the large steel ball instead of the
small one. Once again, each member should make an independent distance measurement
(slightly different from the previous, since the size of the ball has changed), time
measurement, and calculation of the acceleration. Record all of your teammates results in
Table 2, and find the average acceleration.
APS 1030 Laboratory
Table 2:
Experimenter
Names
4
Large Ball
Distance
(meters)
Gravity
Long Distance
Time
(seconds)
Acceleration
(m/sec 2 )
Average
I.7
Within your uncertainty of the measurements, does your experimental data support
Galileo's contention that the acceleration due to gravity is the same for all objects?
Explain.
Since both balls fell nearly equal distances, your measurements haven’t tested whether gravitational
acceleration was a constant value throughout the drop, but merely that it was the same for both
objects (it could have changed identically for both balls during the time that they fell).
I.8
Determine whether the acceleration due to gravity is a constant that is independent of the
drop-distance or duration of the time-of-flight: Reposition the dropping apparatus and
landing pad to perform a shorter drop to the tabletop as shown in the diagram. Using the
large ball, perform independent measurements as before, and record and average your
team's results in Table 3.
Table 3:
Experimenter
Names
Large Ball
Distance
(meters)
Short Distance
Time
(seconds)
Acceleration
(m/sec 2 )
Average
I.9
Within your uncertainty of the measurements, do your results support the contention that
all objects experience the same, constant acceleration due to Earth's gravitational
attraction? Why or why not?
I.10
Average all nine of your separate measurements together (or equivalently, average the
three averages) to come up with a final, best estimate of its value.
APS 1030 Laboratory
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Gravity
Part II. The Mass of the Earth
Isaac Newton showed that the acceleration g experienced by an object due to the gravitational
attraction of a second body is directly proportional to the mass m of the attracting body, and
inversely proportional to the square of the distance r between the bodies (one form of Newton's
universal law of gravitation):
m
g = G 2 .
r
(4)
Here G is the constant of proportionality, called the gravitational constant. (The value of G
was determined very carefully by measuring the gravitational force between two large blocks of
metal hung very close together from wires.) Essentially, G converts the mass and the distance to
an amount of gravitational acceleration. In meters-kilograms-second (mks) units,
G = 6.670 x 10-11
m3
kg· sec 2
(5)
Important: When equation (4) is used with the above value for G, the mass m must be in
kilograms and r in meters; otherwise, any numerical values you obtain will have nonsense units.
Identical
Gravitational
Acceleration
One
Earth
Mass
M
E
Earth
Mass = M E
One
Earth
Radius
R
E
Using calculus (which he invented to enable him to solve his mathematical problems), Newton also
showed that the mass of a spherically-symmetric object, such as the Earth, gravitationally attracts
other objects as if all of its mass were concentrated at its center. Thus, the acceleration produced
by Earth's gravity is the same as that produced by a tiny object having the same mass as the Earth
(ME), and which is located at the center of the Earth a distance of one Earth radius (RE) away.
The Earth's radius is RE = 6,400 km = 6.4 x 106 meters. Since you have already measured the
acceleration due to gravity, g, we can re-write equation (4) to solve for the mass of the Earth:
2
g RE
ME = G
.
(6)
II.1
Calculate the mass of the Earth using equation (6) and your team's average value for the
acceleration of gravity.
II.2
Compare your result with the accepted value for the mass of the Earth: ME = 5.976 x
1024 kg. Is your result accurate to 50%? 10%? 1%? Is your result too large or too
small?
II.3
If you were to repeat this experiment on a mountaintop, would the acceleration due to
Earth's gravity be less than, equal to, or greater than that which you have measured?
Explain (hint: which parameters would be different in equation 4)?
APS 1030 Laboratory
II.4
6
Gravity
Can you think of any reason(s) why your measurement made in Boulder might have
given results that were higher or lower than the "official value"?
Part III. Gravity on Other Solar Sytem Objects
Newton's law of gravity (Equation 4) applies to everything and everywhere in the universe, not
just on the Earth (that’s why they call it the “universal” law of gravitation!).
III.1
Calculate the gravity (ie. acceleration) felt by the Apollo astronauts as they walked on the
Moon. To do this you'll need the mass of the Moon, 7.35 x 1022 kg, and its radius,
1738 km. Compare this acceleration to that here on Earth (don’t forget to convert from
kilometers to meters!). Does this explain why the astronauts "bounced" as they walked?
III.2
Now, find the “surface” gravity for Jupiter. Jupiter's mass is 1.9 x 1027 kg, and its
radius is 71,398 km. Aside from being squashed into a pancake in mere seconds, what
other difficulties would we face if we tried to stand on Jupiter's “surface”?
III.3
Find the “surface” gravity at the outermost visible layer of the Sun, the photosphere. The
Sun’s mass is 1.99 x 1030 kg, and its radius is 6.96 x 105 kilometers.
Part IV. Escape Velocity
When vehicles are launched into space, the rocket must have a minimum speed at launch to escape
the Earth's gravity. We call this velocity the “escape velocity” (vesc). If the rocket is moving
slower than this velocity, it will fall back to the Earth. From the equations given above, we can
calculate the escape velocity from any object, given by the following equation:
vesc =
2GM
R
(7)
“G” is the gravitational constant given in equation (5), and M and R are the mass and radius of the
object you are trying to “escape.”
IV.1
Calculate the velocity needed for a spacecraft to escape from the Earth. (Note that the
escape velocity does not depend on the mass of the spacecraft itself!)
IV.2
Now calculate the escape velocity for the Moon. (This should explain why the Apollo
astronauts needed only a small vehicle to leave the Moon’s surface and return to lunar
orbit, while the Saturn-V rocket used to lift everything off of the face of the Earth was so
huge!)
IV.3
How fast would your spacecraft have to travel to escape from Jupiter?
Individual atoms and molecules of gas in a planet’s atmosphere also have to reach these same
escape velocities in order to “leak off” into space; otherwise, they stay trapped close to the planet’s
surface, just like us.
IV.4
Use the concept of escape velocity to explain why the Moon has virtually no atmosphere,
Earth has a moderate amount, and Jupiter has so much.
IV.5
Streams of charged particles, known as the solar wind, are constantly escaping from
the Sun and blow out into interstellar space. How fast must these particles be travelling
in order to escape the gravitational grip of the Sun?
APS 1030 Laboratory
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Gravity
Part V. Black Holes
When a star begins to die (ie. when it runs out of hydrogen to convert into helium), it cools off and
begins to collapse due to its self-gravity. If it is more massive than our Sun, it will eventually
explode in what is called a supernova. In the explosion, the outer layers of the star are blown off
and form a supernova remnant like the Crab Nebula. The inner core of the star does not explode
outward but instead implodes, collapsing into a very small sphere.
Einstein's theory of special relativity tells us that nothing can go faster than the speed of light: it is
the ultimate speed limit. Imagine that our star has collapsed into a sphere so small (ie., its R is
small) and is massive enough (ie. M is large) such that its escape velocity is greater than the speed
of light. Nothing would then be able to escape its gravitational pull, not even light! This dark star
is the ubitiquous “black hole.” Obviously such an object deserves quite a bit of respect and we
must make sure we do not get too close to it, otherwise we will be uncontrollably pulled in.
So, let's find out how close we can get. If we substitute the speed of light “C” in for vesc in
equation (7) and rearrange the terms, we find:
RSch = 2GM
C2
(8)
Rsch is called the “Schwarzschild radius.” It is the radius at which nothing can escape from the
black hole; if anything gets closer than Rsch to a black hole, no force in the universe can save it
from being pulled in, because the necessary escape velocity would be greater than the speed of
light. Rsch is not the actual radius of the black hole; it is the “event horizon,” so-called because
nothing inside this radius can be seen and we can only speculate on what is happening inside. No
one really knows for certain, but it is speculated that the black hole itself collapses into what is
mathematically known as a singularity - a point that has no radius, no volume, and an infinite
density.
V.1
Imagine that the Sun collapsed into a black hole (which, incidentally, will not happen).
Using equation (8), what would its Schwarzschild radius (Rsch) be? How does this
compare to the actual radius of the Sun? Compare the Sun's Rsch to the radius of the
Earth. If the Sun did actually collapse into a black hole, would the Earth and the other
planets be sucked in? Why or why not?
V.2
Suppose that the Earth collapsed into a black hole. What would its Rsch be? Suppose
that you collapsed into a black hole! How does this Rsch compare to the radius of a
proton (which is around 10-15 m) ?
Now, imagine that you are an astronaut that is going to bravely (i.e., foolishly) fly into a black
hole. When you are 1.5 Rsch away from the black hole, you glance to your side and notice a
strange sight. At this distance, beams of light orbit the black hole much like a satellite orbits the
Earth. So, when you look to the side you see the back of your head!
You continue inward, oblivious to the fact that you have crossed the Schwarzschild radius and
now can never return. Once inside this radius, we have no idea what has happened to you since no
light can escape to tell us - though we can envision what kind of gruesome death awaits you! You
are now about to experience the effect of “differential gravity” - also known as the tidal effect.
In plain English, your body will be stretched out and pulled apart. Your feet are closer to the black
hole are therefore are being pulled towards the black hole faster than your head, causing you to be
stretched apart!
APS 1030 Laboratory
8
Gravity
You
Black
Hole
Schwarzschild
Radius
R
Sch
V.3
Suppose that for a brief moment that you are actually standing on the Schwarzschild
radius of the black-hole Sun calculated earlier; ie. your feet are a distance Rsch from the
black hole. Using equation (4), what acceleration would your feet feel?
V.4
Assuming that you are 1.5 meters tall (that is, your head is Rsch+1.5 m away from the
black hole), what acceleration would your head feel?
V.5
Now, calculate how far this acceleration will take you in one second (presuming that
you're not moving at first) by using equation (2). To get a feel for how far this really is,
compare this distance to the size of Pluto's orbit, 5.9 x 1012 m. (You may be wondering
how this is possible since the Schwarzshild radius of our black hole is much smaller than
this. It isn't - you'll be pulled into the black hole much faster than a second!)
V.6
Now, calculate the difference between the two accelerations; this is the differential
acceleration. This is how quickly your head and feet are moving apart from each other!
This should be on the order of 1010 m/sec2 .
V.7
Finally, calculate how far your head and feet would be separated after one second.
Again, use equation (2) to get a distance in meters. Compare this to the distance from the
Earth to the Moon (3.8 x 108 m).