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The method used to add or take vectors includes drawing a diagram using the head-to-tail method then using Pythagorean theorem and trigonometric functions to find out the figures.
The head-to-tail method involves drawing a vector diagram; where the
head of this vector ends the tail of the next vector begins (thus, head-to-tail
method). The process is repeated for all vectors which are added. Once
all vectors have been added head-to-tail, the resultant is drawn from the
tail of the first vector to the head of the last vector; i.e., from start
to finish. Once the resultant is drawn, its length can be calculated and
converted to real units using the given scale. The magnitude and direction
of vector R in the diagram above can be determined by use of trigonometric
functions.
Stawa Set 11+12
Physics Text book pg 129-130
Physics Study Guide 74-75
2. The Equations that are used to describe the horizontal and vertical
motion:
s = v t (assume constant velocity)
s = ut + ½ at2
v = u + at
v2= u2 + 2as
a= (v-u)/t
s=(u+v)/2 * t
where s is the displacement (m)
v is the final velocity (m/s)
u is the initial vertical velocity (m/s)
t it time (s)
a is accelleration (m/s/s)
In Projectile Motion We Assume:
1. Air resistance is negligible.
2. The Earth's rotation is negligible.
3. The acceleration is downward and constant
at g near the Earth's Surface.
Stawa Set 13
Physics Text book pg 135
Physics Study Guide 77-80
3. There are a few conceptual characteristics of free fall motion.
An object in free fall experiences an acceleration of -9.8 m/s/s. (The
- sign indicates a downward acceleration.) If an object is merely dropped
(as opposed to being thrown) from an elevated height to the ground below,
then the initial velocity of the object is 0 m/s.
If an object is projected upwards in a perfectly vertical directly, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (V V) after traveling to the peak would be assigned a value of 0 m/s. If an object is projected upwards in a perfectly vertical directly, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity which it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height.
So in summery a projectile is any object upon which the only force is gravity, projectiles travel with a parabolic trajectory due to the influence of gravity, there are no horizontal forces acting upon projectiles and thus no horizontal acceleration, the horizontal velocity of a projectile is constant (never changing in value), there is a vertical acceleration caused by gravity; its value is ~10 m/s/s, down, the vertical velocity of a projectile changes by ~10 m/s each second, the horizontal motion of a projectile is independent of its vertical motion.
Example 1
Suppose that the cannonball is launched horizontally (with no upward
angle whatsoever) with an initial speed of 20 m/s. If there were no gravity,
the cannonball would continue in motion at 20 m/s in the horizontal direction.
Gravity causes the cannonball to accelerate downwards at a rate of ~10
m/s/s. This means that the vertical velocity is changing by ~10 m/s every
second. If a vector diagram (showing the velocity of the cannonball at
1-second intervals of time) is used to represent how the x- and y-components
of the velocity of the cannonball is changing with time, then x- and y-
velocity vectors could be drawn and their magnitudes labeled.
The length of the vector arrows are representative of the magnitudes
of that quantity. Such a diagram is shown below.
The important concept depicted in the above vector diagram is that
the horizontal velocity remains constant during the course of the trajectory
and the vertical velocity changes by ~10 m/s every second.
If the projectile is launched upward at an angle to the horizontal, How would the horizontal and vertical velocity values change with time? How would the numerical values differ from the previously shown diagram for a horizontally-launched projectile? The diagram below reveals the answers to these questions. The diagram depicts an object launched upward with a velocity of 40 m/s at an angle of 30 degrees above the horizontal. For such an initial velocity, the object would initially be moving 20 m/s, upward and 34.6m/s, rightward.
Again, the important concept depicted in the above diagram is that the horizontal velocity remains constant during the course of the trajectory and the vertical velocity changes by ~10 m/s every second.
Stawa Set 13
Physics Text book pg 130-147
Physics Study Guide 79
4. A pool ball leaves a 0.60-meter high table with an initial horizontal
velocity of 2.4 m/s. Predict the time required for the pool ball to fall
to the ground and the horizontal distance between the table's edge and
the ball's landing location.
Solution
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Once the time has been determined, a horizontal equation can be used
to determine the horizontal displacement of the pool ball. Recall from
the given information, UH = 2.4 m/s and AH = 0 m/s/s.
The first horizontal equation (sH = uHt + ½
aHt2) can then be used to solve for "s H."
By substitution of known values, the equation takes the form of
SH = (2.4 m/s)*(0.3464 s) + 0.5*(0 m/s/s)*(0.3464 s)2
Since the second term on the right side of the equation reduces to
0, the equation can then be simplified to S H= (2.4 m/s)*(0.3464
s)
Thus, SH = 0.83 m (rounded from 0.8313 m)
The answers to the stated problem is that the pool ball is in the air for 0.35 seconds and lands a horizontal distance of 0.83 m from the edge of the pool table.
Example 2
A football is kicked with an initial velocity
of 25 m/s at an angle of 45-degrees with the horizontal. Determine the
time of flight, the horizontal displacement, and the peak height of the
football.
The solution of any non-horizontally launched projectile problem (in
which U H and Theta are given) should begin by first breaking
the initial velocity into horizontal and vertical components using trigonometric
functions. Thus,
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UH = 25 m/s*cos(45) UH = 17.7 m/s |
UV = 25 m/s*sin(45) UV = 17.7 m/s |
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As indicated in the table, the final x-velocity (VH) is the same as the initial x-velocity (UH). This is due to the horizontal velocity of a projectile is constant; there is no horizontal acceleration. The table also indicates that the final y-velocity (VV) has the same magnitude and the opposite direction as the initial y-velocity (UV). This is due to the symmetrical nature of a projectile's trajectory.
The unknown quantities are the horizontal displacement, the time of flight, and the height of the football at its peak. The solution of the problem now requires the selection of equations and the known information to solve for the unknown quantities. There are a variety of possible strategies for solving the problem. A careful listing of known quantities (as in the table above) provides cues for the selection of a useful strategy.
From the vertical information in the table above and the second equation listed among the vertical kinematic equations (V V =UV + AVt), it becomes obvious that the time of flight of the projectile can be determined. By substitution of known values, the equation takes the form of
With the time determined, information in the table and the horizontal kinematic equations can be used to determine the horizontal displacement (x) of the projectile. The first equation (sH = uHt + ½ aHt2) listed among the horizontal equations is suitable for determining "S" By substitution of known values, the equation takes the form of
Finally, the problem statement asks for the height of the projectile
at is peak. This is the same as asking "what is the vertical displacement
(S V) of the projectile when it is halfway through its trajectory?"
In other words, find S V when t = 1.77 seconds. To determine
the peak height of the projectile (S V with t = 1.77 sec), the
first equation (sV = uVt + ½ aVt2)
listed among the vertical kinematic equations can be used. By substitution
of known values into this equation, it takes the form of SV
= (17.7 m/s)*(1.77 s) + 0.5*(-10 m/s/s)*(1.77
s)2
Using a calculator, this equation can be simplified to SV
= 31.3 m + (-15.7 m)
And thus, SV = 15.6
m
The solution to the problem statement yields the following answers: the time of flight of the football is 3.54 s, the horizontal displacement of the football is 62.6 m, and the peak height of the football 15.6m.
Stawa Set 13
Physics Text book pg 140-149
Physics Study Guide 80
5. Time of flight is the time, Range is the total horizontal distance covered, Maximum height occurs when A V =0
Example
A long jumper leaves the ground with an initial velocity of 12 m/s
at an angle of 28-degrees above the horizontal. Determine the time of flight,
the Range, and the peak height of the long-jumper.
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UH = 12 m/s*cos(28) UH = 10.6 m/s |
UV = 12 m/s*sin(28) UV = 5.6 m/s |
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sH = uHt + ½ aHt2
sH = 10.6(1.1) + ½ (0)(??)
sH = 11.9m
sV = uVt + ½ aVt2
sV = 5.6(0.55) + ½ (-10)(0.55)2
sV = 3.08 -1.5125
sV=1.56
sV =1.6m
Time = 1.1sec
Range = 11.9m
Peak Height 1.6m
Stawa Set 13
Physics Text book pg 140-149
Physics Study Guide 80
6. Air resistance affects the trajectory of a projectile by reducing its range, lowering its maximum height and making the flight path non-symmetrical.
As an object falls through air, it usually encounters some degree of air resistance. Air resistance is the result of collisions of the object's leading surface with air molecules. The actual amount of air resistance encountered by the object is dependent upon two common factors which have a direct effect upon the amount of air resistance are the speed of the object and the cross-sectional area of the object. Increased speeds result in an increased amount of air resistance. Increased cross-sectional areas result in an increased amount of air resistance.
Physics Text book pg 146-147
Physics Study Guide 181
7. Accelleration is the change in velocity over a given time period. Since velocity is a vector quantity, and direction is to be taken into account, it can be said to be the amount of curvature. Since a circle is curved and the curve is constant hence meeting the other point, constant acceleration can be assumed.
So it can also be said that acceleration involves a change in speed and/or direction; it is caused by an unbalanced force. In circular motion, the object moves at constant speed but is accelerating because its direction is constantly changing
Uniform Circular Motion
An object moving in a circle of radius r with constant speed v has
an acceleration whose direction is toward the center of the circle and
whose magnitude is a=v2/r. Acceleration depends upon speed and
radius. The greater the speed, the faster the velocity changes direction;
the larger the radius, the less rapidly the velocity changes direction.
Since the acceleration is directed toward the center of the circle, the
net force must be directed toward the center of the circle too. The net
force must be applied by other objects.
Stawa Set 14
Physics Text book pg 150-152
Physics Study Guide 82
8. The centripetal force is the net force directed towards the centre
of a circular path. An object travelling in a circle at constant speed
has an acceleration called centripetal acceleration, directed towards the
centre of the circular path calculated using Ac =
v²/r.
The centripetal force is the net force directed towards the centre
of a circular path. And since force equals mass*acceleration
Fc = ma = m v²/r.
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Stawa Set 14
Physics Text book pg 152-155
Physics Study Guide 83-84
9. Calculate the centripetal acceleration of our moon. It is in a nearly circular orbit of radius r = 3.84x10 8 m and has a period of 27.32 days. Our moon moves in a circle so its direction changes continually; that means it has an acceleration because its velocity changes; this is the centripetal acceleration. From its radius and period we can find its speed and then calculate its centripetal acceleration. a = v2/r.
First we must find its speed v from the total distance traveled D divided
by the time required T.
D= 2Pr
v = D/T
v = 2Pr/T
v = [2(3.14)( 3.84 x 108 m )]/(27.32 days)]
1 day = 24 hours
1 hour = 3600
v = 1 022 m / s
a = v2/r
a =(1 022 m / s)2/3.84 x 108 m
a = 2.72 x 10-3 m/s/s = 2.72 mm/s/s
Example 2
A 900-kg car moving at 10 m/s takes a turn around a circle with a radius
of 25.0 m. Determine the acceleration and the net force acting upon the
car.
Known Information:
m = 900 kg
v = 10.0 m/s
R = 25.0 m
a = ????
Fnet = ????
To determine the acceleration of the car, use the equation a = (v 2)/R.
a = (v2)/R
a = ((10.0 m/s)2)/(25.0 m)
a = (100 m2/s2)/(25.0 m)
a = 4 m/s/s
To determine the net force acting upon the car, use the equation Fnet
= m*a.
Fnet = m*a
Fnet = (900 kg)*(4 m/s2)
Fnet = 3600 N
Example 3
A 95-kg halfback makes a turn on the football field. The halfback sweeps
out a path which is a portion of a circle with a radius of 12-meters. The
halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine
the speed, acceleration and net force acting upon the halfback.
Known Information:
m = 95.0 kg
R = 12.0 m
Traveled 1/4-th of the circumference in 2.1s
v = ????
a = ????
Fnet = ????
To determine the speed of the halfback, use the equation v = d/t where
the d is one-fourth of the circumference and the time is 2.1 s.
v = d/t
v = (0.25 * 2 * P * R)/t
v = (0.25 * 2 * 3.14 * 12.0 m)/(2.1 s)
v = 8.97 m/s
To determine the acceleration of the halfback, use the equation a =
(v 2)/R.
a = (v2)/R
a = ((8.97 m/s)2)/(12.0 m)
a = (80.5 m2/s2)/(12.0 m)
a = 6.71 m/s/s
To determine the net force acting upon the halfback, use the equation
Fnet = m*a.
Fnet = m*a
Fnet = (95.0 kg)*(6.71 m/s/s)
Fnet = 637 N
Example 4
A 1.5-kg bucket of water is tied by a rope and whirled in a circle
with a radius of 1.0 m. At the bottom of the circular loop, the speed of
the bucket is 6.0 m/s. Determine the acceleration, the net force and the
individual force values when the bucket is at the bottom of the circular
loop.
v = 6 m/s
m = 1.5 kg
a = ??? m/s/s
Fnet = ??? N
Ag = 10 m/s/s
Fgrav = m*g = 15 N
a = (v2)/R = (42)/1
a = 16 m/s/s
Fnet=m*a=1.5 kg*16 m/s/s
Fnet = 24 N, down
Fnet = Fgrav + Ftens, so Ftens = Fnet - Fgrav
Ftens = 24 N - 15 N = 9 N
Stawa Set 14
Physics Text book pg 156
Physics Study Guide 82-84
10. A ball is on a string, traveling in a vertical circle. The gravitational force mg has a component ( m g cos q ) toward the center, and tension T is toward the center so T + mgcosq = mv2/r. The ball could be traveling in either direction. This could be a roller coaster on one of those loops; then replace T with the normal force N or FN.
A pendulum is similar to the above except that it does not go all the way around, just back and forth at the bottom:
Note that q in this picture
is not the same as in the previous picture. S
Fc = mv2/r, so T - mgcosq
= mv2/r.
In the direction of motion S
f = ma becomes mgsinq = ma.
Stawa Set 14
Physics Text book pg 153-156
Physics Study Guide 83-86
11. For a level curve, the centripetal force will be supplied by the friction force between the tires and roadway.A banked curve can supply the centripetal force by the normal force and the weight without relying on friction.
Stawa Set
Physics Text book pg 151, 154,
155, 161
Physics Study Guide 84
12.
The net force on a car traveling around
a curve is the centripetal force, Fc = m v2
/ r, directed toward the center of the curve.
For a level curve, the centripetal force
will be supplied by the friction force between the tires and roadway.
A banked curve can supply the centripetal
force by the normal force and the weight without relying on friction.
Level Curves
For a level curve, the centripetal force will be supplied by the friction force between the tires and roadway.
Example
What is the friction required between the tires and the level roadway
to allow a car to make a curve of radius r = 350 m at a speed of 80 km/h?
For a level curve, the force of friction is the only horizontal force
on a car and provides the centripetal force. This can be seen from the
free-body diagram:
The net force must be horizontal--pointing toward the center of the
circle--and only the friction force is available to provide it. The normal
force and the weight simply cancel each other.
v = 80kmh/3.6
v = 22.2 m/s
Now we can calculate the necessary centripetal force,
Fc = M v2 / r
Fc = M(22.2 m/s)2/350 m
Fc = 1.41 M (m/s2)
FN = M g = (M) (9.8 m/s2)
For this flat curve, the centripetal force is supplied by the friction
force, F f
Ff = FN
Ff = M 9.8 m/s/s
Ff = Fnet = Fc
Ff = 1.41 N
Banked Curves
Some curves are banked to compensate for slippery conditions like ice
on a highway or oil on a racetrack. Below is a car making a banked turn.
Without friction, the roadway still exerts a normal force n perpendicular
to its surface. And the downward force of the weight w is present. Those
two forces add as vectors to provide a resultant or net force Fnet
which points toward the center of the circle; this is the centripetal force.
Note that it points to the center of the circle; it is not parallel to
the banked roadway.
Resolve the forces into their components. Since we are interested in the force that points toward the center of the circle, we choose a coordinate axis that lies along that direction. There is no acceleration in the y-direction so the sum of the forces in the y-direction must be zero.
- w = 0
=
w
= [w / cos ]
sin
/
cos ]
= v 2
/ g r
Below is a car coming toward you on a banked turn going just the right
speed such that there is no friction force perpendicular to the direction
of motion. There is a normal force on each wheel, and the F N shown
is the sum of the four. If the banking angle is q
from horizontal, FN is tilted q
from vertical. If the turn is in a horizontal plane, choose a horizontal
x axis and make it positive to the left, so the car has positive acceleration
to the left (toward the center of curvature).
S Fy = 0 or FN
cosq - mg = 0, and
S Fx = ma or
FN sinq = mv2/r.
Solve the first for FN and plug it into the second, and
you will find that m cancels out, showing that big trucks and little cars
should travel the same speed on the turn.
Example
At what angle should a curve of 200 m radius be banked so that no friction
is required when a car travels at 60 kilometers per hour around the curve?
For a banked curve with no friction, the only forces acting on a car
are the normal force and the weight. We have just looked at this situation.
The centripetal force must be supplied by the horizontal component of the
normal force. We have just analyzed this situation in arriving at tan =v2/gr
While it is easier to think of an automobile travelling at 60 km/h,
it will be easier to use velocity in our calculations in terms of m/s,
so we begin by changing units,
v = 60kmh * 3.6
v = 16.7 m/s
Now we can directly apply
tan = v
2
/ g r = (16.7 m/s) 2 / (10 m/s2)(200 m)
= 7.91o
Stawa Set 14
Physics Text book pg 151-156
Physics Study Guide 82-84
13. Just as in the accelerating elevator problems we did for apparent weight there is an apparent weight felt by objects when the accelerate due to circular motion. Imagine you are standing on a scale that is moving in a vertical circle (such as in a Ferris Wheel.)
Because you are moving in a circle the acceleration must point toward the center of the circle. At the bottom of the circle the acceleration (and hence the net force) points up while at the top it points down. The force diagram for the bottom of the circle is
Notice that in this case the normal force is smaller than the weight
force so the apparent weight is less than the true weight. In the extreme
case the net force might need to be so large that the normal force might
be equal to zero. In this case you would just barely touch the scale. You
would have zero apparent weight. This is what people mean when they say
that someone is "weightless". If the net force needed to be larger still
you would not be able to stay on the circle and would fly off the wheel.
For calculations
Forces at top of circle: F = mv²/r = T + Fw or T = Fc - Fw
Forces at side of circle: F = mv²/r = T
Forces at bottom of circle: F = mv²/r = T - Fw or T = F + Fw
Minimum speed to stay in orbit at top of vertical circle:
vmin = Ögr
or (gr)½.
Maximum force or tension is at bottom of vertical circle. T = Fc +
Fw
Stawa Set 14
Physics Text book pg 157-163
Physics Study Guide 85+86
14. Centre of mass is the term used to define the point of in an object where all mass vectors in all directions are equal. This allows us to simplify an objects motion as well as allowing us to simplify the complex mass into one single point.
Physics Text book pg 132
Physics Study Guide 88, 111
15. Cavendish and the Value of G
Isaac Newton's law of universal gravitation proposed that the gravitational
attraction between any two object is directly proportional to the product
of their masses and inversely proportional to the distance between their
centers. In equation form, this is often expressed as follows:
The constant of proportionality in this equation is G - the universal gravitation constant. The value of G was not experimentally determined until nearly a century later by Lord Henry Cavendish using a torsion balance.
Cavendish's apparatus for experimentally determining the value of G
involved a light, rigid rod which was 6-feet long. Two small metal spheres
were attached to the ends of the rod and the rod was suspended by a wire.
When the long rod becomes twisted, the torsion of the wire begins to exert
a torsional force which is proportional to the angle of rotation of the
rod. Cavendish had calibrated his instrument to determine the relationship
between the angle of rotation and the amount of torsional force. A diagram
of the apparatus is shown below.
Cavendish
then brought two large lead spheres near the smaller spheres attached to
the rod. Since all masses attract, the large spheres exerted a gravitational
force upon the smaller spheres and twisted the rod a measurable amount.
Once the torsional force balanced the gravitational force, the rod and
spheres came to rest and Cavendish was able to determine the gravitational
force of attraction between the masses. By measuring m1, m2, d and Fgrav,
the value of G could be determined. Cavendish's measurements resulted in
an experimentally determined value of 6.75 x 10-11 N m2/kg2. Today, the
currently accepted value is 6.67259 x 10-11 N m2/kg2.
The value of G is an extremely small numerical value. Its smallness accounts for the fact that the force of gravitational attraction is only appreciable for objects with large mass. While two students will indeed exert gravitational forces upon each other, these forces are too small to be noticeable. Yet if one of the students is replaced with a planet, then the gravitational force between the other student and the planet becomes noticeable.
Isaac Newton compared the acceleration of the moon to the acceleration of objects on earth. Believing that gravitational forces were responsible for each, Newton was able to draw an important conclusion about the dependence of gravity upon distance. This comparison led him to conclude that the force of gravitational attraction between the Earth and other objects is inversely proportional to the distance separating the earth's center from the object's center. But distance is not the only variable effecting the magnitude of a gravitational force. In accord with Newton's famous equation Fnet=ma
Newton knew that the force which caused the apple's acceleration (gravity) must be dependent upon the mass of the apple. And since the force acting to cause the apple's downward acceleration also causes the earth's upward acceleration (Newton's third law), that force must also depend upon the mass of the earth. So for Newton, the force of gravity acting between the earth and any other object is directly proportional to the mass of the earth, directly proportional to the mass of the object, and inversely proportional to the square of the distance which separates the centers of the earth and the object.
But Newton's law of universal gravitation extends gravity beyond earth. Newton's law of universal gravitation is about the universality of gravity. All objects attract each other with a force of gravitational attraction. This force of gravitational attraction is directly dependent upon the masses of both objects and inversely proportional to the square of the distance which separates their centers. Newton's conclusion about the magnitude of gravitational forces is summarized symbolically as F=G[(m1m2)/(r2)]
The constant of proportionality (G) in the above equation is known as the universal gravitation constant. The precise value of G was determined experimentally by Henry Cavendish in the century after Newton's death. (This experiment will be discussed later in Lesson 3.) The value of G is found to be G = 6.67 x 10-11 N m2/kg2 The units on G may seem rather odd; nonetheless they are sensible. When the units on G are substituted into the equation above and multiplied by m1*m2 units and divided by d2 units, the result will be Newtons - the unit of force.
Knowing the value of G allows us to calculate the force of gravitational attraction between any two objects of known mass and known separation distance. As a first example, consider the following problem.
Definition: Gravitional Potential Energy
Due to the gravitational force of attraction, any two objects with
masses m1 and m2 located a distance r apart have the ability to do work.
Hence they have potential energy. The gravitational potential energy of
such objects is: PE grav = -G.
Differences in potential energy are physically relevant. In the above,
the zero of gravitational potential energy has been arbitrarily chosen
to be zero at r = . i.e. when the objects are infinitely far apart. The
negative sign is a consequence of the attractive nature of the gravitational
force. When the
objects are far apart, the gravitational force naturally moves them
closer, decreasing their potential energy (i.e. making it more negative).
Stawa Set 15
Physics Text book pg 164-169
Physics Study Guide 87-90
16. Sample Problem #1
Determine the force of gravitational attraction between the earth (m
= 5.98 x 1024 kg) and a 70-kg physics student if the student is standing
at sea level, a distance of 6.37 x 106 m from earth's center.
The solution of the problem involves substituting known values of G
(6.67 x 10-11 N m2/kg2), m (5.98 x 1024
kg ), m2 (70 kg) and d (6.37 x 106 m) into the universal
gravitation equation and solving for Fgrav. The solution is as follows:
FGrav = (6.67 x 10-11Nm2/kg2)(5.98
x 1024kg)(70kg)/(6.37 x 106m)2
FGrav = 688.1N
Sample Problem #2
Determine the force of gravitational attraction between the earth (m
= 5.98 x 1024 kg) and a 70-kg physics student if the student is in an airplane
at 40000 feet above earth's surface. This would place the student a distance
of 6.38 x 106 m from earth's center.
The solution of the problem involves substituting known values of G
(6.67 x 10-11 N m2/kg2), m (5.98 x 1024
kg ), m2 (70 kg) and d (6.38 x 106 m) into the universal
gravitation equation and solving for Fgrav. The solution is as follows:
FGrav = (6.67 x 10-11Nm2/kg2)(5.98
x 1024kg)(70kg)/(6.38 x 106m)2
FGrav = 685.9N
Two general conceptual comments can be made about the results of the
two sample calculations above. First, observe that the force of gravity
acting upon the student (a.k.a. the student's weight) is less on an airplane
at 40 000 feet than at sea level. This illustrates the inverse relationship
between separation distance and the force of gravity (or in this case,
the weight of the student).
The student weighs less at the higher altitude. However, a mere change
of 40 000 feet further from the center of the Earth is virtually negligible.
This altitude change altered the student's weight changed by 3 N which
is less than 1% of the original weight. A distance of 40 000 feet (from
the earth's surface to a high altitude airplane) is not very far when compared
to a distance of 6.37 x 106 m (equivalent to approximately 21
000 000 feet from the center of the earth to the surface of the earth);
this alteration of distance is like a drop in a bucket. As shown in the
diagram below, distance of separation becomes much more influential when
a significant variation is made.
The second conceptual comment to be made about the above sample calculations
is that the use of Newton's universal gravitation equation to calculate
the force of gravity (or weight) yields the same result as when calculating
it using the equation Fgrav = m*g
Fgrav = m*g = (70 kg)*(9.8 m/s/s) = 686 N
Both equations accomplish the same result because (as we will study later in Lesson 3) the value of g is equivalent to the ratio of (G*MEarth)/(REarth)2.
Gravitational interactions do not simply exist between the earth and other objects; and not simply between the sun and other planets; gravitational interactions exist between all objects with an intensity which is directly proportional to the product of their masses. So as you sit in your seat in the physics classroom, you are gravitationally attracted to your lab partner, to the desk you are working at, and even to your physics book. Newton's revolutionary idea was that gravity is universal - ALL objects attract in proportion to the product of their masses.
Stawa Set 15
Physics Text book pg 164-174
Physics Study Guide 87-90
17. We know the equation for determining the force
of gravity (Fgrav) with which an object of mass m was attracted
to the earth is Fgrav = mg
The force of gravity with which an object is
attracted to the earth can be found using:
To understand why the value of g is so location dependent, we will use the two equations above to derive an equation for the value of g. First, set both expressions for the force of gravity are set equal to each other.
Now observe that the mass of the object - m1
- is present on both sides of the equal sign. Thus, m1
can be canceled from the equation. This leaves us with an equation
for the acceleration of gravity. g=G[(m2)/(r2)]
The above equation demonstrates that the acceleration of gravity is
dependent upon the mass of the earth (approx. 5.98x1024 kg)
and the distance (r) that an object is from
the center of the earth. If the value 6.38x106 m (a typical
earth radius value) is used for the distance from Earth's center, then
g will be calculated to be 9.8 m/s2. And of course, the value
of g will change as an object is moved further from Earth's center. For
instance, if an object were moved to a location which is two earth-radii
from the center of the earth - that is, two times 6.38x106 m
- then a significantly different value of g will be found. As shown below,
at twice the distance from the center of the earth, the value of g becomes
2.45 m/s2.
The table below shows the value of g at various locations from Earth's
center.
Location |
Distance fromEarth's center (m) |
Value of gm/s2 |
| Earth's surface | 6.38 x 106 m | 9.8 |
| 1000 km above surface | 7.38 x 106 m | 7.33 |
| 2000 km above surface | 8.38 x 106 m | 5.68 |
| 3000 km above surface | 9.38 x 106 m | 4.53 |
| 4000 km above surface | 1.04 x 107 m | 3.70 |
| 5000 km above surface | 1.14 x 107 m | 3.08 |
| 6000 km above surface | 1.24 x 107 m | 2.60 |
| 7000 km above surface | 1.34 x 107 m | 2.23 |
| 8000 km above surface | 1.44 x 107 m | 1.93 |
| 9000 km above surface | 1.54 x 107 m | 1.69 |
| 10000 km above surface | 1.64 x 107 m | 1.49 |
| 50000 km above surface | 5.64 x 107 m | 0.13 |
As is evident from both the equation and the table above, the value of g varies inversely with the distance from the center of the earth. In fact, the variation in g with distance follows an inverse square law where g is inversely proportional to the distance from earth's center. This inverse square relationship means that as the distance is doubled, the value of g decreases by a factor of 4; as the distance is tripled, the value of g decreases by a factor of 9; and so on. This inverse square relationship is depicted in the graphic at the right.
The same equation used to determine the value of g on earth can also be used to determine the acceleration of gravity on the surface of other planets. The value of g on any other planet can be calculated from the mass of the planet and the radius of the planet. The equation takes the following form:
Using this equation, the following acceleration of gravity values can
be calculated for the various planets.
Planet |
Radius (m) |
Mass (kg) |
g (m/s2) |
| Mercury | 2.43 x 106 | 3.2 x 1023 | 3.61 |
| Venus | 6.073 x 106 | 4.88 x1024 | 8.83 |
| Mars | 3.38 x 106 | 6.42 x 1023 | 3.75 |
| Jupiter | 6.98 x 107 | 1.901 x 1027 | 26.0 |
| Saturn | 5.82 x 107 | 5.68 x 1026 | 11.2 |
| Uranus | 2.35 x 107 | 8.68 x 1025 | 10.5 |
| Neptune | 2.27 x 107 | 1.03 x 1026 | 13.3 |
| Pluto | 1.15 x 106 | 1.2 x 1022 | 0.61 |
The acceleration of gravity of an object is a measurable quantity. Yet emerging from Newton's universal law of gravitation is a prediction which states that its value is dependent upon the mass of the earth and the distance the object is from the earth. The value of g is independent of the mass of the object and only dependent upon location - the planet the object is on and the distance from the center of that planet.
Stawa Set 15
Physics Text book pg 164-186
Physics Study Guide 87-90
18. A satellite is any object which is orbiting the earth, sun or other massive body. Satellites can be categorized as natural satellites or man-made satellites. The moon, the planets and comets are examples of natural satellites. Accompanying the orbit of natural satellites are a host of satellites launched from earth for purposes of communication, scientific research, weather forecasting, intelligence, etc. Whether a moon, a planet, or some man-made satellite, every satellite's motion is governed by the same physics principles and described by the same mathematical equations.
The fundamental principle to be understood concerning satellites is that a satellite is a projectile. That is to say, a satellite is an object upon which the only force is gravity. Once launched into orbit, the only force governing the motion of a satellite is the force of gravity. Newton was the first to theorize that a projectile launched with sufficient speed would actually orbit the earth. Consider a projectile launched horizontally from the top of the legendary "Newton's Mountain" - at a location high above the influence of air drag. As the projectile moves horizontally in a direction tangent to the earth, the force of gravity would pull it downward. If the launch speed was too small, it would eventually fall to earth. Paths A and B in the diagram at the right illustrate the path of a projectile with insufficient launch speed for orbital motion. But if launched with sufficient speed, the projectile would fall towards the earth at the same rate that the earth curves. This would cause the projectile to stay the same height above the earth and to orbit in a circular path. And at even greater launch speeds, a cannonball would once more orbit the earth, but now in an elliptical path. At every point along its trajectory, a satellite is falling toward the earth. Yet because the earth curves, it never reaches the earth.
So what launch speed does a satellite need in order to orbit the earth?
The answer emerges from a basic fact about the curvature of the earth.
For every 8000 meters measured along the horizon of the earth, the earth's
surface curves downward by approximately 5 meters. So if you were to look
out horizontally along the horizon of the Earth for 8000 meters, you would
observe that the Earth curves downwards below this straight-line path a
distance of 5 meters. So for a projectile to orbit the earth, it must travel
horizontally a distance of 8000 meters for every 
5
meters of vertical fall. It so happens that the vertical distance which
a horizontally launched projectile would fall in its first second is approximately
5 meters (0.5*g*t2). For this reason, a projectile launched
horizontally with a speed of 8000 m/s will be capable of orbiting the earth
in a circular path; this assumes that it is launched above the surface
of the earth and encounters negligible atmospheric drag. As the projectile
travels tangentially a distance of 8000 meters in 1 second, it will drop
approximately 5 meters towards the earth; yet, the projectile will remain
the same distance above the earth due to the fact that the earth curves
at the same rate that the projectile falls. If shot with a speed greater
than 8000 m/s, it would orbit the earth in an elliptical path.
The motion of an orbiting satellite can be described by the same motion characteristics as any object in circular motion. The velocity of the satellite would be directed tangent to the circle at every point along its path. The acceleration of the satellite would be directed towards the center of the circle - towards the central body which it is orbiting. And this acceleration is caused by a net force which is directed inwards in the same direction as the acceleration.
This centripetal force is supplied by gravity - the force which universally acts at a distance between any two objects which have mass. Were it not for this force, the satellite in motion would continue in motion at the same speed and in the same direction. It would follow its inertial, straight-line path. Like any projectile, gravity alone influences the satellite's trajectory such that it always falls below its straight-line, inertial path. This is depicted in the diagram below. Observe that the inward net force pushes (or pulls) the satellite (denoted by blue circle) inwards relative to its straight-line path tangent to the circle. As a result, after the first interval of time, the satellite is positioned at position 1 rather than position 1'. In the next interval of time, the same satellite would travel tangent to the circle in the absence of gravity and be at position 2'; but because of the inward force the satellite has moved to position 2 instead. In the next interval of time, the same satellite has moved inward to position 3 instead of tangentially to position 3'. This same reasoning can be continued to explain how the inward force causes the satellite to fall towards the earth without actually falling into it.
Occasionally satellites will orbit in paths which can be described as ellipses. In such cases, the central body is located at one of the foci of the ellipse. Similar motion characteristics apply for satellites moving in elliptical paths. The velocity of the satellite is directed tangent to the ellipse. The acceleration of the satellite is directed towards the focus of the ellipse. And in accord with Newton's second law of motion, the net force acting upon the satellite is directed towards in the same direction as the acceleration - towards the focus of the ellipse. Once more, this net force is supplied by the force of gravitational attraction between the central body and the orbiting satellite. In the case of elliptical paths, there is a component of force in the same direction as (or opposite direction as) the motion of the object. Such a component of force can cause the satellite to either speed up or slow down in addition to changing directions. So unlike uniform circular motion, the elliptical motion of satellites is not characterized by a constant speed.
In summary, satellites are projectiles which orbit around a central massive body instead of falling into it. Being projectiles, they are acted upon by the force of gravity - a universal force which acts over even large distances between any two masses. The motion of satellites, like any projectile, are governed by Newton's laws of motion.
Stawa Set 16
Physics Text book pg 175-186
Physics Study Guide 90-92
19. The motion of objects are governed by Newton's laws. The same simple laws which govern the motion of objects on earth also extend to the heavens to govern the motion of planets, moons, and other satellites.
Consider a satellite with mass Msat orbiting a central body with a mass of mass MCentral. The central body could be a planet, the sun or some other large mass capable of causing sufficient acceleration on a less massive nearby object. If the satellite moves in circular motion, then the net centripetal force acting upon this orbiting satellite is given by the relationship
Since Fgrav = Fnet, the above expressions for centripetal force and gravitational force are equal. Thus,
Observe that the mass of the satellite is present on both sides of the equation; thus it can be canceled by dividing through by Msat. Then both sides of the equation can be multiplied by R, leaving the following equation.
Taking the square root of each side, leaves the following equation for the velocity of a satellite moving about a central body in circular motion
Similar reasoning can be used to determine an equation for the acceleration of our satellite that is expressed in terms of masses and radius of orbit. The acceleration value of a satellite is equal to the acceleration of gravity of the satellite at whatever location which it is orbiting. The equation for the acceleration of gravity was given as g = (G*Mcentral)/R2
Thus, the acceleration of a satellite in circular
motion about some central body is given by the following equation
a = (G*Mcentral)/R2
where G = 6.67 x 10-11 N m2/kg2,
Mcentral = the mass of the central body about which the satellite
orbits, and R = the average radius of orbit for the satellite.
The final equation which is useful in describing the motion of satellites is Newton's form of Kepler's third law. Since the logic behind the development of the equation has been presented elsewhere, only the equation will be presented here. The period of a satellite (T) and the mean distance from the central body (R) are related by the following equation:
There is an important concept evident in all three of these equations - the period, speed and the acceleration of an orbiting satellite are not dependent upon the mass of the satellite.
None of these three equations has the variable Msatellite in them. The period, speed and acceleration of a satellite is only dependent upon the radius of orbit and the mass of the central body which the satellite is orbiting. Just in the case of the motion of projectiles on earth, the mass of the projectile has no effect upon the acceleration towards the earth and the speed at any instant. Whenever air resistance is negligible and all forces but gravity are nonexistent, the mass of the moving object becomes a non-factor. Such is the case of orbiting satellites.
Practice Problem #1
A satellite wishes to orbit the earth at a height of 100 km (approximately
60 miles) above the surface of the earth. Determine the speed, acceleration
and orbital period of the satellite. (Given: Mearth = 5.98x1024
kg, Rearth = 6.37 x 106 m)
Like most problems in physics, this problem begins by identifying known
and unknown information and selecting the appropriate equation capable
of solving for the unknown. For this problem, the knowns and unknowns are
listed below.
|
R = Rearth + height = 6.47 x 106 m Mearth = 5.98x1024 kg G = 6.67 x 10-11 N m2/kg2 |
v = ??? a = ??? T = ??? |
Note
that the radius of a satellite's orbit can be found from the knowledge
of the earth's radius and the height of the satellite above the earth.
As shown in the diagram at the right, the radius of orbit for a satellite
is equal to the sum of the earth's radius and the height above the earth.
These two quantities can be added to yield the orbital radius. In this
problem, the 100 km must first be converted to 100 000 m before being added
to the radius of the earth. The equations needed to determine the unknown
are those which are listed above. We will begin by determining the orbital
speed of the satellite using the following equation: v2
= [ (G*MCentral ) / R ]
The acceleration can be found from either one of the following equations:
(1) a = (G*Mcentral)/R2
(2) a = v2/R
Equation (1) was derived. Equation (2) is a general equation for circular
motion. Either equation can be used to calculate the acceleration; the
use of equation (1) will be demonstrated here.
a = (G*Mcentral)/R2
a = (6.67 x 10-11 N m2/kg2)*(5.98x1024
kg) / (6.47 x 106 m)2
a = 9.53 m/s2
Observe that this acceleration is slightly less than the 9.8 m/s2 value expected on earth's surface. The increased distance from the center of the earth lowers the value of g.
Finally, the period can be calculated using the following equation:
T2/R3
= 4P2/ GMcentral
The equation can be rearranged to the following form
T = Ö[(4*pi2 * R3)
/ (G*Mcentral)]
The substitution and solution are as follows:
T = Ö [(4*(3.1415)2 * (6.47
x 106 m)3) / (6.67 x
10-11 N m2/kg2)*(5.98x1024
kg) ]
T = 5177 s = 1.44 hrs
Practice Problem #2
The period of the moon is approximately 27.2 days (2.35x106
s). Determine the radius of the moon's orbit and the orbital speed of the
moon. (Given: Mearth = 5.98x1024 kg, Rearth
= 6.37 x 106 m)
Like Practice Problem #1, this problem begins by identifying known and
unknown values. These are shown below.
|
T = 2.35x106 s Mearth = 5.98x1024 kg G = 6.67 x 10-11 N m2/kg2 |
R = ??? v = ??? |
The substitution and solution are as follows:
R3 = [ ((2.35x106 s)2 * (6.67
x 10-11 N m2/kg2)*(5.98x1024
kg) ) / (4*(3.1415)2) ]
R3 = 5.58 x 1025 m3
By taking the cube root of 5.58 x 1025 m3, the radius can be determined as follows: R = 3.82 x 108 m
The orbital speed of the satellite can be computed from either of the
following equations:
(1) v = Ö [ (G*MCentral
)
/ R ]
(2) v = (2*pi*R)/T
Equation (1) was derived. Equation (2) is a general equation for circular
motion. Either equation can be used to calculate the orbital speed; the
use of equation (1) will be demonstrated here. The substitution of values
into this equation and solution are as follows:
v = Ö [ (6.67
x 10-11 N
m2/kg2)*(5.98x1024 kg) / (3.82
x 108 m) ]
v = 1.02 x 103 m/s
Practice Problem #3
A geosynchronous satellite is a satellite which remains above the same
point on the earth. A geosynchronous satellite orbits the earth with an
orbital period of 24 hours, thus matching the period of the earth's rotational
motion. If a satellite wishes to orbit the earth at the equator in 24 hours
(86400 s), then how high above the earth's surface must it be located?
(Given: Mearth = 5.98x1024 kg, Rearth
= 6.37 x 106 m)
Just as in the previous problem, the solution begins by the identification
of the known and unknown values. This is shown below.
|
T = 86400 s Mearth = 5.98x1024 kg Rearth = 6.37 x 106 m G = 6.67 x 10-11 N m2/kg2 |
h = ??? |
The
unknown in this problem is the height of the satellite above the surface
of the earth. Yet there is no equation with the variable h. The solution
then involves first finding the radius of orbit and using this R value
and the R of the earth to find the height of the satellite above the earth.
As shown in the diagram at the right, the radius of orbit for a satellite
is equal to the sum of the earth's radius and the height above the earth.
The radius of orbit can be found using the following equation: T2/R3
= 4P2/ GMcentral
By taking the cube root of 7.54 x 1022 m3, the radius can be determined as follows: R = 4.23 x 107 m
The radius of orbit indicates the distance which the satellite is from the center of the earth. Now that the radius of orbit has been found, the height above the earth can be calculated. Since the earth's surface is 6.37 x 106 m from its center (that's the radius of the earth), the satellite must be a height of 4.23 x 107 m - 6.37 x 106 m = 3.59 x 107 m above the surface of the earth. So the height of the satellite is 3.59 x 107 m.
Stawa Set 16
Physics Text book pg 178-187
Physics Study Guide 90-93
20.
Contexts
Astronauts who are orbiting the earth often experience sensations of
weightlessness. These sensations experienced by orbiting astronauts are
the same sensations experienced by anyone who has been temporarily suspended
above the seat on any amusement park ride. Not only are the sensations
the same (for astronauts and roller coaster riders), but the causes of
those sensations of weightlessness are also the same. Unfortunately however,
many people have difficulty understanding the causes of weightlessness.
The cause of weightlessness is quite simple to understand. However, the
stubbornness of one's preconceptions on the topic often stand in the way
of one's ability to understand. Consider the following multiple choice
question about weightlessness as a test of your preconceived notions on
the topic:
Before understanding weightlessness, we
will have to review two categories of forces - contact forces and action-at-a-distance
forces. As you sit in a chair, you experience two forces - the force
of the earth's gravitational field pulling you downwards toward the earth
and the force of the chair pushing you upwards. The upward chair force
is sometimes referred to as a normal force and results from the contact
between the chair top and your rear end; this normal force is categorized
as a contact force. Contact forces can only result from the actual touching
of the two interacting objects - in this case, the chair and you. The force
of gravity acting upon your body is not a contact force; it is often categorized
as an action-at-a-distance force. The 
force
of gravity is the result of your center of mass and the earth's center
of mass exerting a mutual pull on each other; this force would even exist
if you were not in contact with the earth. The force of gravity does not
require that the two interacting objects (your body and the earth) make
physical contact; it can act over a distance through space. Since the force
of gravity is not a contact force, it cannot be felt through contact. You
can never feel the force of gravity pulling upon your body in the same
way that you would feel a contact force. If you slide across the asphalt
tennis court (not recommended), you would feel the force of friction (a
contact force). If you are pushed by a bully in the hallway, you would
feel the applied force (a contact force). If you swung from a rope in gym
class, you would feel the tension force (a contact force). If you sit in
your chair, you feel the normal force (a contact force). But if you are
jumping on a trampoline, even while moving through the air, you do not
feel the earth pulling upon you with a force of gravity (an action-at-a-distance
force). The force of gravity can never be felt. Yet those forces which
result from contact can be felt. And in the case of sitting in your chair,
you can feel the chair force; and it is this force which provides you with
a sensation of weight. Since the upward normal force would equal the downward
force of gravity when at rest, the strength of this normal force gives
one a measure of the amount of gravitational pull. If there were no upward
normal force acting upon your body, you would not have any sensation of
your weight. Without the contact force (the normal force), there is no
means of feeling the non-contact force (the force of gravity).
Weightlessness is simply a sensation experienced by an individual when there are no external objects touching one's body and exerting a push or pull upon it. Weightless sensations exist when all contact forces are removed. These sensations are common to any situation in which you are momentarily (or perpetually) in a state of free fall. When in free fall, the only force acting upon your body is the force of gravity - a non-contact force. Since the force of gravity cannot be felt without any other opposing forces, you would have no sensation of it. You would feel weightless when in a state of free fall.
These feelings of weightlessness are common at amusement parks for riders
of roller coasters and other rides in which riders are momentarily airborne
and lifted out of their seats. Suppose that you were lifted in your chair 
to
the top of a very high tower and then your chair was suddenly dropped (such
as is the case on The Giant Drop at Great America). As you and your chair
fall towards the ground, you both accelerate at the same rate - "g."
Since the chair is unstable, falling at the same rate as you, it is unable
to push upon you. The force of gravity is the only force acting upon your
body. There are no stable external objects touching your body and exerting
a force. As such, you would experience a weightless sensation. You would
weigh as much as you always do (or as little), yet you would not have any
sensation of this weight.
Weightlessness is only a sensation; it is not a reality corresponding to an individual who has lost weight. As you are free-falling on a roller coaster ride (or other amusement park ride), you have not momentarily lost your weight. Weightlessness has very little to do with weight and mostly to do with the presence or absence of contact forces. If by "weight" we are referring to the force of gravitational attraction to the earth, a free-falling person has not "lost their weight;" they are still experiencing the earth's gravitational attraction. Unfortunately, the confusion of a person's actual weight with one's feeling of weight is the source of many misconceptions.
Technically speaking, a scale does not measure one's weight. While we
use a scale to measure one's weight, the scale reading is actually 
a
measure of the upward force applied by the scale to balance the downward
force of gravity acting upon an object. When an object is in a state of
equilibrium (either at rest or in motion at constant speed), these two
forces are balanced; the upwards force of the scale upon the person equals
the downwards pull of gravity (also known as weight). And in this instance,
the scale reading (which is a measure of the upwards force) equals the
weight of the person. However, if you stand on the scale and bounce up
and down, the scale reading undergoes a rapid change. As you undergo this
bouncing motion, your body is accelerating. During the acceleration periods,
the upward force of the scale is changing. And as such, the scale reading
is changing. Is your weight changing? Absolutely not! You weigh as much
(or as little) as you always do. The scale reading is changing, but remember:
the scale doesn't measure your weight. The scale is only measuring the
external contact force which is being applied to your body.
Now suppose you stand on the bathroom scale and ride an elevator up and down. As you are accelerating upwards and downwards, the scale reading is different than when you are at rest and traveling at constant speed. For when you are accelerating, the upwards and downwards force are not equal. But when your are at rest or moving at constant speed, the opposing forces balance each other. Knowing that the scale reading is a measure of the upwards normal force of the scale upon your body, its value could be predicted for various states of motion. For instance, the value of the normal force (Fnorm) on a 50-kg person could be predicted if the acceleration is known. This prediction can be made by simply applying Newton's second law as discussed in Unit 2. As an illustration of the use of Newton's second law to determine the varying contact forces on an elevator ride, consider the following diagram. In the diagram, a 50-kg physics student is traveling with constant speed (A), accelerating upward (B), accelerating downward (C), and free-falling (D) after the elevator cable snaps.
In each of these cases, the upward contact force (Fnorm)
can be determined using a free-body diagram and Newton's second law. The
interaction of the two forces - the upwards normal force and the downwards
force of gravity - can be thought of as a tug-of-war. The net force acting
upon the person indicates who wins the tug-of-war (the up force or the
down force) and by how much. A net force of 100-N,
up indicates that the upwards force "wins" by an amount equal to 100 N.
The gravitational force acting upon the rider is found using the equation
Fgrav = m*g. (An approximated g value of 10 m/s/s is
used to simplify some of the mathematics and in turn make the conceptual
principles more obvious.)
 |
 |
 |
 |
|
Fnet = 0 N |
Fnet = 100 N, up |
Fnet = 100 N, down |
Fnet = 500 N, down |
|
Fnorm = 500 N |
Fnorm = 600 N |
Fnorm = 400 N |
Fnorm = 0 N |
Earth-orbiting astronauts are weightless for the same reasons that riders of a free-falling amusement park ride or a free-falling elevator are weightless. They are weightless because there is no external contact force pushing or pulling upon their body. In each case, gravity is the only force acting upon their body. Being an action-at-a-distance force, it cannot be felt and therefore would not provide any sensation of their weight. But for certain, the orbiting astronauts weigh something; that is, there is a force of gravity acting upon their body. In fact, if it were not for the force of gravity, the astronauts would not be orbiting in circular motion. It is the force of gravity which supplies the centripetal force requirement to allow the inward acceleration which is characteristic of circular motion. The force of gravity is the only force acting upon their body. The astronauts are in free-fall. Like the falling amusement park rider and the falling elevator rider, the astronauts and their surroundings are falling towards the earth under the sole influence of gravity. The astronauts and all their surroundings - the space shuttle with its contents - are falling towards the earth without colliding into it. Their tangential velocity allows them to remain in orbital motion while the force of gravity pulls them inward.
Many students believe that orbiting astronauts are weightless because they do not experience a force of gravity. So to presume that the absence of gravity is the cause of the weightlessness experienced by orbiting astronauts would be in violation of circular motion principles. If one presumes that the absence of gravity is the cause of their weightlessness, then that person is hard-pressed to come up with a reason for why the astronauts are orbiting in the first place.
One might respond to this discussion by adhering to a second misconception: the astronauts are weightless because the force of gravity is reduced in space. The reasoning goes as follows: "with less gravity, there would be less weight and thus they would feel less than their normal weight." While this is partly true, it does not explain their sense of weightlessness. The force of gravity actin upon an astronaut in a typical space shuttle orbit is certainly less than on earth's surface; but how much less? Is it small enough to account for a significant reduction in weight? Absolutely not! If a typical space shuttle orbit is approximately 400 km above the earth's surface, then the value of g at that Laotian will have been reduced from 9.8 m/s/s (at earth's surface) to approximately 8.7 m/s/s. This would cause an astronaut weighing 1000 N to be reduced in weight to approximately 890 N. While this is certainly a reduction in weight, it does not account for the absolutely weightless sensations which astronauts experience. Their absolutely weightless sensations are the result of having "the floor pulled out from under them" (so to speak) as they are free-falling towards the earth.
Still other physics students believe that weightlessness is due to the
absence of air in space. Their misconception lies in the idea that there
is no force of gravity when their is no air. According to them, gravity
does not exist in a vacuum. But this is not the case. Gravity is a force
which acts between the earth's mass and the mass of other objects which
surround it. The force of gravity can act across large distances and its
effect can even penetrate across and into the vacuum of outer space. Perhaps
students who own this misconception are confusing the force of gravity
with air pressure. Air pressure is the result of surrounding air particles
pressing upon the surface of an object in equal amounts from all directions.
Air pressure is not related to the force of gravity. While air pressure
reduces to zero in a location void of air (such as space), the force of
gravity does not become 0 N. Indeed the presence of a vacuum results in
the absence of air resistance; but this would not account for the weightless
sensations. Astronauts merely feel weightless because there is no external
contact force pushing or pulling upon their body; they are in a state of
free fall.
Kepler's Laws of Planetary Motion
Kepler's First Law: The path of each planet about the sun is an ellipse
with the sun at one focus.
Kepler's Second Law: Each planet moves so that an imaginary line drawn
from the sun to the planet sweeps out equal areas in equal periods of time.
Kepler's Third Law: The ratio of the squares of the periods (the time
needed for one revolution about the sun) of any two planets revolving about
the sun is equal to the ratio of the cubes of their mean distances from
the sun.
The following diagram shows how the moon causes tides on earth:
The gravitational forces between the Earth and the Moon cause tides. The Moon's gravitational attraction is stronger on the side of the Earth nearest to the Moon and weaker on the opposite side. Since the Earth, and particularly the oceans, is not perfectly rigid it is stretched out along the line toward the Moon. From our perspective on the Earth's surface we see two small bulges, one in the direction of the Moon and one directly opposite. The effect is much stronger in the ocean water than in the solid crust so the water bulges are higher. And because the Earth rotates much faster than the Moon moves in its orbit, the bulges move around the Earth about once a day giving two high tides per day.
In this diagram you can see that the moon's gravitational force pulls water in the oceans so that there are "bulges" in the ocean on both sides of the planet. The moon pulls water toward it and this causes the bulge toward the moon. The bulge on the side of the earth opposite to the moon is caused by the moon "pulling the earth away" from the water on that side. If you are one the coast and the moon is directly overhead, you should experience a high tide. If the moon is directly overhead on the opposite side of the planet, you should also experience a high tide.
The pull of gravity from the Moon and the Sun are the primary cause for tides. The effect is the greatest when the Moon and Sun are in a straight line with the Earth.
During the day the earth rotates 180 degrees in 12 hours. The moon, meanwhile, rotates 6 degrees around the earth in 12 hours. The twin bulges and the moon's rotation mean that any given coastal city experiences a high tide every 12 hours and 25 minutes or so.
Stawa Set 16
Physics Text book pg 175-193
Physics Study Guide 86, 90, 92
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