www.lsmsa.edu teacher Robert Dalling's Physics Lectures (see also www.ushumans.net)
The iMax movie The Dream is Alive is a good way to start the study of gravity. Gravity is the force of attraction that pulls two masses toward each other. We all know that the Earth’s gravity makes things fall. (NASA shows what life around the Earth is like when gravity is “turned off.”) In addition, the gravitational force keeps the Moon in orbit around the Earth, the Earth in orbit around the Sun, and it keeps us from falling off the surface of the Earth. All masses are attracted to all other masses, but the gravitational force is so tiny that it takes a lot of mass before the force amounts to much. This fore acts along the line joining the two masses.
The orbital motion of the Earth and the Moon, due to the gravitational attraction between them, is similar to the way in which two persons can hold hands while facing each other and lean backwards to sort of spin around each other. If the two persons let go of each other then they will fly apart. They can continue in this circular motion because of the mutual pull of each other's hands. Two skaters sort of orbit each other in this way. If one skater is 100 times heavier than the other then you will hardly notice the motion of the larger person. The Earth and the Moon are held in their circular orbit by their mutual gravitational pull. They orbit each other but the Earth moves little since it is about 100 times as massive as the Moon. The planets orbit the sun because the gravitational force pulls the planet around.
In the year 1680, Newton obtained the equation describing the gravitational force between two objects by considering which radial power rule rn would produce the observed, elliptical orbits that take the planets around the sun. He determined that n = -2 produces the observed orbits:
F = Gm1m2/r2,
where the Gravitational constant G = 6.67x10-11 Nm2/kg2.
We see that two 1-kg masses separated by 1 meter will experience a gravitational attraction of 6.67x10-11 N.
Two persons of mass 75 kg and 100 kg who are 1.5 cm apart feel a gravitational attraction between them that is equal to the gravitational attraction of the moon on one of them.
Fperson = Gm1m2/r2 = ( 6.67x10-11 Nm2/kg2 )( 75 kg )( 100 kg ) / ( 0.015 m )2 = 0.002 N
Fmoon = Gm1m2/r2 = ( 6.67x10-11 Nm2/kg2 )( 75 kg )( 7.35x1022 kg )/( 3.74x108 m)2 = 0.002 N.
CQ: Calculate the gravitational force between the Sun and the Earth which have mass 2x1030 kg and 6x1024 kg and are separated by a distance r = 1.5x1011 meter.
G = mg = GmMe/Re2 so g = Gme / Re2 = 9.8 m/s2.
The gravitational force among three bodies is shown in this simulation. This simulation shows several light masses being pulled toward a heavy mass that is shown as the brown object. Here you can see the orbits of masses around objects.
If a building were 100 miles (160 km high), people standing on the top floor would weigh W’ = mg’ = GmMe / (Re + 160 km)2 = 0.94 mg, or 94% of their weight when standing on the surface. What is g for an astronaut orbiting 100 miles above the surface of the Earth? Same as it is at the top of that building: 0.94 g. This meas that if a space shuttle were not “falling around the Earth,” the astronaut’s weight would be 94% of its value on the ground. The astronauts are weightless because they are falling just as you are weightless when falling from a tree. Suppose you are standing on a scale, weighing yourself in an elevator when the elevator’s supporting cable breaks. You will then be experiencing free fall, as will the elevator and the scale. Your feet are still in contact with the scale but you will be exerting no force against it and will be “weightless.”
When a mass is gravitationally pulled by a number of objects, then the net force must be calculated. Gravity can pull together nearby galaxies. If Newton’s equation F=ma is written for each of 10,000 gravitationally interacting point particles that comprise the galaxy then the motion of colliding galaxies can be calculated and displayed,see also here. Here is a an animation of the collision of two planets. Such a collision resulted in the capture and formation of our moon.
At the surface of the Earth, the gravitational force between you (mass m) and the Earth (mass mE) is given by
F = GmmE/r2 = mg.
Cancelling m, we see that at the surface of the Earth
g = GmE/r2 = ( 6.67x10-11 Nm2/kg2 )( 6x1024 kg ) / ( 6.38x106 m )2 = 9.8 m/s2.
What is the value of g for a neutron star of a given mass and radius? A neutron star is what remains after a star has gone supernova. It has a typical mass 1.4 times that of the sun. This is the same as having the mass of 470,000 Earths squeezed into a 100 km radius. The neutron star’s density = mass/vol =1.4 x 1030 kg / [ 4π(1x105 m)3/3 ] = 7x1014 kg/m3. One cubic centimeter of this material has a mass of 7x108 kg. It has the density of nuclear matter. We get
g = GM / r2 = ( 6.67x10-11 Nm2/kg2 )( 1.4x2x1030 kg ) / ( 1.0x105 m )2 = 2x1010 m/s2.
The value of g on the sun is 274 m/s2.
An object of mass m placed between the Earth and the Moon will experience two gravitational forces that point in opposite directions. At what point are these forces balanced? (These are called Lagrangian points. See here for an example involving Neptune’s Trojan asteroids.) Let R = 3.8x108 meters be the distance between the Earth and the Moon, and let x be the distance between the mass m and the Earth. Then R - x is the distance between the mass m and the Moon. The ratio of the mass of the Moon to that of the Earth is Mm/ME = 7.36x1022 / 5.98x1024.
When the two gravitational forces are equal, we have
GmME/x2 = GmMm / ( R - x )2.
Cancel the common factor Gm to get
( R - x )2 = ( Mm / ME ) x2
R2 - 2Rx + x2 = ( Mm / ME ) x2
( 1 - Mm/ME) x2 - 2Rx + R2 = 0
x = 2R ± √[ 4R2 - 4( 1 - Mm / ME ) R2 ]
2 ( 1 - Mm / ME )
Inside the square root is
4R2 - 4R2 + 4R2Mm / ME = 4R2 Mm / ME..
The twos in the numerator and denominator cancel the 4 within the square root. Bring the R2 out of the square root to get
x = R[ 1 ± √( Mm / ME) ] / (1 - Mm / ME ) = 0.9 R = ( 0.9 )( 3.8x108 m ) = 3.42x108 m.
We take the negative root because the positive root gives x > R, which is beyond the Moon. The gravitational forces from the Earth and Moon are equal at that more distant spot but the mass will be pulled toward the Moon rather than sitting still at a location of balanced forces. Taking the negative root, we see that x is 90% of the way to the Moon.
For the Earth-Sun system, x is even closer to the Earth. We have
x = r[ 1 ± √( ME / MS) ] / (1 - ME / MS ) = 0.9982 R = 2.56x108 m
because ME / MS = 5.98x1024 / 1.99x1030 and R = 1.496x1011 meters. This x-location is closer to the Earth than is the moon. The Earth and Moon are “locked” together while falling around the Sun.
Three stars placed in an equilateral triangle have a net force that is inward and could rotate as a unit. Three objects having equal mass m are placed in the corners of an equilateral triangle having sides of length r. What is the net force on mass A due to masses B and C?
· · · · · · · · · · · · A █ →┼x
Place the x-y coordinate system at mass A. Each corner of the equilateral triangle has a 60° angle. Since half of that lies between A and either B or C, we have θ = 30°. Mass A is equally pulled toward both masses B and C. We see that the y-component of the forces toward masses B and C cancel. The x component of both forces is twice that of one alone.
Fx = -2 F cos( θ ) = -2 Gm2/r2 cos( θ ),
where cos( 30 ) = ½ √3.
This net force is inward and can provide the centripetal force that would make the system move in a circle. The distance R from each point to the center of the circle joining the three masses follows R = r / √3. The centripetal force on one mass in an orbit of radius R is mv2/R and this is equal to the net force just calculated. We then have
mv2/R = √3 mv2/r = 2 Gm2/r2 cos( 30 ) = ½ √3 Gm2/r2.
This gives v = √( Gm/r ) and a period of T = 2πR / v = 2π √[ r3/ ( 3Gm ) ]
A good portion of stellar systems are seen to consist of two stars orbiting about their common center of mass. It would be fun to have two differently sized suns rising and setting with perhaps incommensurate schedules. The universe has lots of triple star systems.
As one moves around the surface of the Earth, the local acceleration due to gravity varies from the equator to the pole because of three effects: changes in local land forms, the spinning Earth attempts to throw a mass off the surface and the spin of the Earth causes the radius of its material to be larger at the equator than at the pole. The international gravity formula adjusts g for latitude. The latitude of Natchitoches, LA is 31.760N, its longitude is 93.086W, and its elevation is 115 feet. The formula gives a local value of g = 9.79464 m/s2. The value is also calculated from Helmert's equation.
The value of g is modified by 3.3% because the Earth is spinning: g = Gme / Re2 - v2 / r, where v = 2πr/period of 24 hours at a radius r determined by latitude. We launch satellites from the equator where the spin of the Earth has the greatest speed.
What is the period of a pendulum while in free fall? How does a candle flame behave in zero g? What is life like at zero g?
Here is an online textbook of rocketry, a space flight simulator, and a discussion of space flight.
Kepler’s Laws of planetary motion
The gravitational force supplies the centripetal force causing a planet of mass m to move in a circular orbit around its star of mass M:
F = mv2 / r = mrω2,
where ω = rv. Since ω = 2πf = 2π / T, we get
F = 4π2mr / T2
T2 = 4π2mr / F = 4π2mr / ( GmM/r2 ) = 4π2r3 / ( GM ).
This is Kepler’s Law. The period squared goes as the radius cubed. Click here to try to launch satellites into stable orbits or send a probe from Earth to Mars. Click here for animations of Kepler’s Laws. Click here for the orbital parameters of the planets.
Hewitt calculates that if gravity were turned off, a steel cable 700 km in diameter would be needed to withstand the force needed to supply the centripetal force that makes the Moon orbit the Earth. Lots of mass makes lots of gravitational force.
A satellite makes an elliptical orbit around a central mass. Its distance from the central mass increases from its point of closest approach–its perigee–to its farthest, or apogee, point. The speed of the satellite increases as it nears the closest point and decreases as it moves farther away. Its stored, gravitational energy changes in the opposite manner.
As a consequence of the conservation of angular motion, a satellite sweeps out equal areas in equal times as it revolves around its parent. A triangle from the sun to the planet’s first and second positions has differential area ΔA = ½ r Δs, where s is the arc length. But rΔs = rv Δt. Multiplying by mass m gives mΔA = ½ mvr Δt. Since L = mvr = constant, we get
ΔA / Δt = L / (2m) = constant.
The period of Uranus is 84 times that of the Earth. What is ratio of the orbital distances for Uranus and the Earth?
The ratio of periods is
( Tu / Te )2 = [ 4π2ru3 / ( GM ) ] / [ 4π2re3 / ( GM ) ] = ru3 / re3 = (84)2,
so the radius of the orbit of Uranus is
ru = (84)2/3 re = 19 re.
What is the period of a geosynchronis satellite that is orbiting the Earth? The satellite has a matching period of 24 hours = 86400 seconds. With this period, we get r = 4.2x107 meters = 26,000 miles. A satellite with a 24-hour period will hover above one spot on the ground at the equator. The motion of the satellite is within an equatorial plane and has a speed of v = 2πr/period = 3000 m/s.
Kepler’s period-radius equation is used to determine the mass of the central object. Knowing the period T = 27.3 days = 2.36x106 sec and distance r = 3.85x108 meters for the Moon’s orbit about the Earth, what is the mass of the Earth?
M = (4π2r3) / ( GT2 ) = 6.3x1024 kg.
The average density of the Earth is its mass divided by its volume 4/3πr3. This gives a value that is 5.5 times the density of water. Since the density of material at the surface of the Earth is found to be less than this, we conclude that the Earth’s density must be higher in its interior. When the Earth formed and was mostly molten, the heavier material sunk to its center in a matter or a few million years. In the coming chapters, we will see that an object the size of the Earth would cool of in a few tens of millions of years if it were not for the heat continually being generated by radioactivity within its material.
A star is pulled toward the center of the galaxy by the net force of attraction between that star and the others. As a star rotates around the center of the galaxy, the net gravitational force supplies the centripetal force. However, there is not enough visible star-mass to supply the centripetal force needed to make the galaxy rotate about its center. There must be more mass that is not lit up. This conclusion led astronomers to calculate the portion of “Dark Matter” in the universe. The portion might be 50% to 90%.
Gravitational formation of galaxies, stars, and planets
We want to understand how planets and stars form into spherically shaped objects due to the mutually attractive force of gravity between each of their pieces. For example, I am standing at a particular spot on the surface of the Earth. There is a gravitational attraction between my mass and the mass of each piece of matter throughout the Earth, including the pieces that make up the Earth's core, the pieces that make up China and Africa, and the pieces right under my feet. The total force of gravity between me and the entire Earth comes from adding together each of these attractions. It turns out that calculus shows that you get the same total force when you replace the spread-out Earth with a single point-sized mass. This single mass is located at the center of the Earth, and it has the same mass as the entire Earth.
To see how this can be true, imagine two persons, John and Jeff, pulling equally hard on ropes tied to Greg. Both John and Jeff are four steps in front of Greg but while John is three steps to Greg's right, Jeff is three steps to the left. We can believe that Greg will move in the vertical direction shown because the horizontal portions of the two pulls cancel each other. If we add two (or even 2 million) more equally-hard pullers, Kari and Bryan, placed right next to John and Jeff, the net or equivalent pull will still be in the same vertical direction as before.
Imagine now that John and Jeff are replaced with two equal-sized portions of Earth which are below its surface. Greg is still being pulled equally hard by two symmetrically paired pulls that are pieces of the Earth. For example, the attraction between Greg and a piece of the Earth that is to his right but twenty miles down into the Earth can be combined to the attractive pull from a piece that is to his left and down twenty miles. If we think of these two pieces to his right and to his left, it is like a pair of ropes pulling him equally to the right and to the left but also downward. These right and left pulls cancel each other's affect. Greg is pulled equally by each so he will not move toward either; instead, he moves along the center line between those two pulls. The two downward pulls add up to a single "gravity rope" pulling him straight toward the Earth's center.
We can divide the entire Earth's material into similarly paired blocks to see that they all add up to a single pull toward the Earth's center. In the same way there are pieces right under his feet that can be compared with far away pieces that are near the opposite surface. The nearby point pulls more strongly than the far-away point. The far away and nearby pulls add up to a single "puller" placed right at the center. This fact was first worked out by Isaac Newton in 1685. (Whenever one of us humans figures out a new fact, it is recorded and made available to everyone– and used again and again.)
By the way, we feel our own weight because that is how strongly all of the pieces of the Earth are pulling on the own matter of our body. If you stand on a scale to weigh yourself but have somebody pull down on your feet then the scale will read as if your weight has increased. When you stand on the scale, the Earth is pulling on your feet with the force indicated by the meter. It is an attractive force between each piece of your body and each piece of the Earth. The more pieces you have the larger will be the force and the higher the scale reading. And, the more massive the planet and the closer you are to its center, the larger will be the gravitational force. We feel heavy (our own weight) because that is how strongly the matter of the Earth is pulling on the matter of our body. How much does your weight change when the Moon is directly overhead compared to it being on the opposite side of the Earth?
The point of this discussion is that each piece of the Earth thinks that it is simply being attracted toward that single central point of the Earth's sphere. The gravitational force between you and the entire volume of the Earth is identical to the gravitational force between you and a point-mass, whose mass equals that of the entire Earth, placed at the position of the center of the Earth.
In interstellar space, a clump of gaseous debris can become gathered together through a mutual attraction toward its center and result in the formation of a spherical star. As an initially spherical cloud of dust rotates, it elongates and flattens into a spinning disk, as shown here. (Visualize a spinning water drop that elongates as it spins.) To conserve angular momentum, it spins more quickly as it condenses to lesser volume. Each piece of the star feels an attraction toward the star's center, and the weight of that material crushes and fuses hydrogen into helium. Once the star begins fusing, outwardly moving products push loose material outward.
A planet can also form in this way, or it might form as bits of material accumulate through their mutual gravitational attraction.
Another example is that all of the mass of the universe is attracted toward its center, and this attraction slows the rate of expansion of the universe.
The gravitational force of extended objects
The equation F = Gm1m2/r2 describes the gravitational force between two point-sized masses. We have to add forces when finding the total attraction between extended objects.
Let’s determine the total force on a mass m due to a bar ■■■■■■■■ of length L and mass M. The mass is a distance a from the nearest end of the bar that lies along the x axis.
· · · · · · · · ·■■■■■■■■
Each mass element dm ■ comprising the extended bar exerts a force dF = Gm dm / x2 on mass m that is a distance x away. The linear mass density of the bar is λ = M/L, which is constant here, so we have dm = λ dx. The total force is found by adding up the little forces from each of the mass elements.
F = ∫ dF = Gm ∫ λ dx / x2 = Gm M/L [ - 1/x].
We evaluate the integral from x = a to x = a + L to get
F = Gm M/L [ -1/( a + L ) - -1/a ] = GmM / [ a( a + L ) ].
A bar lies along the x axis with its left edge at x = -a and its right edge at x = -a + L. A point mass m is located at x = 0 and y = -b.
· · · · · · ·■■■■■■■■■■■■■■■■■■
................................ mass m at (x,y) = (0,-b)
We see that r = √( x2 + b2 ), cos( θ ) = b /r, and sin( θ ) = x / r. We still have dm = λ dx, where λ = M / L. The force dF between m and dm lies along r and has components in both the x and y directions. The total x-component is found by adding up all the little dFx pieces. We will have to integrate x from -a to -a + L.
Fx = ∫dFx = Gm∫λ( sinθ ) dx / r2 = Gm λ∫ x dx / [ x2 + b2 ]3/2
= ( GmM/L )( 1 / [ a2 + b2 ]½ - 1 / [ ( L - a )2 + b2 ]½ )
Fy = ∫dFx = Gm∫λ( cosθ ) dx / r2 = Gm λ∫ dx / [ x2 + b2 ]3/2
= [ GmM/(Lb) ][ a / [ a2 + b2 ]½ + ( L - a ) / [ ( L - a )2 + b2 ]½ ).
Where we have used
∫ x dx / [ x2 + b2 ]3/2 = -1 / [ x2 + b2 ]½,
∫ dx / [ x2 + b2 ]3/2 = 1 / ( b2 [ x2 + b2 ]½ ).
This might represent the gravitational force between a person and a tall building or tree. The direction of the net force is found from tan-1( Fy / Fx ).
For b >> L and b >> a, we get
Fx = GmM / ( bL ) and Fy = GmM / b2.
While the force between two point masses falls with distance as 1/r2, we see that Fx goes down with separation distance as 1/b. The gravitational force between a point mass and a bar is less than the force between two point particles because portions of the bar’s mass lie at greater distances from the point mass. See here for an simulation of the gravitational force between two bars.
Suppose that we were to cut away a sphere of material with radius b = R/2 from just below the Earth's surface. What would then be the force of attraction between an object of mass m that is a distance r from the center of the remaining Earth? The Earth has radius R and mass M before hollowing.
This is equivalent to an attraction between the object and two spherical masses, one of positive mass and radius R, the other with negative mass and radius b = R/2. The ratio of mass contained in spheres of equal, constant density and of radii R and b is mb / M = ( b / R )3 = ½3 = 1/8. We also have 3/2 b = 3/4 R.
F = Fr - Fb = GmM/r2 - Gmmb/( r - 1.5 b )2 = GmM[ 1/r2 - (1/8)/( r - 1.5 b )2.
The gravitational force on an object below the surface of the Earth
The radius of the Earth is R = 6.37 x 106 m. For a point r < R, within the Earth the gravitational force grows linearly with r. The mass density of the Earth is ρ = M / V = M / ( 4/3 π R3 ). The mass contained in the shell above point r produces a net force of zero. (To see that this is true, consider a mass located at the center of a hollow sphere. It will be equally pulled in every direction and so experience a net force of zero.) The mass m enclosed in a sphere of radius r < R is given by
m = ρV = 4/3 π r3 ρ = 4/3 π r3 M / ( 4/3 π R3 ) = r3 M / R3.
The gravitational attraction of this distributed mass m is the same as that of a point mass m located at the center of the sphere. The force between this mass m and an object of mass m’ located at point r < R is then
F = -Gm’m/r2 = -Gm’mr / R3.
Since F = m’a = m’ d2r / dt2, we have
d2r / dt2 + ( Gm / R3) r.
This is of the form
d2r/dt2 + ω2r = 0,
where ω= √( GM / R3) . A mass released at a point 0 < r < R will oscillate between r = -R and r = +r with period T = 2π/ω. For r = R, we have
T = 2π / √(GM / R3) = 2π/ √[ ( 6.67x10-11 Nm2/kg2 )( 5.98x1024 kg ) / (6.37 x 106 m )3 ] = 5058 s
which is 84 minutes and means that an object would take 42 minutes to fall through a tube drilled through the entire Earth. Such a tube can be drilled between, say, San Francisco and New York City and used to assist in travel. Does gravity help and then hinder movement along the tunnel connecting England to the Continent? Here is the entrance to an experimental transit tunnel dug through the Earth at a secret location.
As r goes from 0 to the radius R of the planet, the gravitational force on m’ grows linearly from 0 to Gmm’/R2 and then the gravitational force falls off as 1/r2.
Astrology and the gravitational force
What is the gravitational force of Saturn on you? Will this force affect you? If so then does every building you pass have a force on you? Are these forces included in astrology? You might conduct a calculation to show that the gravitational force between mother and newborn is more than the force between the baby and "astrologically significant" Saturn. How does the position of Saturn affect newborn?
Every 26,000 years, the Earth's axis precesses around a complete circle. As this occurs, the skies are rotating through 12 astrological symbols. We have a rotation rate of 26,000 years / 12 signs = 2000 years per astrological sign. In the 4,000 years or so since these constellational months were defined and named, the Earth has precessed by one astrological sign. For example, those persons born between March 20 and April 20 are said to have been born in the constellation of "Aries." If you were born 4,000 years ago on March 27, then the sun's position at noon on your birthday was indeed within the constellation of Aries. But if you were born on March 27 during some year within the nineteenth century then the sun's position at noon on your birthday was not in Aries but actually in Taurus. If you think that your personality has been that of an Aries, you might be surprised to learn that you are actually a Taurus. Since astrology and its calculations are even older than our realization (by Copernicus in the year 1543 ad) that the Earth and the other planets orbit around the Sun, its calculations are done as if the planets and the Sun orbit around the Earth.
On the family farm, our grandparents planted, harvested, smoked, pickled, and hogged at the astrologically correct time. When we started working in the factories we no longer had lee way in timing our activities. Astrology is less guiding today than it was 150 years ago, but it is still not as ignored as is warranted. Our species has tried for 100,000 years to observe the motion of birds, stars, smoke, and tea leaves to predict the future, but it has not yet worked once. If the wind blows to the right, will politics then move to the right?
Paul Hewitt suggests that we consider pulling on a spherical blob of jello. The jello will remain spherical if we pull equally hard on every part of it. If we instead pull more on its right side than on its left, it will elongate into an oval shape. The gravitational force of the Moon is 6.7% greater on objects located on the near side of Earth than it is for objects located on the side farther away. This difference in force is the cause of tides. The gravitational force of the Sun diminishes by 3% when comparing objects on the near and far sides of the Earth. Just as happened to the jello, the Earth's oceans are elongated into an oval shape that follows beneath the Moon in its 28-day orbit around the Earth. (In fact, the ground is also pulled upward several centimeters as the Moon moves overhead.) The Earth spins once every 24 hours beneath its oval-shaped oceans, causing two high tides and two low tides per day. As the Moon travels around the Earth, the tides move in time by 24/28 = 24-hour earth rotation period / 28-day moon orbital period = 0.857 hours per day = 51 minutes per day. There are no tides in a lake because no part of it is farther away from the moon or sun than is any other part. By the way, the gravitational pull of the half moon is identical to the pull of the full moon.
If the moon had oceans would there be tides moving around the moon? No. The moon does not rotate under the Earth–it keeps one face toward it. The ocean would bulge but the bulge would remain in nearly in one spot. The Moon’s position wiggles a bit such that, over time, we see about 58% of its surface.
As a child in a bathtub, you may have enjoyed sliding back and forth at just the right frequency to cause the bath water to slosh back and forth as a single unit. It takes a certain amount of time for the water to move within the tub. In the same way, a certain amount of time is needed for tidal water to flow into or out of the long, narrow Bay of Fundy in Nova Scotia. The Moon appears over this bay at just the right time to tug on the water and produce 15-meter tides. For photos of the tidal heights, click here.
When a comet, moon, or planet gets too close–within the so-called Roche Limit–to its parent, it is vertically torn apart because of the difference in gravitational pulls experienced by its near and far sides. (Here is a video of a comet colliding with the Sun.) As an object nears a Black Hole, it will be similarly stretched vertically.
The tidal force of Saturn on its moon Titan is about 4000 times larger than the tidal forces of the Moon on the Earth’s ocean. This force drives sufficient wind currents to push up 100-meter high sand dunes.
The Earth has mass ME and radius R. The work done to raise a mass m from the surface of the Earth to a point r > R is found by integrating from R to r:
ΔUg( r ) = - ∫F · dr.
For a mass m located at r > R, the gravitational force of the Earth points back toward the origin and is F = - GmME / r2. The change in gravitational potential energy is then
ΔUg( r ) = GmME ∫dr/r2 = GmME [ -1/r | = GmME ( -1/r - -1/R )
where integration is done from an initial r = R to a final r > R. We get
ΔUg( r ) = Ufinal - Uinitial = U( r ) - U( R ) = GmME ( 1/R - 1/r ).
If we define the zero of potential to be at the surface of the Earth, then U( r = R ) = 0, and
U( r > R ) = GmME ( 1/R - 1/r ).
Notice that an object moving in a circular orbit experiences a zero change in potential energy. Gravity does no work on an object unless r changes. The total work done depends only on the change in r. The integral is independent of the path since there is no contribution to the integral whenever the path is moving along an arc of constant radius.
The total energy is
E = K + U = ½ mv2 + GmME( 1/R - 1/r ).
You might sometime toss an object up to someone who is at the top of a ladder. We try to toss the object such that it barely reaches their height, allowing them to simply reach out and grab the tossed object. We try not to throw the object against their chest or throw it with too little initial speed to reach them.
An object on the surface of the Earth can be tossed upward with sufficient speed vesc such that it leaves the gravitational field of the Earth, escaping to r = infinity. As the object rises, its speed decreases. (The object does not have an engine. The initial throw sends it upward.) The initial potential energy is zero at the Earth’s surface and the kinetic energy is zero when infinitely far from the Earth. The total energy of the object there and at infinity is
E = K + U = ½ mvesc2 + 0 = 0 + GmME( 1/R - 1/∞ ).
Since 1/∞ = 0, we have
vesc = √( 2GME / RE ).
For the Earth, we have vesc = 11.2 km/s = 7 miles per sec = 25,000 mph. By symmetry, an object dropped from infinity will have this speed when it reaches the surface of the Earth. To throw an object out of the solar system requires an initial speed of vesc = 42.5 km/s (at the Earth's orbit) to leave the Sun's gravitational pull.
Compare the escape velocity for a planet having 4 times the mass and twice the radius of the Earth. The ratio of escape speeds for the planet and the Earth is given by
vp / vE = √( 2GMp / Rp ) / √( 2GME / RE ) = √[ ( Mp / ME ) ( RE / Rp ) ] = √( 2 / 4 ) = √2.
The circumference of the Earth is 24,000 miles and it takes 24 hours for the Earth to spin once on its own axis. This means that the Earth's equator moves at a speed of 1000 mph, but the rotational speed of the pole is zero. To take advantage of the Earth’s rotational speed, we launch space ships from as close to the equator as possible. The latitude of Cape Canavaral is 28°, French Guiniea is at 5°, Vandenburg Base is at 34°, and Kaputsin Russia is at 48°. The latitude of Hawaii of 20° is less than that of Cape Canavaral but it is not as near to industry.
NASA is studying the feasability of a mass launcher that could throw objects from the surface of the Earth all the way to the moon. The thrown material would be used to build a moon base. The launcher consists of a track lined with magnets that accelerate the mass.
If a mass enters a solar system, approaching with a certain speed. The trajectory of the mass will bend as it passes the sun(s). If the approaching speed is too high then the mass passes the sun and keeps on moving away. A hyperbolic orbit results.
It is nearly always more convenient to instead set the zero of potential energy at r = ∞. This means that the potential is zero at an infinite distance and becomes increasingly negative for decreasing r. This is similar to defining the zero of potential at the ceiling of a room and having U = -mgy become increasingly negative toward the ground. In this case, an object that moves from ∞ to a distance r, has a change in gravitational potential energy of
ΔUg( r ) = Ufinal - Uinitial = U( r ) - U( ∞ ) = GmME ( 1/r - 1/∞ ) = -GmME/r.
We then have
U( r ) = -GmME/r,
and total energy
E( r ) = K + U( r ) = ½ mv2 - GmME/r
If the total energy is greater than zero, E > 0, then the object will escape the Earth with a nonzero final speed. If E < 0, then the object will never reach r = ∞ and so will be forever bound to the Earth.
When an object moves in a circular orbit of radius r, the gravitational force is supplying the centripetal force.
GmME/r2 = mv2/r,
½ mv2 = ½ GmME/r.
This makes the total energy of this object is equal to half its potential energy.
E = K + U = ½ GmME/r - GmME/r = - ½ GmME/r.
The gravitational self-energy of a spherical mass is the energy needed to disassemble it by moving each mass element away to infinity. The mass m of a sphere of radius r and density ρ is m = 4/3 π ρ r3 and the mass of a differential, spherical shell is dm = 4 π ρ r2 dr. Taking U( r = ∞ ) = 0, the gravitational potential energy of dm with respect to the entire sphere is dU = - m dm / r because the shell is attracted to all of the mass enclosed within the shell–as if it were a point particle having all the mass. The total energy is found by integrating dU from r = 0 to r = R.
U = - ∫m dm / r = - ∫( 4/3 π ρ r3 )( 4 π ρ r2 dr ) / r = - 16/3 π2 ρ2 G ∫ r4 dr = - 16/15 π2 ρ2 G R5.
Since ρ = M / V, this is
U = -3 G M2 / ( 5R ).
The Ancient Greeks discovered the properties of the circle and decided that circles represent perfection in nature. The Greeks saw that the daily and annual motion of the stars followed along perfect circles. The planets are those objects whose path "wanders" among the fixed pattern of stars. Ptolemy (he died in 180 ad) tried to fit circles within circles (epicyles) to match the motion of planets. (The Greeks also said that mathematics was something new in that the wind and thunderstorm was explained only in terms of gods but there was no god of geometry. )
In the year 1543, Copernicus (1473 - 1543) proposed that the planets orbit the Sun rather than the Earth and he explained that the daily motion of the stars is due to the 24-hour rotation of the Earth about its own axis. He explained the retrograde motion of the inner planets. The retrograde motion of inner planets is similar to what occurs when riding in a car rounding a curve and looking at the landscape beyond the posts on the side of the roadway.
Tyco Brahe (1546 - 1601) made nightly measurements of the positions of stars and planets by carefully pointing “sticks” at them and then measuring the angles between the stick and the ground. He worked in an observatory on Uraniborg island off Copenhagen. When his nose was shot off in a duel, he replaced it with a mesh of gold and silver. His staff enjoyed daily feasts, wine, and song all paid for by the king.
Kepler (1571 - 1630) used Tycho Brahe’s measurements to deduced the laws of planetary motion. He concluded that the orbit of Mars follows an ellipse that differs from a circle by 7%. (Tycho’s measurements were accurate to 3%.) He also deduced that orbits sweep out equal areas in equal times. This is a consequence of the conservation of angular momentum.
Galileo was the first person to understand the meaning of acceleration and to distinguish acceleration from speed. He developed the science of mechanics and died in the same year (1642) that Isaac Newton was born.
By building on the understandings obtained by Galileo, Newton deduced that it takes a force to change the velocity of an object. Before Newton realized otherwise, the celestial objects were believed to be unlike Earthly materials. Nobody imagined that celestial objects were comprised of the vary same matter as we encounter on Earth or that the celestial objects obeyed the same laws of nature as does matter here on Earth. Newton realized that the same force that pulls an apple toward the Earth also pulls the Moon toward the Earth. In the year 1687, people were surprised that an equation could describe the motions of heavenly bodies. (Patrick Moore explains this history in Watchers of the Stars.) From the observed similar motions of distant galaxies we know that the gravitational force is the same in the most distant reaches of the universe. We find that F = Gm1m2/r2 for people, for apples on the earth, and for the moon, solar system, galaxies, clusters of galaxies, and the matter comprising the universe. We now understand that all of these motions are explained by writing Newton’s equation for the case of the gravitational force.
When standing on the surface of the Earth and looking straight out at the horizon, how far away is the horizon? Draw a line from the center of the Earth of radius r to the location of a person’s eyeball that is a height h above the ground. The eyeball is a distance r+h from the center of the Earth. Continue the line a distance d toward the horizon by drawing a line tangential to the surface of the Earth. From this intersection, continue the line back to the center of the Earth. The sides of this triangle are related by the Pythagorean theorem. We have (r+h)2 = r2 + d2. For a person whose eyes are 5 feet above the ground, the horizon is about 5 miles away.
Suppose we shoot a cannon parallel to the ground. Since the Earth's surface is curved, the surface sort of drops away from a tangential by 4.9 (about 5) vertical meters for every 8000 horizontal meters. We saw that due to gravity (y = ½ gt2), an object falls about 5 meters in 1 second. For the trajectory of the fired canon ball to match the Earth's curvature, which drops one meter for each 8000 meters along the surface, the cannonball must have a speed of 8000 m/s. If the cannon is launched horizontally at a speed of 8000 meters per second then its trajectory will be a curved path that sort of falls around the Earth. The canon ball will then be a satellite. This is the same way that the Moon falls around the Earth and that the Earth falls around the Sun. Inertia wants straight line motion but gravity keeps the changing the direction of the satellite’s motion. Paul Hewitt explains that if an elevated roadway is built that circles the Earth, then a car on this ramp can pass over roadway gaps if the car moves at 8000 m/s.
Consider the circle that represents the path of the Moon about the Earth. How far does the Moon fall from a tangent in one second: Newton calculated 1/20 inch, or 1.4 mm. We obtain this result in the following manner. The Earth's radius is 6.38 x 106 m and the distance from the Earth to the moon is 3.84x108 meters, which is 60 times the radius of the Earth. (The Apollo astronauts left a mirror on the surface of the Moon so that the Earth-Moon distance could be independently measured in terms of the reflection time needed for a laser beam to travel from the Earth to the Moon and back.) The surface of the Earth is a distance r away from the center of the Earth, but the Moon is a distance of 60r from the center of the Earth. Above the surface of the Earth, the acceleration due to gravity, g, decreases as the square of the distance. We have
(Earth's radius)2 / (moon's distance)2 = 1/602 =
g at the moon's orbital radius) / g at the Earth's surface =
(moon's drop) / (Earth's drop of 4.9m)
so that the moon's drop = Earth's drop / 602 = 0.0014 m = 14 mm
Newton realized that if an object's direction of motion changes then there is a force acting on it, no matter if that object is an apple or the moon. Who'd ever imagine that the moon and planets are just other objects and not something special or magical. Newton was the first person to understand the role of force in nature. (Just as explorers want to walk where nobody has walked before, scientists want to know and understand what nobody has known or understood before. Just as engineers hope to make buildings and roads that last for decades or centuries, scientists hope to build understandings that might be useful for decades or even centuries.)
If a projectile is tangentially launched from the Earth with a speed of 8000 m/s then it falls around the Earth in a circular orbit having constant speed. The force of gravity points straight toward the center of the Earth. This force is perpendicular to the motion. No work is done by gravity because its perpendicular. If projectile is launched from the Earth with a speed greater than 8000 m/s then it instead follows an elliptical orbit. Now F points toward Earth's center and there are differing angles between the F and v directions. As the mass is moving outward it is slowing and it is speeding up when moving inward.
What is the speed of the moon as it orbits the Earth?
v = distance / time = 2πr/period
= 2π(384,000,000m) / [ (27.3 days)(24 hr/day)(3600 s/hr)]
Since gravity supplies the centripetal force, GMmMe /d2 = ½ Mmv2, we also have
v = √( GMe / d ) = √( 6.67x10-11 Nm2/kg2 x 5.97 x 1024 / 3.84x108 m)=1018 m/s,
which is the same result.
Considering the perigee and apogee points in an orbit, at which point is the force of gravity largest and smallest? How about v, kinetic energy, potential energy, total energy, and angular moment? Both the angular momentum (L = mvr) and the total energy are constant throughout the motion. The potential energy is greatest at the apogee but the speed and kinetic energy are greatest at the perigee.
Rockets are used to put satellites into orbit around the Earth. Sputnik was the first artificial satellite that we humans placed into space. It emitted a signal as it orbited the Earth.
Dr. Van Allen once talked at our school about the beginnings of rocket research at White Sands Missile Range in New Mexico. He explained that early designs resulted in rocket failures that once caused a rocket to crash near Juarez, Mexico. To avoid this problem, the engineers placed tall, vertical wires on either side of the launch pad. One engineer visually gauged whether or not the rocket was moving in a purely vertical direction between the guide wires. If not, the engineer pressed a button to detonate the rocket that would have otherwise strayed.
There are currently about 600 satellites in orbit around the Earth. (Click here for tools to model satellite motion) They perform a variety of functions and make continual measurements of many aspects of the Earth, including weather, TV, telephone, monitoring plant types, GPS, and ocean levels and bottoms, see here and here (sorry, no Atlantis), and make astronomical observations with the Hubble Space Telescope. Satellites measure the height of crops and send the information to the farmer’s computer controlled tractors and implements so that each square meter of farm land can receive individualized care. Satellites measure the speed of continental drift, finding a speed of 2.5 cm/year, which is about the speed with which fingernails grow.
No satellite orbit is 100% stable: instead, they need occasional adjustments to keep them in the right place. The orbit of a satellite is sensitive to variations in the gravitational field of the Earth. The Earth is slightly pear-shaped with its smaller, more pointed end located in the northern hemisphere. The pair of satellites comprising the GRACE project are sensitive enough to detect the evolution of the gravitational field due to the climate-driven movement of water.
The Earth spins in a 24-hour period about its own axis while satellites move around the planet. Since the Earth’s surface moves 1500 miles during a satellite's 90 minute period, the location of a satellite is seen to in the figure is displaced 1500 miles.
Einstein's General Relativity
In the year 1915, Einstein's published his paper on General Relativity as a set of 29 coupled, non-linear partial differential equations describing the curvature of space. Each x, y, z, ct point of spacetime has a slope in each of the x, y, z and ct directions, giving 3x3x3+2 = 29 partial differential equations. Einstein figured this out by considering the equivalence between the acceleration of a mass and the free fall motion of a mass.
Einstein showed that mass curves space and that space tells mass how to move. The greater the mass that is concentrated in one place, the greater the curvature. Mass moves by falling along within the curved fabric of space. At the moment of the Big Bang, all the mass of the universe was at a single point and all of the space of the universe was in that same, single point. The Big Bang consisted of all the matter, all the space, all the energy, and all the mass of the universe existing together in one point–all of these then expanded outward together. (By the way, this means that the energy comprising your body was part of the Big Bang.) The universe is not expanding outward into previously existing, empty space; instead, space is expanding outward with the mass of the universe. Space and mass are unfolding together as they expand outward. Space does not exist beyond the edge of the expanding universe because space has not yet got there.
The universe might consist of countless Big Bangs on the edges of other Big Bangs like bubbles on bubbles, or there may be endless Big Bang and Big Crush (Big Gnab) cycles. Recently, astronomers likely ruled out the scenario of cycles when they determined that the universe is expanding outward at an increasing rate. Physicists are now searching for the force that is causing the repulsion. (By the way, one physicist says that people should move all the mass in the universe into a flat, rotating pancake so that it will forever remain stable.)
Einstein's equations predict that a light beam’s trajectory will bend as it passes near a mass because light also travels within space. Physicists have measured the bending of light and of spaceship signals passing near the edge of our sun.
The orbit of Mercury precesses around the sun in a manner explained by Einstein’s theory of Gravity but not by Newton’s gravitational equation. Einstein's equations predict that Mercury's orbital plane will rotate by 1/5 arc second per century and measurements match. In the year 1918, a measured bending through 4 arc seconds confirmed the theory. (More recently, the equipment was reused to repeat the experiment and the scientists found that the results and experimental error of the device was 4 + 8 (!) arc seconds, making the earlier confirmation more of a stroke of luck.) One pair of binary stars orbit each other once every 8 hours and have a precession of four degrees per year, as Einstein says should be found. Sometimes this precession can be used to measure the total mass of the star system.
The gravitational force between a star and a passing mass or “visitor” varies as the inverse square of the distance between them. The maximum size of this force occurs when the visitor is as close as possible to the star, as would occur if it were possible to land on the surface of the star. For two stars of equal radius, the smaller star has a larger maximum force because the visitor can get closer to the star. The closer the visitor gets, the greater will be the gravitational force. When the visitor is equally distant from two stars of equal mass but differing radius, then the visitor feels equal gravitational forces to both of them. Since the visitor can get closer to the star of smaller radius, the maximum gravitational force is greater for the star of smaller radius. The radius of the Sun is about 109 meters, a neutron star has a radius of 104 meters, and a Black Hole has a radius of essentially zero. At a distance of about 100 kilometers, even light can not escape from a Black Hole having a mass 30 times as great as the Sun.
If you have mass m and are a distance r away from a star of mass M, you feel a gravitational force F = GmM/r2. If that star becomes a black hole and you are the same distance r away then you'll still feel the same F = GmM/r2 as before. The reduced radius of the collapsed star means that a person can get much close to the edge of that star. You can not get closer than 109 meters from the Sun without running into its outer edge. But you can get just a few meters away from a Black Hole, and if you do then the gravitational force becomes humongous. You could not throw anything upward and out of it, not even light. A visitor might be approaching feet first when getting near the Black Hole. Since the head and feet of the visitor will be different distances from the Black Hole, those body parts will feel differing gravitational forces. The difference in force is so great that the body of the visitor will be stretched and torn vertically.
Only a few solvable situations have been found for the equations of general relativity. Some solutions show that time travel is possible between two different space-time points but mathematical boundary conditions always seem to prohibit the trajectories. A wormhole connects two space-time points in the manner that occurs when folding a sheet and then drilling a hole between the adjacent layers. But to produce a worm hole requires the same energy per second as contained within an entire star. Would you build a machine whose operation consumed one star per second. The equations show that if the worm hole was maintained for a period of time, say 100 years, then every person who ever crossed through would be found in there together even though they entered at times separated by 100 years. Strange and fun stuff.
We don't yet have any measurements at all concerning gravity on an atomic scale. It might take us 10 years–or 500–before we can. There is no telling what sorts of machines will result from being able to manipulate the vary fabric of space-time. (The "star-trek" propulsion engines were imagined by these researchers.)
Many physicists hope to unite gravity with the four other fundamental forces, resulting in a single equation that describes the interactions of all the particles of the universe in a “Theory of everything.” This is called the Grand Unified Theory. One such attempt is given by string theory.
An object A feels a force toward another mass B. If the position of mass B is changing, then the direction of the force on A is also changing. If mass B is shifted upward to a new location then the direction of the force on A shifts also. If mass B is wiggled back and forth then the direction of the force on A also wiggles back and forth: "A wiggling mass sends out gravitational waves."
If a gravity wave passed through a hoop-shaped mass then the hoop will be alternatingly stretched front-to-back and then left-to-right. In the 1960s, the physicist Weber tried to detect these distortions in an object. The oscillating distortion causes the cylinder's length to stretch by a fraction of a proton's diameter, which is 10-10 meters. Today's measurements are still too coarse to detect the distortions, but sometime soon, we will successfully make gravitational wave detectors to measure motions in the universe. Could we also communicate with such a device?
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