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"[T]he greater the velocity with which [a stone] is projected, the farther it goes before it falls to the earth. We may therefore
suppose the velocity to be so increased, that it would describe an arc of 1, 2, 5, 10, 100, 1,000 miles before it arrived
at earth, till at last, exceeding the limits of the earth, it should pass into space without touching." So wrote a scientist
describing how a projectile could move in orbit around the Earth, or even escape the Earth’s gravity.
Today, satellites that orbit the Earth are a commonplace (but still wondrous) reality. However, the quote above came from
a day when the human launching of a satellite was only dreamed of by a few. It was written by Isaac Newton, in 1687. He theorized
that celestial bodies in orbit, such as the Moon, are high-speed projectiles, moving fast enough that they never fall back
to the surface of the body they are orbiting. Some 270 years later, his theory was validated by the launch of the first artificial
satellite, Sputnik.
There are two elements required for an orbit. One is velocity − Newton conceived of a hypothetical cannon, able to supply a high initial velocity to an object. The other is an attractive,
centrally-directed force such as gravity: If there were no gravity, the projectile would continue in a straight line out into
space, even if fired at only 0.1 m/s. (We will neglect the effects of air resistance throughout this lab, which in the airlessness
of space, is an excellent approximation.) In this lab, starting with the work of Newton, you will see how these two elements
combine to create the conditions necessary for orbits.
You will learn about or observe:
· What is meant by “an orbit"
· How projectile motion relates to orbits
· The relationship between satellite speed and the size of an orbit
· The role of mass, both of the satellite and of the body being orbited, in determining orbital parameters
· How velocity and the force of gravity define orbital motion
· Kepler's law of periods
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Imagine it is the year 1687, and you are Sir Isaac Newton, staring up at the heavens. You see the Moon passing by overhead
night after night. Many scientists of your era think that planetary motion is unrelated to the types of motion that are observed
on the Earth, and is governed by different rules.
However, perhaps your knowledge of the motion of Earth-bound objects can help you understand the motion of the Moon. You know
the faster you throw a stone, the farther it travels before it hits the ground. You also know the Earth is round. If you could
make the stone move very fast − for instance, by firing it out of an extremely powerful cannon − would there be a velocity at which the stone would never hit the ground?
Can the stone be fired fast enough that instead of falling back to the Earth, it will circle around the planet, passing overhead
again and again like the Moon? It will still fall toward the Earth, but its great speed and the Earth’s curvature means it
will never land. Such a path, when one body moves around another body due to its gravitational influence, is called an orbit.
The “moving” body is called a satellite.
You can perform this experiment in the simulation pictured below, using what is today called Newton's Cannon.
Exercise 1. Fire Newton’s cannon.
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In the exercise you will fire Newton’s hypothetical cannon. The cannon launches the cannonball horizontally, from just above
the Earth’s surface. Like Newton, we ignore any mountains and such that might impede the cannonball’s progress, as well as
air resistance. These factors will not be relevant when you study orbiting bodies like the Moon. For the first exercise, we
also treat the Earth as stationary, as he did.
If the cannonball does not travel far, then the effect of Earth’s curvature is minor, and you can treat the Earth as essentially
flat. This is likely how you treated the Earth when you studied projectile motion. At these lesser velocities, the vertical
acceleration due to the force of gravity causes the ball to soon collide with the ground. What do you think will happen if
you fire the cannon horizontally (or more properly, tangentially to the Earth’s surface) at a still relatively slow velocity of, say, 100 m/s? (This means the cannonball is moving at a fairly brisk 360 km/h.) Try it and see. You will
have to adjust the velocity a fair amount from its default value in the simulation. When you want to fire the cannon another
time, press RESET and then GO again. Pressing PAUSE will freeze the ball in flight, and a second press will resume the action.
Now, fire the cannonball much, much faster. Is it possible to fire the cannonball at a high enough initial velocity such that
the Earth’s surface “drops away” under the cannonball at a rate so that the projectile moves in endless circles around the
planet? Try to find a velocity at which the cannonball continues to circle the Earth without landing.
Since the cannon is essentially at ground level, the radius of the orbit will be the starting-center-to-center distance between
the cannonball and the Earth’s center. The average radius of the Earth is 6.38×106 meters.
We speed up time in this simulation. It would take Newton’s cannonball about 84 minutes to complete one orbit of the Earth
at that distance from its center. By using the radius stated above to find the path’s circumference, and dividing it by the
velocity you measure in the experiment, you can check that calculation.
When you are done, close the simulation window and return here to answer some follow-up questions.
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1. When you fire the cannon, why does the cannonball not just fly off in a straight line into space?
2. What is the approximate minimum initial velocity that is required to make the cannonball go all the way around the Earth?
3. As you increase the velocity above that minimum value, does the cannonball continue to travel in the same circular path
around the Earth, only faster? If not, can you describe the geometrical shape of its orbit? You will learn more about this
shape later…
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As we stated earlier, there are two requirements for an orbit: velocity, and gravity. You have explored the first of these,
and now we offer a brief refresher on the other. Newton stated that every particle in the universe attracts every other. He
further stated that the force is directly proportional to the product of their masses, and inversely proportional to the square
of the distance between them. In equation form, Newton’s law of gravitation says that the magnitude of the force between mass
M and mass m, whose centers are separated by a distance r, is
In this exercise, you will make measurements of the gravitational attraction between two spheres to verify this law. You will
also measure the gravitational constant, G.
Exercise 2. Measuring gravitational attraction.
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There are two masses in the simulation. One of them is the Earth, with a mass of 5.97×1024 kg. The other is a small spherical test mass, whose default mass is 10 kg, but whose mass is variable.
You may also vary the separation between their centers, either by clicking and dragging, or by using the up-down arrows. Clicking
and dragging lets you alter the distance quickly; the arrows offer precision control.
Note that as far as gravity is concerned, we are treating the objects as though all of their mass was concentrated at one
point. Newton showed using his shell theorem that for spherical objects, or objects like the Earth which can be approximated
as spheres, this is perfectly valid.
The attractive force between the masses is shown in the force readout. Each object exerts the same magnitude of force on the
other, by Newton’s third law. Return here to answer some follow-up questions, which you will want to read before starting
the experiment.
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4. For the attractive gravitational force between a given pair of masses, what does Newton’s law state will happen if the distance
between the two masses is doubled? Try this experiment, and enter your data in the table below. Do your results agree with
his law?
5. Suppose you place a 1.0 kg mass right at the surface of the Earth, and measure the magnitude of the attractive force. Look
at the mathematical form of Newton’s law of gravitation. What do you predict will happen if one of the two masses is doubled?
To check your hypothesis, measure the attractive forces for a 1.0 kg and a 2.0 kg mass at the surface of the Earth. Enter
your data in the table below. Do your results agree with Newton’s equation?
6. The two situations just described correspond to a 1.0 kg and a 2.0 kg mass, respectively, located at the surface of the
Earth. Explain why the resulting force values look familiar.
7. Now that you have confirmed the form of Newton’s law of gravity, use any of your data points to determine the gravitational
constant, G. Make sure to specify the units, too!
The value you calculated should be close to the value we use in the lab for G. The lab uses an approximation for consistency with typical 3-digit calculations.
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Newton described objects in orbit as projectiles that never hit the ground. In other words, he applied the rules of projectile
motion to an orbiting body. As you may recall, a projectile is defined as an object whose motion is solely defined by its
initial velocity and the acceleration due to gravity.
In the Newton’s cannon exercise, Earth’s gravity pulling the projectile toward the center of the planet provided the necessary
centripetal force that kept the cannonball in orbit. One of Newton’s many strokes of genius was recognizing that the same
holds true for the Moon. We will now walk you through a series of steps, applying concepts like centripetal acceleration,
Newton’s second law and Newton’s law of gravitation. You will use these steps to determine the speed required for a projectile
to achieve a circular orbit around a planet.
An object of mass m traveling at uniform speed in a circular path is continually changing direction, and so is always accelerating toward the
center of the circle. The magnitude of its centripetal acceleration is given by:
a = v2/r
Newton also stated that the net force on a body equals its mass times its acceleration. In this case, that means that F = mv2/r.
A planetary body of mass Mplanet exerts a gravitational force on the object:
F = GmMplanet/r2
where
G = the gravitational constant, 6.67×10−11, in m3/kg·s2
m = mass of the satellite, in kg
r = center-to-center distance between satellite and body being orbited, in meters
For that object (which we’ll call a satellite) of mass m circling a planet under the influence of its gravitational force, the attractive force must equal mv2/r. We set these two statements for force equal to one another.
Next we will solve for v. First, the satellite mass cancels from both sides of the equation. The result is:
v2r = GMplanet
This equation was derived assuming a circular orbit, but it lays the foundation for proving Kepler's Third Law, which relates
orbital period to a measure of the orbital size.
To solve for v, re-arrange the equation and take the square root. As promised, here is an expression for the speed required for a satellite
to achieve a circular orbit around a planet:
In this exercise, we will be carrying our analysis a step further − and imagining scenarios which might make a good late-night science fiction feature. Rest assured though, the laws of physics
used for these orbiting bodies are the same as in reality.
Exercise 3. Can you restore the Moon’s orbit to save the Earth?
Imagine that some unknown force or device has stopped the progress of the Moon around the Earth. Without your intervention,
the Moon will plunge into the Earth causing massive devastation, chaos, and the cancellation of final exams. Can you restart
the Moon at the right tangential velocity to restore its orbit before disaster strikes?
In essence, you are using Newton’s cannon again, but on a planetary scale. You are treating the Moon as a projectile, and
determining what velocity it must have to maintain a circular orbit. Newton’s thought experiment proves to be absolutely relevant
here.
(Note: To make the Earth and Moon more visible in the simulation, we have increased their scale by a factor of 10 relative
to the size of the orbit, and again, we have sped up time.)
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In this exercise you rescue the Moon by putting it back into its usual near-circular orbit. The average orbital radius of
the Moon around the Earth is 3.84×108 meters.
Make a hypothesis before starting the simulation. At this radius, do you think the speed will have to be more or less than
it was for the smaller-radius orbit of the cannonball? You may want to consider some of the analysis conducted above.
Use the simulation to test your hypothesis. When the simulation begins, the Moon has just stopped. You can calculate, using
the equations derived above, the correct velocity for the Moon and set it to the nearest 100 m/s using the up-down arrows.
Press GO to restart the Moon’s movement and restore it to the correct orbit. If you need to try again, press the RESET button
to start over.
You are seeing a crucial relationship: the relationship between the speed of an object in orbit around a particular body,
and the radius of that orbit.
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8. What was the tangential speed required for the Moon? Was this faster than the cannonball in its lower orbit, or was it slower?
9. Use Newton’s law of gravitation to find the gravitational force exerted by the Earth on the Moon (m = 7.35×1022 kg) when their centers are 3.84×108 m apart. Also, using the proper orbital velocity that you determined for the simulation, calculate the centripetal force
required to keep the Moon moving at its proper speed in a circular path of that radius. How are these two values related?
10. Using your speed and the known radius of the orbit, determine the approximate period of the orbit, the time it takes the
Moon to travel once around the Earth. Compare this to the period that appears on a calendar, 28 days.
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Your insights are not limited to objects orbiting the Earth. In this exercise, you will investigate natural satellites of
Mars. The two moons of Mars, Deimos and Phobos, are much smaller and closer to their planet than the Earth’s moon is to its
parent planet.
In this exercise Mars is shown to the same scale as the orbits, but its moons are shown 50 times larger than scale to make
them more visible.
Exercise 4. Newton’s cannon on Mars.
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Imagine that Deimos, the smaller of the two moons of Mars, has been knocked from its orbit and fallen to the planet’s surface.
Phobos maintains its orbit undisturbed. The mass of Mars is 6.42×1023kg, and the proper orbital speed for Deimos is 1350 m/s. You can use these facts and the equations you have learned to determine
the proper distance to restore Deimos to a circular orbit.
Enter the orbital radius you calculate (to three significant digits) in the simulation and press GO. The simulation will place
Deimos at your desired distance, and the Martian moon will move at 1350 m/s. If the moon leaves the planet’s surface and commences
a circular orbit, you have succeeded. The fate of Mars is in your hands; if your answer is too large or too small, an orbital
speed of 1350 m/s will probably result in an orbit shaped like an ellipse, and if you are too far from the correct value,
the residents of the planet may need to duck.
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11. Deimos’s orbital speed is 1350 m/s. What must the radius of its orbit be for it to be circular?
12. How long does it take for Deimos to travel once around Mars? You can assume that the orbit is circular for your calculations.
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The first set of exercises has focused on orbital velocity and radius. Another important concept is orbital period. This is a measure of how long it takes for a satellite to make one complete trip around its orbital path. We use earlier
equations derived in this lab and basic tenets of physics to derive an expression for orbital period.
Earlier we derived the relationship
The period, T, is equal to the circumference of the orbit divided by the orbital speed. (This is dividing distance by speed to determine
the elapsed time.) We substitute the expression for speed above.
Using algebra, we can restate the equation above for Kepler's third law as it pertains to circular orbits:
where
T = period, in seconds
G = the gravitational constant 6.67×10−11, in m3/kg·s2
M = mass of body being orbited, in kg (assumed to be much greater than satellite’s mass)
r = center-to-center distance between satellite and body being orbited, in meters.
The 17th century astronomer and astrologer Johannes Kepler calculated that the square of the period is proportional to the cube of
the orbital radius, though his formulation was more general, as it also applied to elliptical orbits.
Implicit in Kepler's third law is a concept you have already experienced in this lab: A satellite's velocity and orbital radius
are linked. A satellite cannot simply "go faster" or "go slower" and stay in the same orbit.
Kepler's third law is useful in launching geosynchronous satellites − so named because the satellite is always “parked” over the same point on the Earth, such as over a point in the Atlantic
Ocean, relaying signals between Europe and North America. Geosynchronous satellites have a period of 24 hours, the same as
our planet does.
There are other reasons one might want to orbit an object over a particular point on the Earth’s or another planet’s surface.
Scientists have proposed designs for “space elevators” which would consist of a cable of carbon nanotubes, stretching from
the surface of the planet to a point far above. It could be possible to lift objects up this elevator with much greater efficiency
than is possible using chemical rockets. Many space elevator designs, such as the one we will consider in this exercise, rely
on a large mass at a height that keeps the elevator in a “stationary” orbit above the planet’s surface in order to keep the
cable under moderate tension.
In this exercise, you will create an areosynchronous satellite to act as the top of a space elevator. The satellite will remain above a fixed point on Mars. (Ares is the Greek
name for Mars; geosynchronous only applies to objects in orbit around Earth.)
You want to place the satellite in areosynchronous orbit around the Red Planet, whose mass is 6.42×1023 kg. Mars completes a rotation on its axis in 24.6 hours, or 8.86×104 seconds.
Exercise 5. Launch the satellite that will “top” a new Martian space elevator.
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Use the controls in the simulation to enter the orbital radius and orbital speed that the satellite must have to achieve an
areosynchronous orbit (a green dot on Mars marks the rocket launch pad).
Pressing GO will launch the satellite to the orbital radius you entered, then it will blast off to the side at the orbital
speed that you set.
After you have achieved synchronous orbit, return here to answer a follow-up question.
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13. How far from the center of Mars must the satellite be so that it orbits with a period of 8.86×104 seconds?
14. In the desired orbit, what is the correct orbital speed?
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In earlier exercises we ignored the fact that the Earth rotates on its axis. Here you must factor that in: Space engineers
take into account the Earth's spin when launching satellites; in fact, they take advantage of it.
You have a very important mission in this section. A rogue communications satellite is circling the Earth, 38,370 kilometers
above the surface. The mad scientist who designed it has vowed that in a matter of minutes, it will take over all communications
channels and broadcast nothing but Barney® videos and music. This cannot be permitted to happen.
One method that space engineers have devised to destroy enemy satellites in a time of war is to launch a counter-satellite
− an object that enters the same orbit, but travels in the opposite direction. The “killer” satellite crashes into the rogue
satellite and destroys it.
Exercise 6. Take down the satellite before it begins broadcasting.
The goal in this exercise is to launch a satellite from the Earth that orbits in a clockwise (as seen from the South Pole)
circular orbit with a specified radius. You launch it from the equator − the latitude on the Earth's surface with the greatest tangential velocity (since the distance from the axis of rotation is
the greatest there) − in a direction so your ship gets the full benefit of the Earth's rotation. The Earth’s tangential velocity should be added
to the tangential velocity supplied by the rocket engine.
To start to succeed in this exercise, first determine the tangential velocity of a stationary (fixed with respect to the ground)
object at the equator. You can calculate this velocity using the period and the circumference of the Earth. It takes 24 hours
for any point on the Earth to complete a revolution.
15. What is the tangential velocity, in m/s, of a location at the Earth’s equator?
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Test your answer in the simulation. Your satellite is resting at the equator. Enter a value in the control panel for the initial
velocity supplied solely by the rocket engine, in m/s, needed to enter the circular orbit at a height of 38,370 kilometers above the surface. (Be careful: Satellite heights
are often given as a distance above the surface − you must add in the radius of the Earth to get the orbital radius.) The tangential velocity supplied by the Earth will be
added to whatever number you input. The sum of the two determines the speed that the engines of the rocket will maintain as
they lift the satellite to its final orbit.
When you are done, return here to answer some follow-up questions.
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16. With what tangential velocity must you launch a counter-satellite in order to crash into a rogue satellite that is orbiting
38,370 kilometers above the surface of the Earth (give the velocity relative to the Earth’s surface)?
17. Based on what you learned and saw in the simulation, explain why NASA rockets typically are launched a) in an easterly direction
and b) from one of the most southerly points in the continental U.S. (Cape Canaveral, Florida).
18. Could you have destroyed the rogue satellite by launching a satellite into a circular orbit in the same direction? How about
an elliptical orbit? Explain your answers.
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In this simulation, you are the pilot of an orbiting spacecraft, and your mission is to dock with a space station. As shown
in the diagram below, your ship is initially orbiting in a smaller orbit than the space station. Both the ship and the station begin the experiment in circular orbits.
Exercise 7. Match orbits with the space station.
In order to dock, your ship must be in the same orbit as the space station, and it must touch the space station. To dock,
your speed and radius must be very close to those of the space station − a high speed collision does not equate to docking!
May the (gravitational and rocket) Forces be with you! |
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To assist in your efforts, the current orbital paths for both ships are drawn on the screen. Your rocket’s orbit is drawn
in yellow, and the space station’s is drawn in white.
You have two controls, Forward thrust and Reverse thrust. Forward thrust fires an engine from the rear of your rocket. Reverse
fires a retro rocket directed in the opposite direction, from the front of your rocket. This simulation requires no direct
mathematical calculations, but some thought and experimentation are necessary. If you get too far off track, you may press
RESET and try again from the beginning.
When you are done, return here to answer a follow-up question.
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19. What happened to your orbital velocity immediately after you used the forward thrust button? Explain.
The results you observe can be explained by the relationship between orbital radius and speed. You may also consider it in
terms of the total mechanical energy of the spacecraft − the sum of its kinetic and potential energy − and we will provide a very terse explanation of how to consider the energy in light of an orbit. As the radius of the craft’s
orbit increases, its PE increases. For energy to be conserved in a given orbit, as the PE increases, KE (and speed) must decrease. The other factor in play is the relationship of PE and KE. Taking the case of circular orbits, for a given orbit, the magnitude of the gravitational PE is twice the KE. Exploring energy in orbits is the topic of another lab.
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