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504

Week 12: Gravity

Etot ,Ueff

 

 

2

 

 

L

 

 

____

 

 

2mr2

 

 

 

2

 

Ueff =

L

 

____

− GMm

 

2mr2

r

 

 

r

− GMm

 

 

r

 

 

Figure 149: A typical energy diagram illustrating the e ective potential energy, which is basically the sum of the radial potential energy and the angular kinetic energy of the orbiting object.

By drawing a constant total energy on this plot, the di erence between Etot and Ue (r) is the radial kinetic energy, which must be positive. We can determine lots of interesting things from this diagram.

~

In figure 150, we show orbits with a given fixed angular momentum L 6= 0 and four generic total energies Etot. These orbits have the following characteristics and names:

a)Etot > 0. This is a hyperbolic orbit.

b)Etot = 0. This is a parabolic orbit. This orbit defines escape velocity as we shall see later.

c)Etot < 0. This is generally an elliptical orbit (consistent with Kepler’s First Law).

d)Etot = Ue ,min. This is a circular orbit. This is a special case of an elliptical orbit, but deserves special mention.

Note well that all of the orbits are conic sections. This interesting geometric connection between 1/r2 forces and conic section orbits was a tremendous motivation for important mathematical work two or three hundred years ago.

12.7: Escape Velocity, Escape Energy

As we noted in the previous section, a particle has “escape energy” if and only if its total energy is greater than or equal to zero, provided that we set the zero of potential energy at infinity in the first place. We define the escape velocity (a misnomer!) of the particle as the minimum speed (!) that it must have to escape from its current gravitational field – typically that of a moon, or planet, or star. Thus:

Etot = 0 =

1

mvescape2

GM m

(1068)

 

 

2

r

so that

r

 

vescape = 2

GM

= p2gr

(1069)

r

Week 12: Gravity

505

E

, U

 

 

tot

eff

 

 

1

E

 

 

tot

 

 

 

 

 

 

Ek

rmin

 

2

E

 

 

tot

 

 

 

 

 

 

 

Ek

r

3

Etot

 

Ek < 0 (forbidden)

4

Etot

 

rmax

 

r0

 

 

 

 

Figure 150: A radial total energy diagram illustrating the four distinct named orbits in terms of their total energy: 1) is a hyperbolic orbit. 2) is a parabolic orbit. 3) is an elliptical orbit. 4) is a circular orbit. Note that all of these orbits are conic sections, and that the classical elliptic orbits have two radial turning points at the apogee and perigee along the major axis of the ellipse.

where in the last form g = GMr2 (the magnitude of the gravitational field – see next item).

To escape from the Earth’s surface, one needs to start with a speed of:

vescape = r

GM

 

2 RE E = p2gRE = 11.2 km/sec

(1070)

Note: Recall the form derived by equating Newton’s Law of Gravitation and mv2/r in an earlier section for the velocity of a mass m in a circular orbit around a larger mass M :

vcirc2 =

GM

(1071)

r

 

 

from which we see that vescape = 2vcirc.)

It is often interesting to contemplate this reasoning in reverse. If we drop a rock onto the earth from a state of rest “far away” (much farther than the radius of the earth, far enough away to be considered “infinity”), it will REACH the earth with escape (kinetic) energy and a total energy close to zero. Since the earth is likely to be much larger than the rock, it will undergo an inelastic collision and release nearly all its kinetic energy as heat. If the rock is small, this is not necessarily a problem. If it is large – say, 1 km and up – it releases a lot of energy.

Example 12.7.1: How to Cause an Extinction Event

How much energy? Time to do an estimate, and in the process become just a tiny bit scared of a very, very unlikely event that could conceivably cause the extinction of us.

Let’s take a “typical” rocky asteroid that might at any time decide to “drop in” for a one-way visit. While the asteroid might well have any shape – that of a potato, or pikachu228 , we’ll follow

228Wikipedia: http://www.wikipedia.org/wiki/Pikachu. If you don’t already know, don’t ask...

506 Week 12: Gravity

the usual lazy physicist route and assume that it is a simple spherical ball of rock with a radius r. In this case we can estimate its total mass as a function of its size as:

M =

4πρ

r3

(1072)

3

 

 

 

Of course, now we need to estimate its density, ρ. Here it helps to know two numbers: The density of water, or ice, is around 103 kg/m3 (a metric ton per cubic meter), and the specific gravity or rock is highly variable, but in the ballpark of 2 to 10 (depending on how much of what kinds of metals the rock might contain, for example), say around 5.

If we then let r ≈ 1000 meters (a bit over a mile in diameter), this works out to M ≈ 1.67 ×1012 kg, or around 2 billion metric tons of rock, about the mass of a small mountain.

This mass will land on earth with escape velocity, 11.2 km/sec, if it falls in “from rest” from far away. Or more, of course – it may have started with velocity and energy from some other source – this is pretty much a minimum. As an exercise, compute the number of Joules this collision would release to toast the dinosaurs – or us! As a further exercise, convert the answer to “tons of TNT” (a unit often used to describe nuclear-grade explosions – the original nuclear fission bombs had an explosive power of around 20,000 tons of TNT, and the largest nuclear fusion bombs built during the height of the cold war had an explosive power on the order of 1 to 15 million tons of TNT.

The conversion factor is 4.184 gigajoules per ton of TNT. You can easily do this by hand, although the internet now boasts of calculators that will do the entire conversion for you. I get ballpark of ten to the twentieth joules or 25 gigatons – that is billions of tons – of TNT. In contrast, wikipedia currently lists the combined explosive power of all of the world’s 30,000 or so extant nuclear weapons to be around 5 gigatons. The explosion of Tambora (see last chapter) was estimated to be around 1 gigaton. The asteroid that might have caused the K-T extinction event that ended the Cretaceous and wiped out the dinosaurs and created the 180 kilometer in diameter Chicxulub crater229 had a diameter estimated at around 10 km and would have released around 1000 times as much energy, between 25 and 100 teratons of TNT, the equivalent of some 25,000 Tambora’s happening all at once.

Such impacts are geologically rare, but obviously can have enormous e ects on the climate and environment. On a smaller scale, they are one very good reason to oppose the military exploitation of space – it is all too easy to attack any point on Earth by dropping rocks on it, where the asteroid belt could provide a virtually unlimited supply of rocks.

12.8: Bridging the Gap: Coulomb’s Law and Electrostatics

This concludes our treatment of basic mechanics. Gravitation is our first actual law of nature – a force or energy law that describes the way we think the Universe actually works at a fundamental level.

Gravity is, as we have seen, important in the sense that we live gravitationally bound to the outer surface of a planet that is itself gravitationally bound to a star that is gravitationally compressed at its core to the extent that thermonuclear fusion keeps the entire star white hot over billions of years, providing us with our primary source of usable energy. It is unimportant in the sense that it is very weak, the weakest of all of the known forces.

Next, in the second volume of this book, you will study one of the strongest of the forces, the one that dominates almost every aspect of your daily life. It is the force that binds atoms and molecules together, mediates chemistry, permits the exchange of energy we call light, and indeed is the fundamental source of nearly every of the “forces” we treated in this semester in collective form: The electromagnetic interaction.

229Wikipedia: http://www.wikipedia.org/wiki/Chicxulum Crater.

Week 12: Gravity

507

Just to whet your interest (and explain why we have spent so long on gravity when it is weak and mostly irrelevant outside of its near-Earth form in everyday a airs) let is take note of Coulomb’s Law, the force that governs the all-important electrostatic interaction that binds electrons to atomic nuclei to make atoms, and binds atoms together to make molecules. It is the force that exists between two charges, and can be written as:

~

 

ke q1

q2

 

 

F 12

=

r2

 

rˆ12

(1073)

 

 

12

 

 

 

Hmmm, this equation looks rather familiar! It is almost identical to Newton’s Law of Gravitation, only it seems to involve the charge (q) of the particles involved, not their mass, and an electrostatic constant ke instead of the gravitational constant G.

In fact, it is so similar that you instantly “know” lots of things about electrostatics from this one equation, plus your knowledge of gravitation. You will, for example, learn about the electrostatic field, the electrostatic potential energy and potential, you will analyze circular orbits, you will analyze trajectories of charged particles in uniform fields – all pretty much the same idea (and algebra, and calculus) as their gravitational counterparts.

The one really interesting thing you will learn in the first couple of weeks is how to properly describe the geometry of 1/r2 force laws and their underlying fields – a result called Gauss’s Law. This law and the other Maxwell Equations will turn out to govern nearly everything you experience. In some very fundamental sense, you are electromagnetism.

Good luck!

Homework for Week 12

Problem 1.

Physics Concepts: Make this week’s physics concepts summary as you work all of the problems in this week’s assignment. Be sure to cross-reference each concept in the summary to the problem(s) they were key to, and include concepts from previous weeks as necessary. Do the work carefully enough that you can (after it has been handed in and graded) punch it and add it to a three ring binder for review and study come finals!

Problem 2.

It is a horrible misconception that astronauts in orbit around the Earth are weightless, where weight (recall) is a measure of the actual gravitational force exerted on an object. Suppose you are in a space shuttle orbiting the Earth at a distance of two times the Earth’s radius (Re = 6.4 × 106 meters) from its center.

a)What is your weight relative to your weight on the Earth’s surface?

b)Does your weight depend on whether or not you are moving at a constant speed? Does it depend on whether or not you are accelerating?

c)Why would you feel weightless inside an orbiting shuttle?

d)Can you feel as “weightless” as an astronaut on the space shuttle (however briefly) in your own dorm room? How?

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Week 12: Gravity

Problem 3.

Physicists are working to understand “dark matter”, a phenomenological hypothesis invented to explain the fact that things such as the orbital periods around the centers of galaxies cannot be explained on the basis of estimates of Newton’s Law of Gravitation using the total visible matter in the galaxy (which works well for the mass we can see in planetary or stellar context). By adding mass we cannot see until the orbital rates are explained, Newton’s Law of Gravitation is preserved (and so are its general relativistic equivalents).

However, there are alternative hypotheses, one of which is that Newton’s Law of Gravitation is wrong, deviating from a 1/r2 force law at very large distances (but remaining a central force). The orbits produced by such a 1/rn force law (with n =6 2) would not be elliptical any more, and r3 =6 CT 2 – but would they still sweep out equal areas in equal times? Explain.

Week 12: Gravity

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Problem 4.

r

ω

M

R

This problem will help you learn required concepts such as:

Newton’s Law of Gravitation

Circular Orbits

Centripetal Acceleration

Kepler’s Laws

so please review them before you begin.

A straight, smooth (frictionless) transit tunnel is dug through a spherical asteroid of radius R and mass M that has been converted into Darth Vader’s death star. The tunnel is in the equatorial plane and passes through the center of the death star. The death star moves about in a hard vacuum, of course, and the tunnel is open so there are no drag forces acting on masses moving through it.

a)Find the force acting on a car of mass m a distance r < R from the center of the death star.

b)You are commanded to find the precise rotational frequency of the death star ω such that objects in the tunnel will orbit at that frequency and hence will appear to remain at rest relative to the tunnel at any point along it. That way Darth can Use the Dark Side to move himself along it almost without straining his midichlorians. In the meantime, he is reaching his crooked fingers towards you and you feel a choking sensation, so better start to work.

c)Which of Kepler’s laws does your orbit satisfy, and why?

510

Week 12: Gravity

Problem 5.

m

N

r0 = R

ρ

0

S

This problem will help you learn required concepts such as:

Newton’s Second Law.

Newton’s Law of Gravitation

Gravitational Field/Force Inside a Spherical Shell or Solid Sphere.

Harmonic Oscillation Given Linear Restoring Forces.

Definitions and Relations Involving ω and T .

so please review them before you begin.

A straight, smooth (frictionless) transit tunnel is dug through a planet of radius R whose mass density ρ0 is constant. The tunnel passes through the center of the planet and is lined up with its axis of rotation (so that the planet’s rotation is irrelevant to this problem). All the air is evacuated from the tunnel to eliminate drag forces.

a)Find the force acting on a car of mass m a distance r < R from the center of the planet.

b)Write Newton’s second law for the car, and extract the di erential equation of motion. From this find r(t) for the car, assuming that it starts at rest at r0 = R on the North Pole at time t = 0.

c)How long does it take the car to get to the South Pole starting from rest at the North Pole? How long does it take to get back to the North Pole? Compare this (second answer) to the period of a circular orbit inside the death star you found (disguised as ω) in the previous problem.

d)A final thought question: Suppose it is released at rest from an initial position r0 = R/2 (halfway to the center) instead of from r0. How long does it take for the mass to get back to this point now (compare the periods)?

All answers should be given in terms of G, ρ0, R and m.

Week 12: Gravity

511

Problem 6.

Ueff

E

3

r

E

2

E1

E0

The e ective radial potential of a planetary object of mass m in an orbit around a star of mass M is:

Ue (r) =

L2

GM m

 

 

2mr2

r

(a form you already explored in a previous homework problem). The total energy of four orbits are drawn as dashed lines on the figure above for some given value of L. Name the kind of orbit (circular, elliptical, parabolic, hyperbolic) each energy represents and mark its turning point(s).

512 Week 12: Gravity

Problem 7.

In a few lines prove Kepler’s third law for circular orbits around a planet or star of mass M : r3 = CT 2

and determine the constant C and then answer the following questions:

a)Jupiter has a mean radius of orbit around the sun equal to 5.2 times the radius of Earth’s orbit. How long does it take Jupiter to go around the sun (what is its orbital period or “year” TJ )?

b)Given the distance to the Moon of 3.84 × 108 meters and its (sidereal) orbital period of 27.3 days, find the mass of the Earth Me.

c)Using the mass you just evaluated and your knowledge of g on the surface, estimate the radius of the Earth Re.

Check your answers using google/wikipedia. Think for just one short moment how much of the physics you have learned this semester is verified by the correspondance. Remember, I don’t want you to believe anything I am teaching you because of my authority as a teacher but because it works.

Week 12: Gravity

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Problem 8.

It is very costly (in energy) to lift a payload from the surface of the earth into a circular orbit, but once you are there, it only costs you that same amount of energy again to get from that circular orbit to anywhere you like – if you are willing to wait a long time to get there. Science Fiction author Robert A. Heinlein succinctly stated this as: “By the time you are in orbit, you’re halfway to anywhere.”

Prove this by comparing the total energy of a mass:

a)On the ground. Neglect its kinetic energy due to the rotation of the Earth.

b)In a (very low) circular orbit with at radius R ≈ RE – assume that it is still more or less the same distance from the center of the Earth as it was when it was on the ground.

c)The orbit with minimal escape energy (that will arrive, at rest, “at infinity” after an infinite amount of time).

Problem 9.

y

 

 

M = 80m

d

 

 

 

D = 5d

m

x

 

The large mass above is the Earth, the smaller mass the Moon. Find the vector gravitational field acting on the spaceship on its way from Earth to Mars (swinging past the Moon at the instant drawn) in the picture above.

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Week 12: Gravity

Problem 10.

M a

Re

Me

This problem will help you learn required concepts such as:

Gravitational Energy

Fully Inelastic Collisions

so please review them before you begin.

A bitter day comes: a roughly spherical asteroid of radius Ra and density ρ is discovered that is falling in from far away so that it will strike the Earth. Ignore the gravity of the Sun in this problem. Determine:

a)If it strikes the Earth (an inelastic collision if there ever was one) how much energy will be liberated as heat? Express your answer in terms of Re and either g or G and Me as you prefer. It is probably very safe to say that Ma Me...

b)Chuck Norris lands on the surface of the asteroid to save the Earth, but instead of screwing around with drills and nuclear bombs Chuck jumps up from the surface of the asteroid at a speed of vcn to deliver a roundhouse kick that would surely break the asteroid in half and cause it to miss the earth – if it knows what’s good for it (this is Chuck Norris, after all). However, if you jump up too fast on an asteroid, you don’t come down again! Does Chuck ever fall down onto the asteroid after his jump?

c)Evaluate your answers to a-b above for the following data:

ρ

=

6 × 103 kilograms/meter3

Ra

=

104 meters

Re

=

6.4 × 106 meters

g= 10 meters/second2

Me

=

6 × 1024 kilograms

 

vcn

=

5 meters/second

(1074)

Express your answer to a) both in joules and in “tons” (of TNT) where 1 ton-of-TNT = 4.2 ×109 joules. Compare the answer to (say) 30 Gigatons as a safe upper bound for the total combined explosive power of every weapon (including all the nuclear weapons) on earth.

d)Find the size of an asteroid that (when it hits) liberates only the energy of a typical thermonuclear bomb, 1 megaton of TNT.

Week 12: Gravity

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Problem 11.

There is an old physics joke involving cows, and you will need to use its punchline to solve this problem.

A cow is standing in the middle of an open, flat field. A plumb bob with a mass of 1 kg is suspended via an unstretchable string 10 meters long so that it is hanging down roughly 2 meters away from the center of mass of the cow. Making any reasonable assumptions you like or need to, estimate the angle of deflection of the plumb bob from vertical due to the gravitational field of the cow.

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Week 12: Gravity

Optional Problems

The following problems are not required or to be handed in, but are provided to give you some extra things to work on or test yourself with after mastering the required problems and concepts above and to prepare for quizzes and exams.

Continue Studying for Finals using problems from the online review!