Universal Gravitation

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© 1998-2008 by Glenn Elert -- A Work in Progress
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the comet

Isaac Newton was born on Christmas Day, 1642 in the village of Woolsthorpe (near Grantham), Lincolnshire, England. In 1661 he enrolled in Trinity College, Cambridge University (about midway between Woolsthorpe and London) where he studied mathematics. In 1665 the Black Plague made it's way to England forcing the closure of Trinity and sending Newton home to Woolsthorpe for a year or two. It was during this time that he formulated most of his important contributions to mathematics and physics including the binomial theorem, differential calculus, vector addition, the laws of motion, centripetal acceleration, optics, and universal gravitation. Upon returning to Cambridge, Newton was made a professor of mathematics and then proceeded to do what professors still do to this day -- teach and publish. Most of these papers Newton submitted for publication were on optics, especially on the theory of colors. Then, eighteen years later in 1684, Edmund Halley (1656-1742) came to Newton with a problem he thought Newton might be able to solve.

Comets are astronomical objects that are visible to the unaided for only a month or so. They were a serious problem for early astronomers as they would appear without warning, hang around in the sky for awhile, and then disappear never to be seen again. Halley was studying historical records of cometary appearances when he noticed four comets with nearly the same orbit separated in time by approximately 76 years. He reasoned that the comets of 1456, 1531, 1607, and 1682 were sightings of a single comet and that this comet would reappear in the winter of 1758. When it did as predicted, sixteen years after his death, it became known as Comet Halley. It should be noted that Halley did not discover the comet that bears his name, he was just the one who identified it as a celestial body with a definite period in orbit around the sun. Halley's comet has probably been seen since the dawn of civilization when humans first looked up and they sky and wondered how it all worked. Historical records from India, China, and Japan record its appearance as far back as 240 BCE (with one appearance not recorded). Its most recent appearances were in 1833, 1909, and 1985 and its next will be in 2061.

Halley also noticed that the comet described an orbit around the sun that was in accordance with Kepler's laws of planetary motion; namely, that the orbit was an ellipse (albeit a highly elongated one) with the sun at one focus and that it obeyed the harmonic law (r3 ∝ T2) as if it was another planet in our solar system. Halley asked Newton in 1684 if he had some idea why the planets and this comet obeyed Kepler's laws; that is, if he knew the nature of the force responsible. Newton replied that he had indeed solved this problem and "much other matter" pertaining to mechanics eighteen years earlier but hadn't told anyone about it. He then proceeded to rummage around looking for his notes from the plague years, but could not find them. Halley persuaded Newton to compile everything he ever knew on mechanics and offered to pay the costs so that his ideas might be published.

In 1687, after eighteen months of effectively non stop work, Newton published Philosophiæ Naturalis Principia Mathematica (The Mathematical Principles of Natural Philosophy). Probably the single most important book in physics and possibly the greatest book in all of science, it is almost always just known as the Principia. It contains the essence of the concepts presented in the chapters on mechanics in every subsequent physics textbook, including this one. Probably the only important concept it misses is energy, but everything else is there: force, mass, acceleration, inertia, momentum, weight, vector addition, projectile motion, circular motion, satellite motion, gravitation, tidal forces, the precession of the equinoxes, ….

Halley:
What kind of curve would be described by the planets supposing the force of attraction towards the sun to be reciprocal to the square of their distance from it?

Newton:
An ellipse.

Halley:
But how do you know?

Newton:
I have calculated it.

In 1684 Dr Halley came to visit at Cambridge [and] after they had some time together the Dr asked him what he thought the curve would be that would be described by the Planets supposing the force of attraction toward the Sun to be reciprocal to the square of their distance from it. Sr Isaac replied immediately that it would be an Ellipsis. The Doctor, struck with joy & amazement, asked him how he knew it. ‘Why,' saith he, ‘I have calculated it,' whereupon Dr Halley asked him for his calculation without farther delay, Sr Isaac looked among his papers but could not find it, but he promised him to renew it, & then to send it to him.
Recollection of Edmond Halley's 1684 visit with Isaac Newton from A Short History of Nearly Everything by Bill Bryson and other online sources

De motu corporum in gyram (On the motion of bodies in orbit)

the law

The Principia contains in it the unification of terrestrial and celestial gravitation. The acceleration due to gravity described by Galileo and the laws of planetary motion observed by Kepler are different aspects of the same thing. There is no terrestrial gravitation for earth and no celestial gravitation for the planets, but rather a universal gravitation for everything.

  1. Every object in the universe attracts every other object in the universe with a gravitational force.
  2. The magnitude of the gravitational force between two objects is …
    1. directly proportional to the product of their masses and
    2. inversely proportional to the square of the separation between their centers

Newton's law works since we live in a universe with three spatial dimension. As gravity extends out into space it spreads itself thinner and thinner, covering an area that expands as the square of the distance from the source. If space wasn't three-dimensional, Newton's law wouldn't work.

Although space appears three dimensional, there's no obvious reason why it has to be. Some advanced theories seem to suggest that there may be additional spatial dimensions. The reason we haven't seen them is that they're curled up rather tightly. If they exist, it should be possible to find deviations in the force of gravity from Newton's inverse square law at extremely small distances. Testing for these deviations is quite difficult. The best experiments currently (2001) show that the inverse square law holds down to 218 μm (2.18 × 10−4 m). Since the size of these hidden dimensions is thought to be on the order of 10−35 m, we've still got a long ways to go.

the moon

The earth-moon separation is approximately sixty times greater than the radius of the earth. The acceleration due to gravity at this distance is 1/3600 the acceleration due to gravity at the surface of the earth.

the apple

This must have been the time of the famous and disputed fall of the apple. One of the records of the apple story is in a biography of Newton written in 1752 by his friend, William Stukeley In it we read that on a particular occasion Stukeley was having tea with Newton; they were sitting under some apple trees in a garden, and Newton recalled that:

he was just in the same situation, as when formerly, the notion of gravitation came into his mind. It was occasioned by the fall of an apple, as he sat in a contemplative mood. Why should that apple always descend perpendicularly to the ground, thought he to himself. Why should it not go sideways or upwards, but constantly to the earth's centre?

Newton once wrote,

I began to think of gravity extending to the orb of the moon, and … from Kepler's rule I deduced that the forces which keep the Planets in their orbs must be reciprocally as the squares of their distances from the centres about which they revolve: and thereby compared to the force required to keep the moon in her orb with the force of gravity at the surface of the earth, and found them to answer pretty nearly. All this was in the two plague years of 1665 and 1666, for in those days I was in the prime of my age for invention, and minded mathematics and philosophy more than at any time since

the formula

Fg =  Gm1m2
r2

 

g =  Gm
r2

the constant

Cavendish experiment

The Great Pyramid is so massive that a plumbline will not hang straight down when near the pyramid but will swing toward the structure. Cf. Tompkins, Secrets of the Great Pyramids, pp. 84-85, where Tompkins, discussing the measurements taken by Piazzi Smyth, writes "To obtain the correct latitude of the Great Pyramid without having his plumb line diverted from the perpendicular by the attraction of the huge bulk of the Pyramid, Smyth made his observations from the very summit; there the Pyramid's pull of gravity would be directly downward". Tompkins, Peter. Secrets of the Great Pyramid (New York: Harper Collins, 1971).

the critics

Action at a distance. Newton's reply to these criticisms was basically, "I don't care. The theory works."

Rationem vero harum gravitatis proprietatum ex phænomenis nondum potui deducere, & hypotheses non fingo…. Et satis est quod gravitas revera existat, & agat secundum leges a nobis expositas, & ad corporum cælestium & maris nostri motus omnes sufficiat.   I have not been able to discover the cause of those properties of gravity from phenomena, and I frame no hypotheses…. And to us it is enough that gravity does really exist, and act according to the laws which we have explained, and abundantly serves to account for all the motions of the celestial bodies, and of our seas.

beyond that …

Summary

Problems

practice

  1. Verify the inverse-square rule for gravitation with the following chain of calculations …
    1. Determine the centripetal acceleration of the moon. (Assuming the moon is held in it's orbit by the gravitational force of the earth, you are then also calculating the acceleration due to gravity of the earth at the moon's orbit.)
    2. Determine the ratio of the radius of the moon's orbit to the radius of the earth.
    3. Use the results of a. and b. to calculate the acceleration due to gravity on the surface of the earth.
    4. How does this value compare to the generally accepted value of g? Are the results of your calculations in close enough agreement with experimental observations to verify the inverse square rule for gravitation? Discuss briefly.

    Solutions …

    1. Start with the centripetal force formula. Recall that the speed of an object moving with uniform circular motion is the circumference of the circle divided by the period. Substitute the relevant values. Watch the units. Also, cary along a few more significant figures than normal for the hell of it.
       
       gearth-at-moon =  ac =  v2  =  ⎛
      ⎝      		 2πr ⎞2
      ⎠      		  ÷ r =  4π2r  =  4π2(384,400,000 m)
      r T T2 (27.321661 × 24 × 3600)2
       gearth-at-moon =  0.002723373 … m/s2 
       
      Although half the population of earth is familiar with the synodic period of the moon (the period between full moons) few people know the sidereal period (the period as measured relative to the stars). The difference between the two is significant, so don't mess them up. The first is 29.530588 days and is nearly the same as the human menstrual cycle, while the second is 27.321661 days and is relevant to this problem.
    2. Look these values up in reference sources. Use the most precise values you can find. Since the earth is a slightly flattened sphere, we should use the mean radius here.
       
      rearth-at-moon  =  384,400,000 m  = 60.2683541 …
      rearth-on-surface 6,378,140 m
       
    3. Use the inverse square law to answer this part of the question. Since a point on the surface of the earth is roughly 60 times closer to the center of the earth than is the moon, the acceleration due to gravity here should be roughly 602 or 3600 times stronger. Let's be a little bit more precise, however.
       
      gearth-on-surface  =  ⎛
      ⎝      		 rearth-at-moon ⎞2
      ⎠
      gearth-at-moon rearth-on-surface
       gearth-on-surface  = 0.002723373 … m/s2 × 60.2683541 … 
       gearth-on-surface  = 9.89 m/s2 
       
    4. This value is within 1% of the generally accepted value for the mean acceleration due to gravity on the surface of the earth. If I were me (which I am) then I would have no problem with saying that these calculations are sufficient to verify the inverse square rule for gravitation. If I were Isaac Newton (which I am most definitely not) then I would have had to use significantly less accurate measurements and would have calculated a value with a higher deviation. Still, if it was me doing this I would not be upset, but such is not the case with Newton. The story I have heard is that Newton (who did a calculation similar to, but not identical to this one) thought the deviation between his calculated value and the experimental value was enough to be significant. Realizing that a part of the deviation was due to inaccurate measurements (in particular, the sidereal period of the moon) he decided to play it safe and wait until some better measurements came along. Little did he know that no matter how accurate the numbers were, there would always be a discrepancy using this method. It has nothing to do with the theory, but rather it's in the simplification of the model. Here are the systematic errors I have identified in Newton's analysis …
      • The earth is not a sphere and the moon does not orbit in a plane perpendicular to the axis of the earth's rotation. Thus, the earth cannot easily be simplified by a point body.
      • The orbit of the moon is not circular, but elliptical. Newton surely knew of this since he was a fan of Johannes Kepler -- the first person to propose that orbits could take on shapes other than the perfect circles of the Ancient Greeks.
      • Not only isn't the orbit circular, it's not even closed. The moon's orbit wobbles around a bit. (In more formal language, the orbit precesses.) Newton surely knew of this fact since it is used to pedict solar eclipses.
      • And the reason it's not a closed path is because of the sun — that great big ball of incandescent gas at the heart of the solar system. The sun contains something between 99.8 and 99.9 per cent of the mass of the solar system. It's effect cannot be ignored if one wants to make predictions with the same degree of precision as astronomical measurements are made.
      Still, if it was me I would have been satisfied with a 1% deviation, but I have the advantage of 400+ years of hindsight.
  2. Estimate the value of the universal gravitational constant from the following approximate measurements taken during the original Cavendish experiment (and converted into SI units for us) …
    • one hundred kilogram fixed and one kilogram rotating masses
    • ten centimeter separation between fixed and rotating masses
    • one millionth newton of force on each of the rotating masses

    Solution …

    Newton's original law of universal gravitation was not stated as an equation, but rather as a proportion. Transforming a proportion into an equation requires a choice of units followed by the measurement of the constant of proportionality. (Picking the constant first and then measuring the units will also work, but that's not the way it was done here.) Let's do it.

    Fg =  Gm1m2 ⇒ 10−6 N =  G(102 kg)(100 kg) ⇒ G = 10−10 Nm2/kg2
    r2 (10−1 m)2

    Now, before you go telling me I've failed because the accepted value for G ends in 10−11 we need to recall that a half power in the mantissa separates one order of magnitude from another. A more complete value of G is 6.67 × 10−11 Nm2/kg2, which to the nearest power of ten is 10−10 Nm2/kg2 since 6.67 is greater than 3.16 or 10½ .

  3. Check it out.
    1. Determine the acceleration due to gravity (g) on the surface of the earth from Newton's law of universal gravitation.
    2. How does this value compare to the standard acceleration due to gravity, g?
    3. Are the results of your calculation in close enough agreement with the standard value to verify the mass-dependent portion of Newton's law of universal gravitation? Discuss briefly.

    Solutions …

    1. This looks quite simple (and it is). So let's just do it.
               
      g =  Gm  =  (6.673 × 10−11 Nm2/kg2)(5.9736 × 1024 kg)  = 9.82069 m/s2
      r2 (6.371 × 106 m)2
               
    2. Hmmm, this doesn't compare all that favorably to the standard value for the acceleration due to gravity.
         
      9.82069 m/s2 − 9.80665 m/s2  = 0.0014 = 0.14%
      9.80665 m/s2
         
    3. Are these results close enough? Of course they are. Well, maybe not. Drat! Why doesn't this work out exactly? Of course it will never work out "exactly". There are no "exact" values based on measurement. There can't be. Still, I'd hoped to do better than this. Astronomical measurements are among the most precise in all of science. The only thing that outdoes it is particle physics. (Notice how it is the very small and very large that have the best measurements.) Believe it or not this tiny discrepancy -- which most students would be more than happy to live with for their own experiments -- this tiny discrepancy is significant.

      First of all, the earth isn't a true sphere. This shouldn't be a surprise. After all, what sphere has mountains and valleys on it? But the earth isn't even a bumpy sphere. The rotation of the earth generates a centrifugal force that flattens it out a tiny bit. As a result, the earth is wider when measured across the equator (r = 6378.1 km) than when measured from pole to pole (r = 6356.8 km). In fancy language, the earth is an oblate spheroid. The value calculated above is for the volumetric mean radius of the earth. So everything should still work out fine. Since it doesn't, there must be another confounding factor.

      The rotation of the earth induces a slight outward acceleration. This centrifugal acceleration might be called "fictitious" but it does have a real, measurable effect. The earth's rotation reduces the apparent pull of gravity on its surface everywhere except the poles. How much do you wanna bet that this will cure everything? All we have to do is subtract the varying centrifugal acceleration from the varying gravitational acceleration and then use a bit of mathematical magic to compute the average value on the surface of the earth.

      Sorry my friend. Even that won't work. You see, the whole notion of a "standard value" for the acceleration due to gravity on the surface of the earth is a legal construct. By the end of the Nineteenth Century, several countries had adopted 9.80665 m/s2 into law. In 1901 the International Bureau of Weights and Measures did the same thing and brought the rest of the world along. There most certainly was a calculation done by somebody somewhere at sometime that resulted in a value of 9.80665 m/s2, but any repeat of that calculation using modern values wouldn't result in that same series of six significant digits. The value caluclated above is totally fine.
  4. Jupiter is about eleven times larger in diameter and three hundred times more massive than the earth. How does the gravitational field on Jupiter compare to that on earth?

    Solution …

    The acceleration due to gravity is directly proportional to the mass of the gravitating body and inversely proportional to the square of its radius.

     gjupiter = gearth  ⎛
    ⎝    		  300  ⎞
    ⎠    		  ≈ 2.5 gearth ≈ 24 m/s2
    112

    What a nice simple problem.

conceptual

  1. What would happen to objects on the earth's surface if …
    1. the earth's gravitational field gradually disappeared?
    2. the earth's gravitational field was fine, but the earth slowly stopped rotating?
  2. What effect, if any, would removing the earth's core have on the gravitational field at its surface?
  3. The earth has a radius about twice as great and a mass ten times greater than the planet Mars. How does the acceleration due to gravity on Mars compare to that on earth?

numerical

  1. Determine the following quantities for a 10 kg frozen turkey …
    1. its mass on the surface of the earth,
    2. its weight on the surface of the earth,
    3. its mass in orbit one earth radius above the surface of the earth,
    4. its weight in orbit one earth radius above the surface of the earth,
    5. its mass on the surface of the moon, and
    6. its weight on the surface of the moon.
  2. Astrology
    1. Calculate the force of gravity between a 3.0 kg newborn baby and a 75 kg doctor standing 0.25 m away.
    2. Calculate the force of gravity between a 3.0 kg newborn baby and the planet Jupiter when it is nearest to the earth.
    3. What is the ratio of the force of gravity from Jupiter on the baby compared to the force of gravity from the doctor on the baby?
    4. What is the likelihood that astrology (assuming it had any validity) could be explained as a result of planetary gravitation at the moment of your birth? (Keep in mind that Jupiter is the largest planet and that it is rarely as far from the earth as its nearest approach.)
  3. Moon Walk
    1. Calculate the acceleration due to gravity on the surface of the moon.
    2. What is the ratio of the acceleration due to gravity on the surface of the moon to the acceleration due to gravity on the surface of the earth?
    3. What effect would the moon's reduced gravity have on your athletic abilities?
  4. Weightlessness
    1. Calculate the weight of a 75 kg astronaut on the surface of the earth.
    2. Calculate the same astronaut's weight aboard the space shuttle as it orbits 3.5 × 105 m above the earth's surface.
    3. According to common wisdom, objects in outer space are "weightless". Why then isn't the answer to the second part of this question zero? What's wrong with the common wisdom?
  5. Geosynchronous, Earth-Orbiting Space Station

    [magnify]
    For a sufficiently advanced human civilization, the occasional trip into outer space may become a reality for the general population. Having large numbers of spacecraft landing and taking off from the surface of the earth would probably not be acceptable, however. One way around this would be to build a ring around the earth that rotates at the same rate as the earth. This ring would be linked to the equator by electrically powered space elevators. No more noisy, dirty rockets. Just hop on the space elevator, press the "up" button, and stare nonchalantly at the door for a couple of hours. Such a massive structure would also house a large population of full-time inhabitants. Since they're the descendants of earth-bound humans, they would probably feel most comfortable in a 1 g environment. Determine the radius of such a megastructure. State your answer in terms of
    1. multiples of earth's radius.
    2. the fraction of the distance from the earth to the moon.
    (When this question appeared in the centripetal force section of this book, the effects of gravity were assumed negligible. This was a very reasonable assumption. The pull of gravity is relatively weak at this distance and, if you have already solved the earlier version of this problem, will only affect your results slightly.)

algebraic

  1. Determine the height h above the surface of a planet of radius r and mass m at which the gravitational field will be one half its surface value.
  2. The purpose of this problem is to determine the possible nature of the planet Krypton. Begin by reading the introduction from the 1950s TV series Superman.
     
    Faster than a speeding bullet. More powerful than a locomotive. Able to leap tall buildings in a single bound. Look, up in the sky! It's a bird! It's a plane! It's Superman! Yes, it's Superman, strange visitor from another planet who came to earth with powers and abilities far beyond those of mortal men. Superman, who can change the course of mighty rivers, bend steel with his bare hands, and who, disguised as Clark Kent, mild-mannered reporter for a great metropolitan newspaper, fights a never-ending battle for truth, justice and the American way.
     
    You should note in this synopsis of the origins of Superman, that all his feats are those of great strength. The first appearance of the character we now recognize as Superman was in Action Comics, Vol. 1, No. 1, 1938. At that time Superman was just like any ordinary man, except he was very, very strong (and also of good character). As the comic book, then radio show, then TV show, then movie (then video game?) evolved Superman became more and more super. He learned to fly, he had x-ray vision, he could hear really quiet sounds from great distances, he could blow very cold air, and so on. In this analysis we will stick to the golden age of Superman -- back when he was just a super man.
     
    Superman's strength was attributed to the gravity of his home planet, Krypton. The people of Krypton evolved great strength so that they could stand, walk, and lift ordinary objects in Krypton's strong gravitational field. When Superman came to earth, he found that his Kryptonian physique was sort of over-designed. This is much like when human go to the moon, they find themselves strong enough to do all sorts of things they couldn't do on earth -- like run effortlessly with long strides while wearing an 83 kg (183 lb.) space suit, for example.
     
    These questions should be solved as proportions. State all answers in comparison to their values on earth ("twice as big as on earth," for example).
    1. Derive an equation that relates height jumped to the acceleration due to gravity. Use this to determine the acceleration due to gravity on the surface of Krypton from the statement, "Able to leap tall buildings in a single bound." [Hint: Think of how high a typical human can jump on earth and how high Superman can jump on earth. How much stronger then must gravity have been on Krypton in proportion to the gravity on the earth?]
    2. Derive an expression that relates the acceleration due to gravity on the surface of a spherical planet to the density and radius of the planet (instead of the mass and radius, which is the usual way it is stated). Use this equation to …
      1. determine the radius of Krypton assuming it has the same average density as the earth, and
      2. determine the average density of Krypton assuming it has the same radius as the earth.
    3. Comment on the physicality of these answers. (Remember, Superman and the other inhabitants of Krypton look and act like humans in nearly every way other than their unusual strength.) How likely is one to find a terrestrial planet with …
      1. the radius you just calculated or
      2. the average density you just calculated?
    4. Is there anything else you would like to say about Superman or Krypton?

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