Universal Gravitation
Practice
practice problem 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.)
 Determine the ratio of the radius of the moon's orbit to the radius of the Earth.
 Use the results of a. and b. to calculate the acceleration due to gravity on the surface of the Earth.
 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.
solution
Start with the equation for centripetal acceleration. 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, carry along a few more significant figures than normal for the hell of it.
g_{earthatmoon} = a_{c} = v^{2} r g_{earthatmoon} = ⎛
⎝2πr ⎞^{2}
⎠÷ r T g_{earthatmoon} = 4π^{2}r T^{2} g_{earthatmoon} = 4π^{2}(384,400,000 m) (27.321661 × 24 × 3,600 s)^{2} g_{earthatmoon} = 0.002723373… m/s^{2} 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 the one you need to solve this problem.
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 probably use the mean radius here.
r_{earthatmoon} = 384,400,000 m r_{earthonsurface} 6,378,140 m r_{earthatmoon} = 60.2683541… r_{earthonsurface} 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 60^{2} or 3,600 times stronger. Let's be a little bit more precise, however.
g_{earthonsurface} = ⎛
⎝r_{earthatmoon} ⎞^{2}
⎠g_{earthatmoon} r_{earthonsurface} g_{earthonsurface} = 0.002723373… m/s^{2} × 60.2683541… g_{earthonsurface} = 9.89 m/s^{2} This value is within 1% of the generally accepted value for the standard acceleration due to gravity on the surface of the Earth. If I were me (which I am) then I would say 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 used significantly less accurate measurements and would have calculated a value with a higher deviation. Once again, I wouldn't be upset, but then I'm not Newton. The story I 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 large 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 inverse square law, 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, it precesses.) Newton surely knew this fact since it is used to predict 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 350+ years of hindsight.
practice problem 2
 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.
F_{g}  = 


10^{−6} N  = 


G  =  10^{−10} Nm^{2}/kg^{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} Nm^{2}/kg^{2}, which to the nearest power of ten is 10^{−10} Nm^{2}/kg^{2} since 6.67 is greater than 3.16 or 10^{½}.
practice problem 3
 Determine the acceleration due to gravity (g) on the surface of the Earth from Newton's law of universal gravitation.
 How does this value compare to the standard acceleration due to gravity (g)?
 Are the results of your calculation close enough to the standard value to verify the distancedependent portion of Newton's law of universal gravitation? Discuss briefly.
solution
This looks quite simple (and it is). So let's just do it.
g = Gm r^{2} g = (6.673 × 10^{−11} Nm^{2}/kg^{2}) (5.9736 × 10^{24} kg) (6.371 × 10^{6} m)^{2} g = 9.82069 m/s^{2} Hmmm, this doesn't compare all that favorably to the standard value for the acceleration due to gravity.
∆g = 9.82069 m/s^{2} − 9.80665 m/s^{2} g 9.80665 m/s^{2} ∆g = 0.0014 = 0.14% g 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/s^{2} into law. In 1901 the International Bureau of Weights and Measures (BIPM) 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/s^{2}, but any repeat of Newton's calculation done using the moon's orbit parameters would certainly not result in that same series of six significant digits. (The BIPM chose the standard value since it pretty nearly agrees with the value at their offices.)
The value caluclated above is totally fine.
practice problem 4
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.
g_{jupiter} =  G (300 m_{e})  
(11 r_{e})^{2}  
g_{jupiter} =  300  g  
11^{2}  
What a nice simple problem.