We had a section in our first chapter on fundamental forces.  If you recall, we looked at Newton's law of universal gravitation.  The magnitude of the gravitational force between two masses is given by [br][br][center][math]F_g=\frac{Gm_1m_2}{r^2}. \\[br]\text{Comparing with } F_g = mg \text{ It is clear that } \\[br]g=\frac{Gm_{earth}}{r_{earth}^2}. \\[br]\text{This, however, is not the value of g we've been using all semester.}[/math][/center][br][br]Using the law of universal gravitation, we can't quite get the right value for earth's gravitational constant.  That's because when we write [math]g=9.8m/s^2[/math] we are not actually giving a value for the strength of the gravitational field.  We are instead giving the [i]effective gravitational constant [/i]that needs a tiny adjustment to compensate for earth's rotation on its own axis. So while we didn't put the "eff" subscript on g, it was implied all along.[br][br]If we first imagine being on the equator, this calculation is easy. The earth rotates roughly once around per 24 hours.  We discussed how the solar day is actually around one degree more than a full rotation, but we'll ignore that right now.  In any case, we know that to stay on earth's surface means to undergo centripetal acceleration directed toward earth's own center as seen in the graphic above if you set latitude to zero.  If we imagine standing on the equator, then gravity pulls downward, and the centripetal acceleration is also downward.  That means that the effective gravitational constant is reduced by an amount equal to the acceleration we are undergoing while spinning along with earth!  It turns out that the reduction is small, and is easy to calculate.  [br][br]Recall that [math]a_c=\omega^2r.[/math]  The rotation rate is just [math]\omega=\frac{2\pi}{86400s}[/math] (there are 86,400 seconds per day) and the radius of earth is [math]r=6.37\times 10^6m.[/math]  This gives [math]a_c=0.0335m/s^2.[/math]  The number we use for [math]g[/math] is less than the actual gravitational field by this amount - which is an easily measurable discrepancy.  Notice two things from the animation above:  The radius of the path depends on location since it is really [math]r=r_{earth}\cos(latitude)[/math] and that the direction of centripetal acceleration is not actually toward the center of the planet, but rather toward the axis of rotation.  The only place where those are the same direction is on the equator. [br][br]The fact that [math]\vec{g}_{effective}[/math] does not point directly toward earth's center indicates that we don't really stand vertically, if what we mean by "vertically" is along a line that would intersect earth's center.  Instead we find that we lean slightly northward (in the northern hemisphere) due to earth's rotation.  This would be significant if earth were to rotate much faster.  In that case we'd feel thrown toward the equator just like how you feel thrown outward on a playground merry-go-round.

The fact that on a rotating planet gravity feels different means that the direction we call "down" is inward and slightly southward if you are in the northern hemisphere and inward and northward in the southern hemisphere.  It's not just you that feels this.  The material that makes up the planet feels this as well.  In fact, the shape of the planet is influenced by this same effective gravitation.  This is the reason that earth is not a perfect sphere.  Gravity alone would tend to make any large masses spherical by pulling inward.  Thus planets, moons and stars are roughly spherical.  However, if the large mass is rotating, its mass will be pulled outward away from the rotation axis a bit due to the centripetal acceleration.[br][br]Since the outward pull is due to [math]a_c=\omega^2r[/math] it grows with distance r from the rotation axis.  The result is that rotating planets and stars become oblate spheroidal in shape, or roughly 3D ellipses.  So the direction we call "downward" is in fact perpendicular to the ground (barring subtler effects) at our latitude of 34 degrees north, but that direction does not intersect the earth's center of mass.  See this [url=https://en.wikipedia.org/wiki/Spheroid]link[/url] for more discussion of oblate spheroids.  [br][br]In the event that mass is rotating rapidly, gravity doesn't bring mass together, but rather causes it to spin about a center of mass and never comes together to look spherical but instead looks like Saturn's rings look, or like our galaxy looks - both of which are full of chunks of mass that are traveling along paths such that the gravitationally induced central force is not sufficient to draw the masses together.  That is just another way to describe an orbit.

If we are being careful, we should ask whether earth's orbital centripetal acceleration is also worth worrying about.  The earth revolves in a roughly circular orbit of radius [math]1.5\times10^{11}m[/math] which of course is the distance from the sun's center of mass to earth's center of mass.  The orbital angular velocity is [math]\omega = \frac{2\pi}{\text{seconds in a year}}.[/math]  Since [math]a_c=\omega^2r[/math] we get a value of 0.00595m/s[sup]2 [/sup]when plugging in values.  [br][br]This is actually only 5 to 6 times less than the acceleration due to earth's axial rotation.  In principle, if this is real, we should be able to measure this orbital-motion-induced acceleration with a carefully conducted experiment. What we find, however is that while there would be a variation in the measured weight dependent on time of day, it is not in agreement with this calculation.[br][br]It's worth noting that this effect would differ during the day when we are roughly facing the sun versus during the night when we are facing directly away from the sun.  For the self test below, draw a diagram of the earth orbiting the sun.  Draw the orbital centripetal acceleration of earth on the diagram.  Use [math]\vec{g}_{effective}=\vec{g}-\vec{a}[/math] to deduce the answer.  It might help to draw two little people on your earth - one at mid-day, one at midnight. Next we will see why this is, however, not correct.

While what we did in the last section may seem correct, there is a subtle problem with the orbital contribution to [math]\vec{g}_{effective}[/math] that may easily go unnoticed. The problem is that we need to consider more than just earth's gravitational field. The sun also pulls on us gravitationally. That pull is just like earth's pull. The gravitational field of the sun is obtained using the same expression:[br][br][center][math]\vec{g}_{sun}=G\frac{m_{sun}}{r^2_{\text{sun to earth}}}.[/math][/center][br]This term must be added to the expression for the effective gravitation. When we do this, we find that the sun's gravitational field and the centripetal acceleration it induces on us cancel one another.[br][br]With even more careful inspection, however, we find that while the cancellation is true for the planet earth as a whole, that objects on its surface move at different rates, and that for them the terms almost cancel one another out, but do not vanish. Because of our retrograde velocity while on the side of earth closer to the sun due to earth's rotation on its axis, the gravitational term wins out during the day, and the prograde motion at night makes the centripetal term dominate. This means that we actually find our weight on a scale is a tiny bit less at mid day and midnight than around sunrise and sunset![br][br]This variation in your own weight is also the reason the oceans have tides. High tides are where you'd be lightest and low tides where you'd feel heaviest.