Our accomplishments in this chapter put us in a position to do a little extra. What follows (being within our reach but beyond the introductory level) is presented in this special section.
Earth's attractive force acts on mass wherever the mass might be. The magnitude of the attractive force is called weight. Every mass has weight everywhere, is not "weightless," so long as Earth exists. The direction of the force is "Earthward." Our familiarity with weight is principally at the surface of Earth. To establish that number (the magnitude of the attractive force) , we sometimes use the weighing process of placing the mass on a calibrated spring scale. At locations where a scale cannot be applied, the magnitude of the gravity force of Earth can be calculated by the above equation.
GRAVITY and ALTITUDE Virtually no engineering systems operate at altitudes higher than the upper boundary of our atmosphere - about 30 kilometers (or 16 miles). The radius of Earth is about 6380 kilometers. Suppose one were to ignore the "inverse-square rule" of gravity [(ro/r)2] and simply use the constant, Earth-surface magnitude, go up to an altitude of 30 kilometers. What maximum error might be incurred?
♦ The correct and approximate expressions for g(r) are:
Entering these into the equation for percent error, we have:
Since the error is less than 1%, it is wise to use this approximation. It is also wise and completely within pencil and paper accuracy to use 9.81, rather than 9.80665, and to use 32.2 rather than 32.1740. Your calculator permits use of π as 3.14592654. But should you have to write it, use 3.14. Have the habit of saving time.
Below are presented further methods for treatment of slightly more involved cases of motion. First we use what we know to calculate the altitude of a geostationary orbit. You need to review derivatives of sine and cosine functions. You will use them here to calculate the actual altitude of those orbits: 22,298 miles. Needs some messy dang algebra - go slow and write carefully. Don't move terms left to right, over the equality until all terms have been cast as numbers.
ALTITUDE of GEOSTATIONARY ORBIT:
Modern rockets routinely place communications satellites in orbits that are called geostationary. The plane of these orbits is the plane of the Earth equator. Once placed in that plane above Earth the proper distance and moving circularly at the proper speed, a geostationary satellite will remain stationary, directly above dome point on the equator of earth. This permits Earth-surface communication satellite dishes to be "pointed" in a specific directions to constantly receive satellite telecasted information. The position of the satellite can be selected about the equator; its altitude cannot. Calculate the altitude of a geostationary orbit.
♦ The figure depicts the satellite orbiting in the plane of Earth's equator. Coordinates and unit vectors have been included. Imagine an origin at the center of Earth and the axis X locked (at mid-night) pointing at a distant star.
The above is the momentum of the satellite, we write Newton's Second Law:
Now enter the momentum and write the gravity force.
Perform the differentiation left of equality. Then gather terms.
The above example describes position and velocity using sines and cosines with the usual, vector basis locked to Earth's axis of totation (K) and I or J directed at a distant star. The only calculus needed is that one be able to differentiate the sine and cosine. This is a good place to show how to write a vector using the simplest unit vector basis that rotates.
VECTORS for CIRCULAR MOTION: Motion of a mass at constant speed constrained by forces to follow a circular path is kinematically described by the circle radius, and system vectors - position, velocity and acceleration.
The graphic shows the geometry with positive change of angle, θ(t) and d(θ(t))/dt being counter-clockwise. We use the I, J vector basis in our development. Position, in terms of the I - J basis is:
P(t) = r [cos(θ(t))I + sin(θ(t))J]
Two derivatives are taken to yield successively the velocity vector, V(t), and A(t), the acceleration vector. Each is written in a "magnitude times unit vector" form where the unit vectors, I and J are fixed in direction.
The above is correct in terms of the constant vector basis: I and J. Problems can be solved with this set - just as they are. However, these results can be used to confirm a simpler, less writing-intensive way of representing P(t), V(t) and A(t).
Compare the directions (unit vector parts) of the vectors, V and A with that of the initial unit vector, the direction of P. Immediately we see the direction of A is opposite the direction of P. Using a simple sketch, we see the direction of velocity, V is perpendicular to the direction of P, and directed toward increasing angle, θ. Thus we define er(t), and eθ(t).
The graphic representation of the unit vectors with analytic statements of position, velocity and acceleration vectors are shown.
P(t) = r er(t) V(t) = r(dθ/dt)eθ(t)
A(t) = - r (dθ/dt)2 er(t)
Extension of this idea to apply to non-circular motion, that is for a radius that varies in time, r = r(t), follows easily since the radius is a scalar. The new vector basis described here is:
er(t), eθ(t) (and K if and when needed).
In this next example we eztablish a new representation for unit vector.
• POLAR VERSUS EQUATORIAL WEIGHT: Suppose we have a certified mass of 1,000 lbm. Demonstrate that the mass will weigh 1,000 pounds force at the North Pole but only 996 pounds force at the equator?
♦ Newton's Second Law will be applied. In each case the system is the thousand pound mass. The forces acting will be gravity and the support force of the spring of the scale. Our inertial reference has the center of Earth as the origin. This means we assume the center of Earth moves in a straight line at a constant speed). The vector triad, er(t), eθ(t) and K will be used. Imagine the mass to rest on a spring scale at the North Pole and later at the Equator. The sketches depict these measurements of weight:At the North pole the position vector of the mass initiates at the center of Earth and extends the distance, r, through the bottom of the scale to the mass. The mass, in constant contact with the scale, rotates or pivots as Earth rotates beneath it.
A precise answer is not needed; let the radius of Earth equal 3960 miles and the surface acceleration of gravity equal 32.2 ft/s2).
For the weighing at the equator, the mass and scale supporting and the place where it rests are rotating with Earth. The position vector of the mass and scale rotates about Earth's center. So we differentiate position correctly as follows:
Now enter there results into Newton's Second Law:
A gravity force will act on the mass on the spring scale whether at the North Pole or equator. At the North Pole the mass has no component of momentum relative to the Earth center. The K-axis is directed vertical from the weing place at either pole. The mass and scale just pivot or rotate about the axis vertically through them. Locations on Earth's equator are not inertial. The mass there has momentum; to weigh a mass there will incur a small error.
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