Read the fifth chapter of Nonsense on Stilts (Pigliucci) and write a one-page reaction to or summary of.
Due: Oct. 13
Monday:
We looked at Newton's law of universal gravitation.
Fgravity = - G M m / r2
The minus sign indicates that the force is an attractive force -- that each mass
experiences an attraction (equal and opposite according to Newton's Third Law) toward the
other. The G is a universal constant given by:
6.67384 x 10-11 N m2 / kg2
The force is proportional to each of the two masses involved in the attraction and
is inversely proportional to the square of the distance between their centers.
We worked on recovering Galileo's notion of terrestrial gravity (the same acceleration for all
freely falling masses with a value of g=9.8 m/s2). Combining Newton's Second Law
F=ma with his law of universal gravitation yields.
F = - G Mearth mobject / Rearth2 =
mobject a = - mobject g
Note that the mass of the object appears on both sides at part of the "cause" (gravitational
attraction) and part of the "effect" -- mass times acceleration. The object mass "cancels"
giving us Galileo's independence of free fall acceleration on mass. Then g should be
g = G Mearth / Rearth2
Inserting values for the mass of the earth (5.972 x 1024 kg) and radius
(6.371 x 106 m).
g = 6.67384 x 10-11 * 5.972 x 1024 / (6.371 x 106 )2
= 9.819 m / s2
Because it is the universal law of gravitation we could also calculate the acceleration of
the moon toward the earth. We alleviated our fears that the moon was going to crash into the
earth by reminding ourselves that an acceleration can come either from a change in speed
(magnitude of velocity) or a change in the direction of speed. And the moon orbiting around
the earth was the latter type of acceleration. Then we started deriving a formula for
(instantaneous) centripetal acceleration v2 / r.
We considered an object taking a circular path of radius r and speed v. If we consider a
semi-circular arc. The velocity will end up in exactly the opposite direction that we
started in and hence the change in velocity will be 2*v. The time it takes to travel a
semicircle can be obtained by using distance = speed * time.
π r = v * t so that t = π r / v
So the average acceleration for a semicircular path is
aave-semi-circular = 2 * v / ( π r / v ) = ( 2 / π ) ( v2 / r)
We have roughly the behavior we want. The acceleration increases if the velocity increased --
we change direction more quickly if the speed is higher. The acceleration increases if the
radius decreases -- we change direction more quickly if the radius is smaller.
We started working toward deriving an expression for the "instantaneous" centripetal acceleration.
Wednesday:
We chose a point in circular motion of radius r about the origin for which the Cartesian coordinates
are (r, 0) -- all x and no y. We said that the velocity at that point was (0, v) -- no x and all v.
Next we chose a point after a small rotation of angle theta, the carteisan coordinate of the position
was (r cos θ, r sin θ) -- mostly x and a little y. We said that the velocity at that point
was (- v sin θ, V cos θ) -- a little bit of negative x and mostly y.
Next we discussed the unit of radians to measure angles. That there are 360 degrees in a circle is an idea we inherit from
the Babylonians. 360 is a nice number with a lot of factors and close to the number of days in the year, but measuring
angles in radians is more useful mathemtically -- possibly even more "real".
(Why
There Are 360 Degrees in a Circle, 60 Minutes in an Hour and 60 Seconds in a Minute…)
We can generalize the formula for the circumference (distance around a full circle)
C = 2 π r to a formula for the distance around any arc to s = θ r provided the angle
θ is measured in radians. Then the time it takes to travel around an arc of angle
θ at a speed v is
s = v t = θ r which solved for t yields t = θ r / v
Putting the pieces together the acceleration is the change in velocity divided by the change in time. The change in velocity is
(vfinal_x - vinitial_x, vfinal_y - vinitial_y) =
(-v sin θ , V cos θ -v). The acceleration associated with moving through an angle θ is then
athrough angle θ = [-v sin θ / ( θ r / v), (v cos θ -v)/ (θ r / v)] =
[-v2 sin θ / ( r θ) , v2( cos θ -1)/ ( r θ))]
We then used Excel to take the limit of these results as θ goes to zero.
Applying these limits we find
ainstantaneous = [-v2 / r, 0 ]
(We just took a derivative. Newton and Leibniz would be so proud.) Also note that the direction of the
centripetal acceleration is in the negative x direction for the point we choice -- or more generally
toward the center of the circle.
Kepler from Newton: If we use Newton's Second Law, Newton's Law of Universal Gravitation and the
expression for centripetal acceleration we can derive Kepler's Third Law. (The square of the orbital
period of a planet is proportional to the cube of the semi-major axis of its orbit.). Let us consider
specifically the earth revolving around the sun.
F = - G Msun mearth / Rsun-earth2 = mearth a
= - mearth v2 / Rsun-earth
G Msun / Rsun-earth = v2
Since Kepler's Third Law is about the period we substitute in v = 2 π Rsun-earth / Tyear
G Msun / Rsun-earth = (2 π Rsun-earth / Tyear)2
Tyear2 = 4 π2 Rsun-earth3 / (G Msun)
Year from Kepler:
Substituting in Rsun-earth 1.496 x 1011 m and
Msun = 1.989 x 1030 kg and
G = 6.67384 × 10-11N m2 / kg2 yields
Tyear = 31555275 s = 525921.25 minutes = 8765.35 hours = 365.22 days
