July 12, 2009
sqrt(2)
sqrt(2)^2
= 2
= 100 / 50
= 100 / 49
= 10^2 / 7^2
= (10/7)^2
= (1.42...)^2
[ 10 sqrt(2) ]^2 = 200
14^2 = 196
sqrt(2) = 1.4...
[14+x]^2 = 196 + 2 * 14 * x
= 200 - 4.00 + 2 * 14 * x
= 200 - 2 * (2.00 - 14 * x )
= 200 - 2 * (200 - 14*x) / 100
= 200 - 2 * (200 - 14^2) / 100
x = .14
sqrt(2) = 1.414...
July 11, 2009
How do we know the world is round?
Lore, mostly (it's what we are taught) but primarily I'm asking what we can observe that supports a round model of the world...
Some ideas
- horizons
- tides
- star altitudes / shadows.
- atmospheric pressure
July 1, 2009
Inverse Square Law
Feynman's bridge between the Kepler's laws (in particular, the relationship between the period of an orbit and the length of the major axis) and Newton's law of gravitation (that the gravitational force goes inversely as the square of the distance) is very terse. He's not proving that these two things are equivalent - only that the they are equivalent for circular orbits.
"The reason for that, is something like this...."
First point - suppose that an orbit were a perfect circle. Because equal areas are swept in equal times, the planet must be traveling at a constant speed around the circle. This means that the velocity has a constant magnitude and a changing direction. We've already learned that the direction of the velocity change is "toward the sun", so what's left is the magnitude of that change per unit time.
Feynman tells us to consider what happens if we consider two points in the orbit separated by some small angle (exaggerated in the image below). If we consider the difference between the initial and final velocities, we see that we have an isosceles triangle, where the speed determines the sides. Euclid would have us bisect the angle between them, and notice that the velocity is now the hypotenuse of a the triangle. The base (change in velocity) is going to be the hypotenuse (the velocity) multiplied by some value which depends on the angle alone.
Therefore Δ v ≈ v.
How long did it take to make this change? Well, the angle is fixed, so we are talking about some fraction of the length of the orbit; so the time taken must be the same fraction of the period.
So if
a Δt = Δv
then it follows that we can substitute the period T times some constant for Δt, and the velocity v times some constant for Δv ...
a Δt = Δv
a T ≈ v
a ≈ v / T
Now, v and T are already related - speed times time equals distance; the velocity times the period is the circumference, which is proportional to the radius.
R ≈ v T
Most familiar at this point is to substutute R/v for T...
a ≈ v / T
a ≈ v / (R/v)
a ≈ v^2 / R
But Feynman prefers for this demonstration to replace v...
a ≈ v / T
a ≈ (R/T) / T
a ≈ R / T^2
... because Kepler's third law relates the square of the time (T^2) to the cube of the radius (R^3)
a ≈ R / T^2
a ≈ R / R^3
a ≈ 1 / R^2
Not that it is only in this last substitution that we finally discover the inverse square relationship - if there had been some other relationship between the period and the distance, then we would get a force law. In fact, Newton in the Principia demonstrates a number of substitutions for the period, just to show that it would work.
If you aren't happy with this approach, and want an "essentially geometric demonstration", you can get to v^2 / R by applying the Pythagorean theorem.
R^2 + (v Δt)^2 = ( R + A Δt ) ^2
R^2 + (v Δt)^2 = R^2 + 2R A Δt + (A Δt) ^2
(v Δt)^2 = 2R A Δt + (A Δt) ^2
At this point we seem to have a mess, but only because we haven't accurately expressed the acceleration. Since we obtain a distance by multiplying A by a time interval, it must be a velocity - what it really represents here is the average velocity caused by acceleration over the time interval. We're assuming that the acceleration is constant in time, so the changed velocity at the beginning of the interval would be zero, and at the end of the interval would be the rate of change of velocity times the interval (a Δt), so we can say that the average changed velocity is ( a Δt - 0 ) / 2. (This is the same result that you would get through calculus, of course - a(t) = dv/dt, v(t) = t dv/dt, x(t) = 1/2 t^2 dv/dt....)
(v Δt)^2 = 2R A Δt + (A Δt) ^2
(v Δt)^2 = 2R (1/2 aΔt) Δt + ((1/2 aΔt) Δt) ^2
(v Δt)^2 = R aΔt ^2 + (1/2 aΔt^2 ) ^2
v^2 Δt^2 = R aΔt ^2 + (1/2 a ) ^2 Δt^4
Taking very small time intervals, we can ignore anything smaller than Δt^2...
v^2 Δt^2 = R aΔt ^2 + O(Δt^4)
v^2 Δt^2 = R aΔt ^2 +0
v^2 = R a
v^2 / R = a
Update: Dot Physics has a similar derivation, using trigonometry.