Schwarzschild Metric Geodesic vs. Inverse Square Law of Gravity (How CLOSE They Are Visually)

I posted this video on youtube on January 28, 2020 showing how the inverse square law really is an extremely good approximation of a geodesic on a Schwarzschild Metric:

https://www.youtube.com/watch?v=KUKcp_2rd-Y


The graphs I use in the video are: (dy/dx)^2 = (2*3.985*10^14)/y for the inverse square law and (dy/dx)^2= (1-0.0089/y)^2*(299,000,000^2 – 299,000,000^2*(1-0.0089/y)) for the Schwarzschild metric geodesic


For the zoom of the graph, I first use the ranges:

X range is -6K to 6K

Y range is 0 to 1.4 x 10^7

Because the distance, Y is measured from the center of the earth instead of the earth’s surface, and earth’s radius is pretty big, 6.371 million meters.

I also zoom WAY in close later in the video, 

X range is -5 to 8

Y range is 0.0079 to 0.0099

To show the graphs aren't EXACTLY the same.

My general strategy was to take the inverse square law d/dx of dy/dx = k/y^2 where x is time and y is distance from the center of gravity, and turn it into a graph.  Then I took the relevant part of the Schwarszchild metric, ds^2=(1-rs/r)c^2*dt^2 – dr^2/(1-rs/r), and tried to figure out how to do a geodesic on it.  

Going to ideas from Sam Lilley’s book, it gave me a kind of formula for what a geodesic on a metric is.  If you draw a graph with a r axis and a t axis, you can think of dr and dt as infinitesimal distances on this graph.  On a graph where r and t would be flat spacetime, dr and dt would be equal everywhere.  But warped space has varying dr and dt at different points.  

We can think of ds as the hypotenuse of a triangle completing the distance moved dt in the t direction and then moved dr in the r direction.  This isn’t entirely accurate with the Schwarzschild metric because I think ds is actually smaller than either dt or dx here, but it’ll at least give us a rough idea of what’s going on.  If we take this picture here, and imagine that ds is ALWAYS proportional to dt, no matter how long or short dt is, we’ll get a geodesic – a line that doesn’t bend to the left or right as you move across it.  So we just need ds to be equal to some constant times dt and plug that into our metric to get a geodesic on that metric.  

So the next part of the strategy was to set ds equal to a constant B times (1-rs/r)dt.  Then plug it into the metric.  

After messy algebra, the equations for the graphs used in the video popped out.