Calculating the fall of the moon in one second

[Silversonic]

Homework Statement

If I know only the circumference of the orbit of the moon, and the time it takes to make an orbit (29 days), how far does the moon fall in one second?

The Attempt at a Solution

I'm failing to understand how certain assumptions can be made in the geometry here. From Figure 1 of this link;

http://www.michaelbeeson.com/interests/GreatMoments/Newton.pdf

It appears that the straight line the moon's tangential trajectory $x$, plus the vertical displacement by gravity $s$ would meet at the exact path of the Moon's actual orbit in one second, point $B$. Normally this wouldn't be the case, but I guess it's indicating it's okay to make this assumption because we're only talking about 1 second of a 29-day orbit. But then again, how can I be sure this assumption won't affect my end result, considering my end result is also going to be VERY small (1/20 of an inch)? It seems like it should only be okay to make this assumption, about the perpendicularity, as long what I was calculating wasn't also very small.

 

[RUber]

I suppose that depends on the accuracy you require on your small estimate. If you want a tolerance within 1% of you estimate, i.e. .05 +/- .0005, then maybe you should not make the assumption. However, for most practical purposes, and looking at the precision of the other constants used in the calculation, you should be fine with this simplification.

Also, you can look at the portion of the arc traversed in the time, 1sec is 2pi/2505600 radians of the orbit. This should be small enough to treat this section of the arc as a straight line.

 

[dean barry]

You might want to look at the moons sidereal orbit time (about 27.3 days), not the synodic time as you have, also don't forget the Earth and moon orbit each other around a common point (barycentre)

 

[dean barry]

Heres the word data sheet on two body systems attached