- #1
Silversonic
- 130
- 1
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 [itex] x [/itex], plus the vertical displacement by gravity [itex] s [/itex] would meet at the exact path of the Moon's actual orbit in one second, point [itex] B [/itex]. 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.