Correct formula for deflection of a falling object

[davidwinth]

Hello,

I found a derivation on NASA and several others that say something different. Can someone tell me which is correct? They differ by a factor of 3. This is confusing because some places say an object falling from 100m should deflect by 3cm but NASA says .16mm. Those are more different than a factor of 3, so I am not understanding what the correct formula is.

https://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/falling_eastward.htm

http://hepweb.ucsd.edu/ph110b/110b_notes/node14.html

 

[anorlunda]

This is not an answer to your question, but the two linked problems are not the same. One is for 32 degrees latitude and the other for 42 degrees.

 

[jbriggs444]

From the Nasa Article: "With

= 7.27 x 10^-7/sec"

What is the angular rotation rate of the Earth in radians/sec?

 

[davidwinth]

From the Nasa Article: "With View attachment 205117 = 7.27 x 10^-7/sec"

What is the angular rotation rate of the Earth in radians/sec?

I get 2*pi/(24*3600) = 7.3E-5. Is NASA wrong on such a basic thing?

 

[jbriggs444]

I get 2*pi/(24*3600) = 7.3E-5. Is NASA wrong on such a basic thing?

Anyone can slip a couple of digits. Especially if they fail to sanity-check their work.

Sanity-check: The difference in velocity of a ball at 100 m altitude and the ground at 0 m altitude is the same as that of a ball moving in a 100 m radius circle once every 24 hours (*). That's 2 pi times 100 m every 86400 seconds. Multiply that by a 4.5 second fall time and you get 3 cm. [Rather than trying to do a double integral of Coriolis acceleration in the rotating frame, I'm just using the inertial frame. We have a moving object falling onto a more slowly moving surface]

You sanity-check the 4.5 second fall time by reasoning that a 5 meter fall time is 1 second and that 100 meters is a factor of 20 farther so the fall time should be $\sqrt{20}$ times longer. 20 is about halfway between 16 and 25, so its square root should be about halfway between 4 and 5.

(*) We're using the back of an envelope and need not worry about sidereal versus solar days. And I'm doing my back of the envelope at the equator.

 

[davidwinth]

Thanks! Do you know where the factor of 3 came from in the non-NASA formulas? I have found several derivations and they all include that factor.

 

[jbriggs444]

Thanks! Do you know where the factor of 3 came from in the NASA formula?

They were integrating $gt^2$. The integral is $\frac{1}{3}gt^3$.

Edit: or perhaps you were referring to this factor of 3...

With h = 1000 m, l = 42 deg, we find

t = 14.3 sec

Multiply height by a factor of 10 and you've multiplied fall time by a factor of approximately 3