Justin A. Parr is an experienced Technologist and Information Technology Leader based in the Dallas/Fort Worth, Texas area, having a diverse background that includes Consulting, Integration, Architecture, Project Management, and (long ago) Software Development.

**I’m the nerd that other nerds call, when they need help.**

Technical Professionals are welcome to connect to me on LinkedIn:

http://linkedin.com/in/justinparr

I do not do Facebook or Twitter.

http://justinparrtech.com/JustinParr-Tech/why-i-dont-facebook-nor-twitter/

**You can reach out to me by submitting a comment on this website.**

### About the JPTech logo

One day, long ago, I noticed that my initials “JP” form a hat and eye patch for a pirate smiley.

I don’t illegally copy nor advocate illegal copying of software and media, nor do I have a fascination with pirates. It’s just a cool looking thing I can do with my initials.

I can only identify with pirates in the most general, Jimmy Buffett sense — I chart my own course, and I probably won’t turn down a margarita.

I like to think that this logo reflects the light-hearted and “uncharted” exploration and sharing of science and technology.

for Mr. Parr. I joined because I saw you were knowledgeable in physics. I hope here is the place to ask a question. If earth were to turn around its axis twice as fast than present, then due to the virtual centrifugal force a 2Lbs weight would weigh how much? If you can answer then I would be able explain how there is a serious reason behind the question. Thanks, Michael

(Ugh… I just realized I posted a new comment rather than replying. Here is the reply, LOL)

Hi, Michael! “Mr. Parr” is my dad – you can call me “Justin” :-)

I did write a post that on the topic of centrifugal force due to the Earth’s spin:

http://justinparrtech.com/JustinParr-Tech/centrifugal-force-at-the-earths-surface/

In order to answer the question, we need to make a few assumptions. The major assumption is based on latitude, because the distance from the Earth’s axis of rotation to its surface varies from around 6,378 km at the circumference to zero at the north and south pole. At either pole, you would be standing directly along the axis of rotation, and thus the centrifugal force would be zero. So in the post above, I estimated that at 39 degrees north latitude, the rotational radius is about 4,951 km.

So I’ll answer the question three ways: a) 39 degrees north latitude at NORMAL rotation, b) 39 degrees north latitude at DOUBLE the angular speed, and c) at the equator (0 degrees latitude)

A: Rotational circumference at 39 degrees = 4,951 * 2 * pi = 31,092 km. Divide by 24 = 1,296 km/h rotational angular speed. Convert to m/s = 1,296 km/h * 1,000 m/km / 3,600 s/h = about 360 m/s. a (acceleration) = v^2/r, so a = 360^2 / 4,951,000 (all in meters) = 0.026 m/s/s. F=m • a, so 1 kg at 39 degrees north latitude experiences a force of 0.026 Newtons. At 4.45 N/lb, 1kg of mass experiences a force of 0.026 / 4.45 = 0.0058 lb. However, that’s the centrifugal force of 2.2 lbs, because 1kg mass has a gravitational force of 2.2lb, so we divide 0.0058 / 2.2 = 0.0026 lb of centrifugal force per lb of gravitational force due to mass. Multiply by 2 = 0.0052 lb, and since centrifugal force acts in the opposite direction of gravity we subtract this from 2 = 1.9948 lb total force.

So at NORMAL rotational speed, the centrifugal force at 39 degrees north latitude acting on a mass whose gravitational force is 2 lb = 0.0052 lb, with a total unbalanced force of 1.9948 lb, give or take some rounding errors, LOL

B: Because of the equation, a = v^2/r, if we double the velocity at a unit radius, we quadruple the acceleration. So 0.026 * 4 = 0.104 m/s/s at a 12 hour day. We can check this by assuming double the ground speed, or 720 m/s ^2 / 4,951,000 = 0.1047 — check, but let’s round up to 0.105 m/s/s acceleration. In essence, we could take our first set of results and multiply by 4, but let’s work it out with 0.105 m/s/s instead. So 1 kg experiences a force in lb of 0.105 / 4.45 = 0.023 lb. 1 kg = 2.2 lb, so a mass whose gravitational force = 1 lb experiences 0.023 / 2.2 = 0.0104. We can check this by dividing by 4 as compared to normal rotational speed = 0.0026 (check). And a mass of 2 lb would have double that, or 0.0208 lb of centrifugal force.

So at DOUBLE rotational speed, the centrifugal force at 39 degrees north latitude acting on a mass whose gravitational force is 2 lb = 0.0208 (4 times that of normal) for a total unbalanced force of 1.9792 lb.

C: At the circumference, which is our extreme case, we use the same process. With a circumference of about 40,075 (If you use 6,378 as a radius you get 40,054, but these numbers constantly change by a small amount, anyway), the Earth’s surface velocity at DOUBLE the rotational speed = 40,075 / 12 (12 hours is now a full day + night cycle) = 3,339.58 or really close to 3,340 km/h. Times 1,000 m/km and divided by 3,600 s/h = 928 m/s – almost 200 m/s faster than at 39 degrees north latitude. A = v^2 / r = 0.135 m/s/s (using 6,378,000 meters as the radius of rotation). Convert Newtons to lb to 1lb = 0.0138 lb of centrifugal force per lb of gravitational force. Double that for 2 lb: 0.0276

So at the equator, at DOUBLE rotational speed, a mass whose gravitational force is 2 lb would have a centrifugal force of 0.0276 lb, with a total unbalanced force of 1.9724 lb.

I may have some rounding errors, so don’t use this to do any life-critical calculations, but as a thought exercise, it should be fairly close.

One additional consideration: If the Earth’s rotational speed had been double as the Earth’s core cooled and its crust formed, it’s quite likely that the Earth’s radius at the circumference would be somewhat longer – it’s impossible to tell. If that’s the case, there would be a larger gradient between the radius at the circumference and the radius at the poles (currently the difference is only about 22km). The calculations above assume that the radius from the center of the Earth is relatively constant, but if it bulged by hundreds of km at the circumference, you would have to take that in to account as well.

Hope this helps!!

Pingback: HOME | Justin A. Parr - Technologist