What is the Linear Velocity of a Location on Earth's Equator?

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Homework Help Overview

The problem involves calculating the linear velocity of a location on Earth's equator based on its rotation. The context includes the Earth's rotation period and the radius at the equator.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss using the formula V=rw for linear velocity, with attempts to clarify the calculations involving angular velocity and the radius of the Earth. Questions arise regarding potential errors in decimal placement and calculator input.

Discussion Status

Some participants confirm the original poster's method as correct, while others suggest checking numerical inputs. There is an exploration of alternative approaches to the problem, including calculating velocity using distance over time.

Contextual Notes

Participants note the absence of specific homework equations and the potential for misinterpretation of numerical values. The discussion reflects a collaborative effort to verify calculations and clarify understanding.

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Homework Statement



Earth rotates in on axis that goes through both the North and South Poles. It makes one complete revolution in 24 hours. If the distance from the axis to any location on the equator is 3960 miles, find the linear speed (in miles per hour) of a location on the equator.

Homework Equations



-None-

The Attempt at a Solution



I used the formula V=rw to find the linear velocity, and for 'r' I used 3960, and 'w' is used 2pi/24 for making one revolution per hour, so I simplified that down to pi/12 as angular velocity.

V=3960 x (pi/12 radians/hour)≈ 1036.73 miles/hour

Is this correct? It got checked in school, but I think I was off in the decimals or something. Is there something I did wrong or?

Thanks for any help!
 
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darshanpatel said:

Homework Statement



Earth rotates in on axis that goes through both the North and South Poles. It makes one complete revolution in 24 hours. If the distance from the axis to any location on the equator is 3960 miles, find the linear speed (in miles per hour) of a location on the equator.

Homework Equations



-None-

The Attempt at a Solution



I used the formula V=rw to find the linear velocity, and for 'r' I used 3960, and 'w' is used 2pi/24 for making one revolution per hour, so I simplified that down to pi/12 as angular velocity.

V=3960 x (pi/12 radians/hour)≈ 1036.73 miles/hour

Is this correct? It got checked in school, but I think I was off in the decimals or something. Is there something I did wrong or?

Thanks for any help!

Your method is correct, but check the numbers. 3690*3.14/12 is way off from your result.

ehild
 
I re-did it, and it came out to the same? Maybe it is something you inputted into your calc, I did pi/12 first then multiplied by 3960, and it came out to this?

I don't know what I am off at? I even did it the way you wrote it out, but it still worked?
 
Mark44 said:
Looks OK to me.

Thanks! :D
 
Sorry, I misread the number 3960:shy:. Your result is correct.

ehild
 
Another way to do this, though essentially the same, is to use "velocity= distance/time". You are given that the radius of the earth, at the equator, is 3960 so the circumference is [itex]2\pi(3960)= 24881[/itex] miles. Divide that by 24 hours to get 1036.7 mph.
 

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