Linear speeds at points on Earth

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balletgirl
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Homework Statement


What is the linear speed of a point:
a) on the equator,
b) on the Arctic Circle (latitude 66.5 degrees N),
c) at a latitude of 45 degrees N, due to the Earth's rotation?

Given
Radius (earth)= 6.38x10^6m
v?

Homework Equations


v=rw
w= 2pi*f
t= 1/f


The Attempt at a Solution


I am not sure how go about doing this problem. I wanted to solve for w with 2pi(1/24) but am not sure how to calculate the speed at different areas on Earth.
 
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balletgirl said:

Homework Statement


What is the linear speed of a point:
a) on the equator,
b) on the Arctic Circle (latitude 66.5 degrees N),
c) at a latitude of 45 degrees N, due to the Earth's rotation?

Given
Radius (earth)= 6.38x10^6m
v?

Homework Equations


v=rw
w= 2pi*f
t= 1/f


The Attempt at a Solution


I am not sure how go about doing this problem. I wanted to solve for w with 2pi(1/24) but am not sure how to calculate the speed at different areas on Earth.

You've got the radius of the sphere and thus the circumference. Use the time period 24 hours. For higher latitudes, the radius decreases as the COS of the latitude as does the velocity because the distance traveled decreases.
 
I don't understand how to make this change though. Is there an equation I can use?
 
Hi balletgirl! :smile:

(have a pi: π and a degree: º and try using the X2 tag just above the Reply box :wink:)

The point goes round a circle (of latitude) …

what is the radius of this circle? :smile:
 
Hello,

The radius is 6.38 x 10^6 m. Would I add [Cos(θ)] to the equation v=rω and make it v=R*Cos(θ)*ω for parts b and c? (I tried it already for part b, but got a negative answer).

By the way, I have v=463.83 m/s for part a.
 
Tiny Tim, Though your posts are usually helpful, your signature is somewhat distracting just now.

Balletgirl, what part don't you understand?
 
Last edited:
balletgirl said:
I don't understand how to make this change though. Is there an equation I can use?

The circumference of the circle is proportional to its radius, which is proportional in this case to its latitude as SIN of the Lat.
 
balletgirl said:
Hello,

The radius is 6.38 x 10^6 m. Would I add [Cos(θ)] to the equation v=rω and make it v=R*Cos(θ)*ω for parts b and c? (I tried it already for part b, but got a negative answer).

By the way, I have v=463.83 m/s for part a.

It would appear we became asynchronous.