Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pl...Asked by heather
The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of 6.38 x 106 m, determine the centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 52.0 ° north of the equator.
Answers
Answered by
Drew
What do we know?
r=6.38x10^6m
T=1 day=86,400s
Part A:
T=2(pi)r/V ==> V=2(pi)r/T
a=V^2/r (Plug & chug)
Part B:
Now, what do we know?
Only T=1 day=86,400s. We no longer know r.
Latitude we'll say is represented by "l".
So r(l) = r cos l
Once you find that radius, you follow the same series of equations as Part A.
r=6.38x10^6m
T=1 day=86,400s
Part A:
T=2(pi)r/V ==> V=2(pi)r/T
a=V^2/r (Plug & chug)
Part B:
Now, what do we know?
Only T=1 day=86,400s. We no longer know r.
Latitude we'll say is represented by "l".
So r(l) = r cos l
Once you find that radius, you follow the same series of equations as Part A.