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# 2 dimensional motion problem

Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis with a period of 24 hours. Consider only the effect of this rotation on the person on the surface. (Ignore Earth's orbital motion about the Sun.)
(a) What is the speed and what is the magnitude of the acceleration of a person standing on the equator?

Express the magnitude of this acceleration as a percentage of g.

(b) What is the direction of the acceleration vector?

(c) What is the speed and what is the magnitude of the acceleration of a person standing on the surface at 35°N latitude?

(d) What is the angle between the direction of the acceleration of the person at 35°N and the direction of the acceleration of the person at the equator if both persons are at the same longitude?

### 1 Answer by Expert Tutors

BRUCE S. | Learn & Master Physics & Math with Bruce SLearn & Master Physics & Math with Bruce...
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Hello Ben,

First some numbers:
Radius of the earth: Re=6.38 * 10^6 meters
Circumference of the Earth: Ce =Pi * 2 * Re
Ce =40.1 * 10^6 meters

The person at the equator travels the circumference of the earth every 24 hours.

Their speed, Se, is Ce/24 hrs:

Se=C/24
=Pi * 2 * R / 24
= 40.1 * 10^6 meters / 24hr
= 1.67 *10^6 meters/hr
In MKS units (ie seconds)
Se=1.67 *10^6 meters/hr * (1hr/3600 sec)
Se = 4.64 * 10^2 m/s

The acceleration of an object traveling in a circle is given by:

A = r * W ^2 where W is the angular speed of the object

For the person at the equator the equation becomes:

Ae = Re * W^2
W = 2*Pi/24 hr (this result in radians per hr)

Ae= 6.38 * 10^6 meters * (2*Pi /24 hr)^2
Ae =6.38 * 10^6 meters * (0.0685 )
Ae = 4.37 * 10^5 meter/hr^2
In MKS units:
Ae= 4.37 * 10^5 meter/hr^2 (1hr^2/3600^2sec^2)
Ae=3.37 * 10^-2 meter/sec^2

This is the Centripetal acceleration and is directed from the equator to the earth’s rotational axis and is perpendicular to the rotational axis.

Earth’s gravitational acceleration is 9.8 meters/sec^2. As a fraction, the centripetal acceleration at the equator is:

{3.37 * 10^-2 m/s^2 } / 9.8 m/s^2 = 3.43 * 10^1

or Ae ~ 34.3% of gravity

Now we move to 35°N latitude. The speed and centripetal acceleration is less because the radius between the earth’s surface and the earth’s rotational axis is now less. What is the effective radius at 35N? It is determined by the complement of 35N and the radius of the earth, Re. Here is a suggestive diagram showing the angles and radii.

*<= Rotational Axis
*
*
********** <=R @ 35N = Re*Sin(55)
*           *
* 55   *
*     * <=Re
*   *
* * 35N
**************** <= Equator

We use the same equations as before but now the effective radius is reduced by a factor equal to Sin(55) or 0.819.

The equator results were:
Se = 4.64 * 10^2 m/s
Ae= 3.37 * 10^-2 m/s^2

The results at 35N are:
S(35N) = 0.819 * 4.64 * 10^2 = 3.80 * 10^2 m/s
A(35N) = 0.819 * 3.37 * 10^-2 = 2.76 * 10^-2 m/s^2

Finally the centripetal force is always perpendicular to the rotational axis. Therefore all objects at the same longitude will have parallel centripetal accelerations. Ae and A(35N) are parallel.
BruceS