Question

Our usual frame of reference is the surface of the Earth, which is actually a non-inertial reference frame since the Earth is spinning! In other words we are constantly accelerating, since we travel in a circle around Earth's rotation axis. For the questions below assume the Earth is spherical with a radius of 3959 miles and the latitude of Los Angeles is approximately 34°.

a) How much lighter will a scale read a 70 kg70 kg person's weight standing at the equator compared to one standing at the poles in lbs?

b) How much lighter will a scale read a 70 kg person's weight standing at the equator compared to one standing in LA in lbs?

c) If you are standing at the equator, what would the period of the Earth have to be in minutes so that the fictitious centrifugal force would make al m=70 kg person fly off the surface?

Answer 1

a)

At the poles it is 9.832m/s²9.832m/s² and at the equator it is 9.780m/s²9.780m/s².

You can use this formula

g(ß)=g−ш²(Rcosß)

Where g(ß) is acceleration due to gravity at a latitude ßß, g=9.832m/s²g=9.832m/s², шш is angular speed of the earth = 7.29 x 10^-5 rad/s, RR is the radius of the earth.

So, at poles

W = 70 x 9.832 = 688.24 N

at equator

W = 70 x 9.78 = 684.6 N

a)

Difference in weight = 3.64 N

= 3.64/9.81 = 0.371 kg = 0.82 lbs

b)

g = 9.832 - (7.29x10^-5)^2 x 6400 x 10^3 cos34

= 9.803

difference in g = 0.03

diff in Wt = 70 x 0.03
diff in kg = 70x0.03/9.81 = 0.214 kg

in lbs = 0.472 lbs

c)

mg = mrw^2

w^2 = g/r = 9.78 / (6400x10^3)

w = 1.24 x 10^-3

T = 2 x 3.14 / (1.24 x 10^-3) = 5080 sec

= 84.67 min