Pocket-calculator Science (1)

 Pocket-calculator Science (1)    

    I enjoy viewing the earth as a sphere, for then I can calculate its volume, surface area, etc using the tricks I learned in middle school. 

    OK, it is not quite spherical, for the circumference through the poles is 39,942.209 km, while the circumference at the equator is 40,074.156 km. (I shall sometimes simplify by assuming it is a sphere with an average diameter of 40,008 km).

    The force of gravity (Fg) was found by Newton to follow the equation 

Fg =G.m1.m2/r^2

where G= the universal gravitational constant, m1 and m2 the mass of the man and the earth respectively, and r is the distance between the two centres of mass.  Imagine a man of mass 72 kg. Clearly, when standing at the north pole he is closer to the centre of the earth than when standing at the equator.  

    Taking the gravitational constant (G) as 6.674 x (10)^-11 ( N. m^2/Kg^2)

and the mass of the earth (m2) as 5.972 x (10)^24, and Newton's relation: F =G.m1.m2/r^2, from these I calculate the gravitational force at the north pole (Fnp) to be 9.8628 Newtons;  while the gravitational force at the equator (Feq) to be 9.798 N. (An average value is often quoted as 9.81 Newtons.)

    Applying this geometrical effect alone (ignoring any centrifugal effect) I would expect the man (of mass 72kg) would weight at north pole 72 x 9.8628 = 710 N; and at the equator 72 x 9.798 = 705 N. 

    However, the earth rotates on its axis once a day, relative to the sun. (Relative to the fixed stars it rotates a bit more because of its rotation round the sun in one day; approximately 361º in a 24 hour day, or 15.04º in one hour.)

    At the equator (circumference 40,074,156 m`), we are rotating eastwards with an angular velocity (𝜔) of 2𝛑 radians per day, but a quasi-linear velocity of 40,074.156 ÷ 24 = 1669.57 km/hr. The inertial centrifugal force (Ff), regarded as acting radially, is usually calculated  as F=m.r .𝜔^2 (where m signifies mass and r radius). A man of mass 72 kg (weight = 710 Newtons at the north pole) would find his weight reduced at the equator not only because of his greater distance from the centre of the earthe, but by a small extra amount due to the centrifugal force acting (as far as he is concerned) vertically upwards.

    If T is the rotational period (1 day, or 86,400 s), we can write the rotational velocity (𝜔) as 2𝛑 ÷ 86,400 radians per second.

    We can then write the centrifugal force (Ff) as:

    Ff = m.r.𝜔^2 = m.r.(2𝛑 /T)^2

Taking r as 40,074.156/2𝛑 km = 6,378 km; or rather 6,378,000 m:

    Ff = 72 x 6,378,000 x (2𝛑 ÷ 86,400)^2 Newtons = 2.4286 Newtons. (Note that at the north pole the man rotates at the same speed, but r= 0, so he experience no centrifugal force.)

    So, combining the geometrical effect of the greater radius at the equator and the spin which is negligible at the north pole, a man that weighed 710 Newtons at the north pole would be expected to weigh only 705 - 2.43 = 702.53 Newtons at the equator. 

(Next post Coriolis, if I succeed in understanding it!)

(Comments are welcome to cawstein@gmail.com)