Question

An object is suspended on a spring balance in a ship sailing

along the equator with a speed v. Show that the scale reading

will be very close to where  is the angular

speed of the Earth and W0 is the scale reading when the ship

is at rest. Explain the plus or minus.

Answer 1

Find the force of gravitational attraction assuming the earth does not rotate:

$$F_1=mg.$$

Find the force of gravitational attraction taking into consideration the fact that the earth rotates with angular speed $\omega$:

$$F_2=m(g-a)=m(g-\omega^2R).$$

Find the force of gravitational attraction assuming that the earth rotates with angular speed $\omega$ and the ship is sailing along the equator at speed v in the direction of rotation of the planet:

$$F_3=m(g-a-a_s)=m\bigg(g-\omega^2R-\frac{v^2}{R}\bigg).$$

Find the ration of numerical values for $F_2$ and $F_3$:

$$\frac{F_2}{F_3}=\frac{g-\omega^2R}{g-\omega^2R-v^2/R},\\\space\\ \frac{F_2}{F_3}=\frac{9.8-[2\pi/(24\cdot3600)]^2\cdot6375\cdot10^3}{9.8-[2\pi/(24\cdot3600)]^2\cdot6375\cdot10^3-v^2/(6375\cdot10^3)}=\\\space\\ =\frac{9.77}{9.77-\frac{v^2}{6375\cdot10^3}}.$$

Since the value of v for ships is quite small (some meters per second), divided by 6375000 the scales will show almost no difference compared to weight of a body resting somewhere at the equator.