Gravitation Near Earth's Surface — Question 4

Question 4

An equatorial scale supplies less force than gravity alone

A spherical planet rotates with angular speed $\Omega$. Its radius is $R=6.4\times10^6\,\mathrm m$ and gravitational surface acceleration is $g_0=10\,\mathrm{m/s^2}$. Take $\Omega=7.3\times10^{-5}\,\mathrm{s^{-1}}$. An object of mass $m$ rests on an equatorial scale.

1. Find the scale reading per unit mass in an inertial frame.

2. Find the rotation period at which the idealized equatorial normal force would vanish.

Question 4 - Solution

An equatorial scale supplies less force than gravity alone

(a) Derivation and calculation

Take inward as positive. The inertial-frame centripetal balance is $$mg_0-N=m\Omega^2R,\quad \Omega^2R=(7.3\times10^{-5})^2(6.4\times10^6)=.0341056\ \mathrm{m/s^2}.$$ Thus $$\boxed{N/m=g_0-\Omega^2R=9.9658944\ \mathrm{m/s^2}}.$$

(b) Result and interpretation

At the limiting rotation rate $N=0$, $$\Omega_c^2R=g_0,\quad \Omega_c=\sqrt{g_0/R}.$$ $$\boxed{T_c=\frac{2\pi}{\Omega_c}=2\pi\sqrt{\frac{6.4\times10^6}{10}} =5026.55\ \mathrm s=83.7758\ \mathrm{min}}.$$ This is the fixed spherical model’s limit for unsupported equatorial contact.

Physical check

At zero rotation, $N=mg_0$; a faster rate than the limit would require a downward restraint.