Irodov 1.219: Gravity, Spin, and the Sun Compared – Amit & Deepali Sinha Foundation

Problem Irodov 1.219

Compare three accelerations felt by a body resting on the Earth’s surface: $w_1$ , produced by the Earth’s own gravitational pull; $w_2$ , the centrifugal acceleration arising from the Earth’s rotation, evaluated at the equator; and $w_3$ , produced by the Sun’s gravitational pull on a body at the Earth’s location. Find the ratio $w_1 : w_2 : w_3$ .

The Earth's rotation about its axis. At the equatorial point shown, w_2 is the outward centrifugal acceleration; a point at another latitude carries the inward gravitational pull w_1, which is the same everywhere on the (uniform, spherical) surface. — The Earth’s rotation about its axis. At the equatorial point shown, $w_2$ is the outward centrifugal acceleration; a point at another latitude carries the inward gravitational pull $w_1$ , which is the same everywhere on the (uniform, spherical) surface.

The Sun's direct gravitational pull w_3 on a body at the Earth (not to scale). — The Sun’s direct gravitational pull $w_3$ on a body at the Earth (not to scale).

Given

$g$
$\gamma M_\oplus$ , $R_\oplus$ , $\omega_\oplus$
$\gamma M_\odot$ , $d$

Find

the ratio

w_1:w_2:w_3

Solution

Step 1

w_1

: gravity at the Earth’s surface

$w_1$ is the Earth’s gravitational pull alone, so it is computed from the Earth’s mass and radius rather than read off as the measured free-fall acceleration: $g$ is smaller than $w_1$ by the small centrifugal correction $w_2$ that Step 2 computes.

w_1 = \frac{\gamma M_\oplus}{R_\oplus^2} = \frac{(6.67\times10^{-11})(5.96\times10^{24})}{(6.37\times10^{6})^2} = 9.80\ \text{m/s}^2 \tag{1}

At the two-figure precision carried here that coincides with the measured $g=9.8\ \text{m/s}^2$ , which is why the same number serves in the ratios of Step 4.

Step 2

w_2

: centrifugal acceleration at the equator

With the Earth’s rotation angular speed $\omega_\oplus = 2\pi/T$ , using the mean solar day $T=8.64\times10^4\ \text{s}$ ,

\begin{align*}\omega_\oplus &= \frac{2\pi}{8.64\times10^4\ \text{s}} = 7.27\times10^{-5}\ \text{s}^{-1} \\ w_2 &= \omega_\oplus^2 R_\oplus = (7.27\times10^{-5})^2(6.37\times10^6\ \text{m})\end{align*}

w_2 = \omega_\oplus^2 R_\oplus \approx 3.37\times10^{-2}\ \text{m/s}^2

(2)

Step 3

w_3

: the Sun’s pull on a body at the Earth

\begin{align*}w_3 &= \frac{\gamma M_\odot}{d^2} = \frac{(6.67\times10^{-11})(1.97\times10^{30})}{(1.50\times10^{11})^2}\end{align*}

w_3 = \frac{\gamma M_\odot}{d^2} \approx 5.84\times10^{-3}\ \text{m/s}^2

(3)

Step 4Form the ratio

Dividing (2) and (3) by (1) gives the ratio in the requested order.

\begin{align*}\frac{w_2}{w_1} &= \frac{3.37\times10^{-2}}{9.8} = 3.4\times10^{-3} \\ \frac{w_3}{w_1} &= \frac{5.84\times10^{-3}}{9.8} = 6.0\times10^{-4}\end{align*}

w_1:w_2:w_3 = 1 : 3.4\times10^{-3} : 6.0\times10^{-4}

(4)

Check

All three are accelerations, m/s $^2$ , so the ratio is dimensionless, as it must be. $w_2/w_1\approx0.3\%$ is the measured equatorial weight deficit: objects weigh a few tenths of a percent less at the equator than they would if the Earth stood still. $w_3/w_1\approx0.06\%$ shows the Sun’s direct pull is three orders of magnitude smaller than the Earth’s own gravity at the surface.

Remark

Unlike $w_2$ , which is felt directly as the equatorial reduction in weight, nobody feels the Sun’s pull $w_3$ at all, and the reason has nothing to do with its being small: the Earth and everything on it are in free fall around the Sun, so in the Earth’s own (orbiting) frame $w_3$ is almost exactly canceled by the orbital motion, and only the much smaller residual, the tidal effect coming from how the Sun’s pull varies across the Earth’s diameter, survives. $w_2$ , by contrast, is not canceled by anything: it is felt directly as a latitude-dependent reduction in weight.