Irodov 1.226: Earth's Spin Splits Satellite Kinetic Energy

Problem Irodov 1.226

A satellite skims the Earth’s surface in a circular orbit lying in the equatorial plane, radius essentially equal to the Earth’s radius $R$ . Two senses of travel are possible: with the Earth’s own rotation, or against it. Working in a reference frame fixed to, and rotating with, the Earth, find the ratio of the satellite’s kinetic energy when it travels against the Earth’s rotation to its kinetic energy when it travels with the Earth’s rotation.

The satellite m on a ring of radius R about the spinning Earth. v_1 is its velocity traveling with the Earth's own spin Ω, v_2 the same orbit run the other way. — The satellite $m$ on a ring of radius $R$ about the spinning Earth. $\vect v_1$ is its velocity traveling with the Earth’s own spin $\vect\Omega$ , $\vect v_2$ the same orbit run the other way.

Given

circular equatorial orbit at $r\approx R$
two senses: with or against the Earth’s rotation $\vect\Omega$

Find

the ratio

K_{\text{against}}/K_{\text{with}}

of the kinetic energies measured in the Earth-fixed frame

Solution

Step 1Orbital speed and the Earth’s own rotation rate

Gravity supplies the centripetal force for the circular orbit at radius $R$ , where $M$ is the Earth’s mass; the Earth itself turns uniformly at rate $\Omega$ about its axis, carrying a surface point at the equator along at speed $u$ .

\frac{\gamma M m}{R^2} = \frac{mv^2}{R} \;\Longrightarrow\; v=\sqrt{\frac{\gamma M}{R}}=\sqrt{gR}, \qquad u = \Omega R = \frac{2\pi R}{T}. \tag{1}

Step 2Speed relative to the ground, in each sense

A point of the Earth-fixed frame at the satellite’s radius moves eastward at $u$ . Traveling with the rotation, the satellite’s own velocity is in that same sense, so the two speeds subtract; traveling against it, the velocities are opposite, so the two speeds add.

\begin{align*} v_{\text{with}} &= v - u, \\ v_{\text{against}} &= v + u. \tag{2}\end{align*}

Step 3The kinetic energy ratio

The satellite’s mass is the same in both cases, so it cancels from the ratio of the two kinetic energies, leaving only the ratio of the relative speeds squared.

\frac{K_{\text{against}}}{K_{\text{with}}} = \frac{\dfrac12 m v_{\text{against}}^2}{\dfrac12 m v_{\text{with}}^2} = \left(\frac{v+u}{v-u}\right)^2.

Step 4Evaluate

Using $g=9.8\ \mathrm{m/s^2}$ , $R=6.37\times10^6$ m, and the Earth’s sidereal rotation period $T=8.6164\times10^4$ s (the period relevant here, since $u$ is the ground’s speed in the same inertial frame the orbit itself is analyzed in), (1) gives

v = \sqrt{gR} = 7.901\ \mathrm{km/s}, \qquad u = \frac{2\pi R}{T} = 0.4645\ \mathrm{km/s}.

Substituting into the boxed ratio,

\frac{K_{\text{against}}}{K_{\text{with}}} = \left(\frac{7.901+0.4645}{7.901-0.4645}\right)^2 = \left(\frac{8.366}{7.436}\right)^2.

\frac{K_{\text{against}}}{K_{\text{with}}} = \left(\frac{v+u}{v-u}\right)^2 \approx 1.27

(3)

Check

With $x=u/v\approx0.0588$ , expanding $(1+x)^2/(1-x)^2\approx1+4x+8x^2$ gives $1+4x\approx1.235$ and, adding the next term $8x^2\approx0.028$ (about $2.2\%$ of the total), $\approx1.263$ , close to the exact $1.265$ to within the still-neglected $O(x^3)$ terms, confirming the algebra. Setting $u=0$ (a non-rotating Earth) collapses the boxed ratio to exactly $1$ , as it must: with no preferred spin direction the two senses would be identical.

Remark

The satellite’s kinetic energy in the inertial frame, $mv^2/2$ , is exactly the same for both senses of travel, since both are circular orbits of the same radius and hence the same speed $v$ . The whole $1.27$ -fold difference this problem asks for is therefore not a difference in the satellite’s actual orbital energy at all: it is purely a statement about which frame is doing the measuring, the Earth-fixed frame subtracting a different ground speed depending on which way the satellite passes overhead.