Irodov 1.223: The Satellite That Never Moves – Amit & Deepali Sinha Foundation

Problem Irodov 1.223

A satellite travels a circular orbit about the Earth and remains motionless with respect to a point on the Earth’s surface, while the Earth itself turns with rotation period $T$ . Given the Earth’s radius $R$ , the free-fall acceleration $g$ at its surface, and $T$ , find the orbital radius $r$ , together with the satellite’s speed $v$ and acceleration $a$ as measured in the inertial reference frame that instantaneously coincides with the Earth’s center.

A satellite riding the ring co-rotates with the Earth's own spin ω, so it always sits over the same point of the equator; its velocity v is tangent to the orbit and its acceleration a points inward, toward the center. — A satellite riding the ring co-rotates with the Earth’s own spin $\vect\omega$ , so it always sits over the same point of the equator; its velocity $\vect v$ is tangent to the orbit and its acceleration $\vect a$ points inward, toward the center.

Given

Find

r

;

v

and

a

in the instantaneous Earth-centred inertial frame

Solution

Step 1The condition for the satellite to hang motionless

For the satellite to stay above one fixed point of the rotating surface, its own angular velocity about the Earth’s center must equal the Earth’s rotation rate, and the orbit must lie in the equatorial plane, since a circular orbit is centered on the Earth’s center and only an equatorial one keeps the sub-satellite point fixed; with $T$ the Earth’s rotation period, that rate is

\omega = \frac{2\pi}{T}. \tag{1}

Step 2Circular orbit condition, written through

\omega

Gravity again supplies the centripetal force, now expressed through $\omega$ rather than a speed, since it is the angular rate that must match the Earth’s; equating the two forms and solving for $r$ gives the orbital radius in terms of $\omega$ .

\begin{align*}\frac{\gamma M m}{r^2} &= m\omega^2 r \\ \gamma M &= \omega^2 r^3 \\ r^3 &= \frac{\gamma M}{\omega^2}\end{align*}

Substituting (1),

r^3 = \gamma M\left(\frac{T}{2\pi}\right)^2.

r = \left[\gamma M\left(\frac{T}{2\pi}\right)^2\right]^{1/3}

(2)

Step 3Evaluate the radius

Using $\gamma M=gR^2$ and the Earth’s rotation period $T=8.6164\times10^4\ \text{s}$ (one revolution relative to the distant stars) in (2) builds up the numerical value of $r^3$ one factor at a time.

\begin{align*}\gamma M &= gR^2 = (9.8)(6.37\times10^6)^2 = 3.98\times10^{14}\ \text{m}^3/\text{s}^2 \\ \frac{T}{2\pi} &= 1.371\times10^4\ \text{s} \\ r^3 &= \gamma M\left(\frac{T}{2\pi}\right)^2 = (3.98\times10^{14})(1.881\times10^8\ \text{s}^2) = 7.48\times10^{22}\ \text{m}^3 \\ r &= (7.48\times10^{22}\ \text{m}^3)^{1/3} = 4.21\times10^7\ \text{m}\end{align*}

r = \left[\gamma M\left(\dfrac{T}{2\pi}\right)^2\right]^{1/3} = 4.2\times10^4\ \text{km}

(3)

Step 4Speed and acceleration in the inertial frame

In the inertial frame instantaneously centered on the Earth, the satellite still traces its circle at the constant rate $\omega$ , so its speed and its centripetal acceleration follow directly from $r$ and $\omega$ .

\begin{align*}v &= \omega r = (7.29\times10^{-5}\ \text{s}^{-1})(4.21\times10^7\ \text{m}) \\ a &= \omega^2 r = (7.29\times10^{-5}\ \text{s}^{-1})^2(4.21\times10^7\ \text{m})\end{align*}

v = \omega r = 3.1\ \text{km/s}

(4)

a = \omega^2 r = 0.22\ \text{m/s}^2

(5)

Check

The acceleration can be found a second, independent way, directly from Newton’s law of gravitation, $a=\gamma M/r^2=(3.976\times10^{14})/(4.213\times10^7)^2=0.224\ \text{m/s}^2$ , matching $\omega^2 r$ to three figures, as it must for any circular orbit. Also $v^2/r = (3.07\times10^3)^2/(4.21\times10^7)=0.224\ \text{m/s}^2$ agrees too.

Remark

The Earth’s rotation period entering $\omega$ must be the sidereal day, $T=8.6164\times10^4$ s, the time for one turn relative to the distant stars, not the $8.64\times10^4$ s mean solar day: a satellite fixed over one ground point keeps pace with the Earth’s rotation in the inertial frame, which the stars mark and the Sun does not. Using the mean solar day instead changes $r$ by only about $0.2$ percent, too small to show up at this problem’s two-figure precision, but the sidereal day is the physically correct choice, and the gap between the two would matter at higher precision.