Determine the radius $r$ of the orbit of an Earth satellite that is above the same point on the Earth's surface at all times.
Decision:
The orbit along which the satellite must travel is in a plane passing through the Earth's equator. Otherwise, the latitude at which the satellite is located will vary. In addition, the orbit must be circular, since in an elliptical orbit the satellite, unlike the Earth, rotates irregularly. Thus, we need to find the radius r of a circular orbit in the equatorial plane with the period of the satellite's revolution $T = 24 hours$. From the equation of motion of the satellite we obtain
$\frac{v^{2}}{r} = \frac{GM}{r^{2}}$,
where $v$ is the velocity of the satellite, $M$ is the mass of the Earth, and $G$ is the gravitational constant. Noting that $g= \frac{GM}{R^{2}_{earth}}$, we obtain for the period of revolution
$T=2 \pi \frac{r}{v}=2 \pi \sqrt{\frac{r^{3}}{ R^{2}_{eaerth}g}}$
From this expression we easily find the radius of the satellite's orbit
$r=R_{earth} \left ( \frac{gT^{2}}{4 \pi^{2}} R_{earth} \right )^{1/3}=6.4R_{earth}=42.24 \cdot 10^{2} km$
Decision:
The orbit along which the satellite must travel is in a plane passing through the Earth's equator. Otherwise, the latitude at which the satellite is located will vary. In addition, the orbit must be circular, since in an elliptical orbit the satellite, unlike the Earth, rotates irregularly. Thus, we need to find the radius r of a circular orbit in the equatorial plane with the period of the satellite's revolution $T = 24 hours$. From the equation of motion of the satellite we obtain
$\frac{v^{2}}{r} = \frac{GM}{r^{2}}$,
where $v$ is the velocity of the satellite, $M$ is the mass of the Earth, and $G$ is the gravitational constant. Noting that $g= \frac{GM}{R^{2}_{earth}}$, we obtain for the period of revolution
$T=2 \pi \frac{r}{v}=2 \pi \sqrt{\frac{r^{3}}{ R^{2}_{eaerth}g}}$
From this expression we easily find the radius of the satellite's orbit
$r=R_{earth} \left ( \frac{gT^{2}}{4 \pi^{2}} R_{earth} \right )^{1/3}=6.4R_{earth}=42.24 \cdot 10^{2} km$