A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800 km. The angular speed of the satellite as it orbits the Earth is (M_E = 5.98 × 10²⁴ kg, R_E = 6.37 × 10⁶ m, G = 6.67 × 10⁻¹¹ N·m²/kg²) A) 9.50 × 10⁻⁴ rad/s. B) 1.04 × 10⁻³ rad/s. C) 1.44 × 10⁻³ rad/s. D) 1.90 × 10⁻³ rad/s. E) 2.20 × 10⁻³ rad/s.

Question

Quizplus · Accepted Answer

The angular speed ($\omega$) of a satellite in orbit can be found using the formula for the gravitational force providing the centripetal force: $F_{\text{gravity}} = F_{\text{centripetal}}$, which simplifies to $G\frac{Mm}{r^2} = m\frac{v^2}{r}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, $m$ is the mass of the satellite (which cancels out), and $r$ is the distance from the center of the Earth to the satellite. The angular speed is given by $\omega = \frac{v}{r}$, and the orbital speed $v$ can be found from the rearranged equation $v = \sqrt{\frac{GM}{r}}$. Given $M = 5.98 \times 10^{24}$ kg, $G = 6.67 \times 10^{-11}$ N·m$^2$/kg$^2$, and the height of the satellite above the Earth's surface is 800 km (800,000 m), the total distance $r$ from the center of the Earth is $r = 6.37 \times 10^6 m + 800,000 m = 7.17 \times 10^6 m$. Substituting these values into the formula for $v$ and then finding $\omega$, we get an angular speed that matches choice B, indicating the calculations align with the given values and the principles of orbital mechanics.

A 5,000 Kg Satellite Is Orbiting the Earth in a Circular