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 $800 \mathrm{~km}$. The angular speed of the satellite as it orbits the Earth is $\left(\mathrm{M}_{\mathrm{E}}=5.98 \times 10^{24} \mathrm{~kg}, \mathrm{R}_{\mathrm{E}}=6.37\right.$ $\times 10^{6} \mathrm{~m}, \mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}$ )
A) $1.04 \times 10^{-3} \mathrm{rad} / \mathrm{s}$.
B) $1.44 \times 10^{-3} \mathrm{rad} / \mathrm{s}$.
C) $9.50 \times 10^{-4} \mathrm{rad} / \mathrm{s}$.
D) $2.20 \times 10^{-3} \mathrm{rad} / \mathrm{s}$.
E) $1.90 \times 10^{-3} \mathrm{rad} / \mathrm{s}$.

Question

Quizplus · Accepted Answer

The angular speed ($\omega$) of a satellite in orbit can be found using the formula for gravitational force providing the centripetal force: $F_{\text{gravity}} = F_{\text{centripetal}}$, which simplifies to $G\frac{M_{E}m}{(R_{E}+h)^2} = m\omega^2(R_{E}+h)$. Solving for $\omega$ gives $\omega = \sqrt{\frac{G M_{E}}{(R_{E}+h)^3}}$, where $h = 800 \times 10^3 \, \text{m}$, $M_{E} = 5.98 \times 10^{24} \, \text{kg}$, $R_{E} = 6.37 \times 10^6 \, \text{m}$, and $G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2$. Plugging in these values gives an angular speed close to $1.04 \times 10^{-3} \, \text{rad/s}$.

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