A satellite orbits the Earth in a circular orbit at an altitude of $500,\text{km}$ above the Earth’s surface. Given that the Earth’s mass is $5.98\times 10^{24},\text{kg}$, determine (i) the orbital radius and (ii) the orbital speed of the satellite.

Step-by-step solution

1. 1. Compute orbital radius $r$ as the sum of Earth’s radius $R_E=6.371\times 10^6,\text{m}$ and the altitude $h=5.00\times 10^5,\text{m}$: $r=R_E+h=6.871\times 10^6,\text{m}$.
2. 2. Use the expression for orbital speed in a circular orbit: $v=\sqrt{\frac{GM}{r}}$ where $G=6.674\times 10^{-11},{\text{N,m}}^2{\text{/kg}}^2$ and $M=5.98\times 10^{24},\text{kg}$.
3. 3. Compute $GM=6.674\times 10^{-11}\times 5.98\times 10^{24}=3.989\times 10^{14},{\text{N,m}}^2\text{/kg}$.
4. 4. Divide by $r$: $\frac{GM}{r}=3.989\times 10^{14}/6.871\times 10^6=5.807\times 10^7,{\text{m}}^2{\text{/s}}^2$.
5. 5. Take the square root: $v=\sqrt{5.807\times 10^7}=7.63\times 10^3,\text{m/s}\approx 7.6\times 10^3,\text{m/s}$.
6. 6. State the results: $r=6.87\times 10^6,\text{m}$ and $v=7.6\times 10^3,\text{m/s}$.

Concept tested: Orbital dynamics

Published 4 October 2026 · How this solution was prepared and checked