Orbital Motion of Satellites

For a satellite of mass $m$ orbiting a planet of mass $M$ at orbital radius $r$ (measured from the planet's center), Newton's law of universal gravitation gives the attractive force: $$F_g = \frac{GMm}{r^2}$$ This force acts as the centripetal force keeping the satellite in its circular orbit: $$F_c = \frac{mv^2}{r}$$ Setting $F_g = F_c$ and cancelling $m$ from both sides: $$\frac{GM}{r^2} = \frac{v^2}{r}$$ Solving for orbital speed: $$v_{\text{orb}} = \sqrt{\frac{GM}{r}}$$ Notice that $m$ disappeared — the orbital speed depends only on the central mass $M$ and the orbital radius $r$, not on the satellite's own mass.

The orbital period $T$ is the time to complete one full revolution. Since the circumference of a circular orbit is $2\pi r$ and the satellite travels at speed $v_{\text{orb}}$: $$T = \frac{2\pi r}{v_{\text{orb}}} = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi\sqrt{\frac{r^3}{GM}}$$ Squaring both sides reveals Kepler's Third Law: $$T^2 = \frac{4\pi^2}{GM}\,r^3$$ This states that the square of the period is proportional to the cube of the orbital radius — a result Kepler discovered empirically decades before Newton derived it from first principles.

The total mechanical energy of a satellite combines kinetic and gravitational potential energy. Using $v_{\text{orb}}^2 = GM/r$, the kinetic energy is: $$K = \frac{1}{2}mv^2 = \frac{GMm}{2r}$$ The gravitational potential energy (taking zero at infinity) is: $$U = -\frac{GMm}{r}$$ The total orbital energy is therefore: $$E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$$ The negative sign indicates a bound orbit. To move a satellite to a higher orbit (larger $r$), energy must be added; $E$ becomes less negative (increases toward zero). A satellite with $E = 0$ has exactly escape velocity and follows a parabolic trajectory, while $E > 0$ corresponds to a hyperbolic (unbound) path.

A geostationary satellite appears stationary over a fixed point on the equator because its orbital period equals Earth's rotation period: $T = 24\text{ h} = 86400\text{ s}$. Using Kepler's Third Law, we can solve for the required radius: $$r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3}$$ Substituting $G = 6.674\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$, $M_E = 5.972\times10^{24}\,\text{kg}$, and $T = 86400\,\text{s}$ gives $r \approx 42{,}164\,\text{km}$ from Earth's center, or about $35{,}786\,\text{km}$ above the surface. All geostationary satellites share this single orbital radius.

• v_{\text{orb}} — Orbital speed (m/s): The tangential speed required to maintain a circular orbit at radius r
• G — Gravitational constant (N m² kg⁻²): Universal constant equal to 6.674 × 10⁻¹¹ N m² kg⁻²
• M — Mass of central body (kg): Mass of the planet or star being orbited (e.g., Earth: 5.972 × 10²⁴ kg)
• r — Orbital radius (m): Distance from the center of the central body to the satellite

Problem. The International Space Station (ISS) orbits Earth at an altitude of approximately 408 km above the surface. Earth's radius is $R_E = 6.371\times10^6\,\text{m}$, Earth's mass is $M_E = 5.972\times10^{24}\,\text{kg}$, and $G = 6.674\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$. (a) Find the orbital speed of the ISS. (b) Find its orbital period in minutes.

Answer. The ISS orbits at approximately 7.67 km/s and completes one orbit every 92.6 minutes.

• Centripetal Force and Acceleration — Orbital motion is maintained by gravitational force acting as centripetal force; understanding $F_c = mv^2/r$ is essential to deriving orbital speed.
• Newton's Law of Universal Gravitation — The gravitational force $F = GMm/r^2$ is the source of the centripetal acceleration in orbital motion and directly determines orbital speed and period.
• Gravitational Potential Energy — Total orbital energy combines kinetic energy and gravitational potential energy $U = -GMm/r$; negative total energy defines a bound orbit.
• Conservation of Angular Momentum — Angular momentum $L = mvr$ is conserved in orbital motion (no torque), which explains why satellites speed up at closer orbital radii in elliptical orbits (Kepler's Second Law).