AP Physics 1- 6.6 Motion of Orbiting Satellites- Study Notes- New Syllabus

Motion of Orbiting Satellites

Satellites orbit planets due to the gravitational force acting as a centripetal force. Their motion is governed by Newton’s laws of motion and universal gravitation.

A satellite of mass \( m \) orbiting a planet of mass \( M \) at a distance \( r \) from the planet’s center experiences a gravitational force:

\( F_g = \dfrac{G M m}{r^2} \)

For a stable circular orbit, this gravitational force provides the centripetal force:

\( F_c = \dfrac{m v^2}{r} \)

Equating centripetal and gravitational forces:

\( \dfrac{m v^2}{r} = \dfrac{G M m}{r^2} \implies v = \sqrt{\dfrac{G M}{r}} \) where \( v \) is the orbital speed.

Time Period of a Satellite

The orbital period \( T \) is the time taken by a satellite to complete one full orbit around the planet. For a circular orbit, it is related to the orbital radius \( r \) and the mass of the planet \( M \) as follows:

From Newton’s law of gravitation and circular motion:

Gravitational force provides the centripetal force:

\( \dfrac{G M m}{r^2} = \dfrac{m v^2}{r} \implies v = \sqrt{\dfrac{G M}{r}} \)

The orbital period is the circumference divided by the orbital speed:

\( T = \dfrac{2 \pi r}{v} = \dfrac{2 \pi r}{\sqrt{\dfrac{G M}{r}}} = 2 \pi \sqrt{\dfrac{r^3}{G M}} \)

Key Point: This formula comes directly from combining Newton’s law of gravitation with the formula for circular motion and the definition of speed as distance over time.

Energy Considerations:

Translational kinetic energy of the satellite:

\( KE = \dfrac{1}{2} m v^2 = \dfrac{1}{2} m \cdot \dfrac{G M}{r} = \dfrac{G M m}{2 r} \)

Gravitational potential energy:

\( U = – \dfrac{G M m}{r} \)

Total mechanical energy:

\( E = KE + U = \dfrac{G M m}{2 r} – \dfrac{G M m}{r} = – \dfrac{G M m}{2 r} \)

Key Insight: The negative total energy indicates that the satellite is bound to the planet. Half of the magnitude of potential energy is converted into kinetic energy in a stable circular orbit.