Encyclopedic Guide to Circular Motion and Gravitational Dynamics

Uniform circular motion is defined as the motion of an object traveling at a constant speed along a circular path.
While the speed remains constant, the velocity is continuously changing because the direction of the object's motion is constantly changing at every point along the path.
The period $T$ is the time taken for one complete revolution of the object.
The frequency $f$ is the number of revolutions completed per unit time, defined by the relationship: $f = \frac{1}{T}$
Angular displacement $\theta$ is the angle swept out by the radius connecting the object to the center of the circle, measured in radians $\text{rad}$ .
Angular velocity $\omega$ is the rate of change of angular displacement with respect to time: $\omega = \frac{\Delta \theta}{\Delta t}$
For one complete revolution, the angular displacement is $2\pi$ radians, leading to the formulas: $\omega = \frac{2\pi}{T}$ $\omega = 2\pi f$
The linear speed $v$ of the object is related to its angular velocity and the radius of the path $r$ by: $v = \omega r$
Substituting the expressions for $\omega$ gives the linear speed as: $v = \frac{2\pi r}{T}$

Because the velocity vector is constantly changing direction, an object in uniform circular motion always experiences acceleration.
This acceleration is directed toward the center of the circular path and is known as centripetal acceleration $a_c$ .
The magnitude of centripetal acceleration is given by: $a_c = \frac{v^2}{r}$
Using the relationship $v = \omega r$ , centripetal acceleration can also be expressed as: $a_c = \omega^2 r$
According to Newton's Second Law ( $F = ma$ ), a net force must be acting on the object to produce this acceleration.
This net force is the centripetal force $F_c$ , which always acts perpendicular to the direction of motion and toward the center of the circle.
The magnitude of the centripetal force is: $F_c = \frac{mv^2}{r}$
Or, in terms of angular velocity: $F_c = m\omega^2 r$
It is critical to note that centripetal force is not a special type of force but is the name given to the net force (provided by tension, friction, gravity, or normal force) that maintains circular motion.

Every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
The formula for the magnitude of this gravitational force $F_g$ is: $F_g = G \frac{M m}{r^2}$
In this equation:
- $G$ is the universal gravitational constant, approximately equal to $6.67 \times 10^{-11}\,\text{N}\,\text{m}^2\,\text{kg}^{-2}$ .
- $M$ and $m$ are the masses of the two objects.
- $r$ is the separation distance between the centers of the masses.
This law follows an inverse square relationship; if the distance between the two masses is doubled, the gravitational force reduces to one-fourth of its original value.
The gravitational field strength $g$ at a distance $r$ from the center of a mass $M$ is defined as the gravitational force per unit mass: $g = \frac{F_g}{m} = \frac{GM}{r^2}$
At the surface of the Earth, where $r$ is equal to the radius of the Earth $R_E$ , the gravitational field strength is approximately $9.81\,\text{m/s}^2$ .

A satellite is an object that orbits a larger body under the influence of gravity. For a satellite in a stable circular orbit, the gravitational force provides the necessary centripetal force.
By equating gravitational force to centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$
Solving for the orbital speed $v$ of the satellite: $v = \sqrt{\frac{GM}{r}}$
This expression shows that the orbital speed of a satellite is independent of its own mass $m$ and depends only on the mass of the central body $M$ and the radius of the orbit $r$ .
To find the orbital period $T$ , we substitute $v = \frac{2\pi r}{T}$ into the orbital speed equation: $\left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r}$ $\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r}$
Rearranging for the period results in Kepler's Third Law (Harmonic Law): $T^2 = \frac{4\pi^2}{GM} r^3$
This demonstrates that the square of the orbital period is proportional to the cube of the orbital radius: $T^2 \propto r^3$

A geostationary satellite is one that remains at a fixed position relative to a point on the Earth's surface.
For a satellite to be geostationary, it must meet three specific criteria:
1. It must orbit in the same direction as the Earth's rotation (West to East).
2. It must have an orbital period exactly equal to the Earth's rotational period ( $T = 24\,\text{hours}$ or $86,400\,\text{seconds}$ ).
3. Its orbit must be directly above the Earth's equator (an equatorial orbit).
By substituting $T = 8.64 \times 10^4\,\text{s}$ into Kepler's Third Law, the required orbital radius for a geostationary satellite is calculated to be approximately $4.22 \times 10^7\,\text{m}$ .
Subtracting the Earth's radius ( $6.37 \times 10^6\,\text{m}$ ) gives the altitude above the Earth's surface, which is roughly $3.6 \times 10^7\,\text{m}$ ( $36,000\,\text{km}$ ).
Geostationary satellites are used extensively for telecommunications and weather monitoring because they provide constant coverage of a specific geographic area.
Polar satellites orbit at a lower altitude and pass over the North and South Poles. These satellites move quickly relative to the Earth's surface and are used for remote sensing, high-resolution Earth imaging, and environmental monitoring.

Astronauts and objects inside an orbiting satellite feel "weightless," but they are not outside the reach of gravity. In fact, gravity is the only force acting on them, providing the centripetal acceleration needed to stay in orbit.
This sensation is called "apparent weightlessness."
Since the satellite and all its contents are in a state of continuous free-fall toward the center of the Earth at the same rate, there is no normal (contact) force between the astronaut and the floor of the spacecraft.
Since scales and human senses rely on the normal force to detect "weight," the absence of this force results in the sensation of having no weight.

Question: Does the mass of a satellite affect its required orbital speed for a specific radius?
Answer: No. Based on the derivation $\frac{GMm}{r^2} = \frac{mv^2}{r}$ , the mass of the satellite $m$ cancels out. The required speed depends only on the mass of the planet being orbited and the distance from its center.
Question: What happens to the orbit if the speed of a satellite is suddenly increased?
Answer: If the speed increases, the gravitational pull will no longer be sufficient to keep the object in its current circular path at that radius. The satellite will move into an elliptical orbit with a larger average radius or may reach escape velocity if the speed increases enough.
Question: Why must geostationary satellites be over the equator?
Answer: If a satellite were at an angle to the equator, it would move North and South relative to a ground observer during its orbit. To remain perfectly stationary relative to a point on the ground, the orbital plane must coincide with the plane of the Earth's rotation, which is the equatorial plane.