# Characteristics of Circular Orbits

**Circular orbits** are the simplest kinds of orbits in celestial mechanics, where an orbiting body remains at constant radius as it travels around a gravitating mass. In reality, no orbits of astronomical bodies are perfectly circular, since every object is constantly perturbed by the gravity of a very large number of other nearby objects like the planets in the solar system. However, when a small object like a satellite, asteroid, or small moon orbits a large object like a planet or star, it is a good approximation to treat the system as a two-body system with the larger body fixed. Below, the characteristics of a small satellite orbiting a massive planet at uniform speed in perfectly circular orbit are derived.

## The Orbital Velocity

In the two-body approximation, holding the larger mass fixed, it is possible to compute the velocity needed to maintain a perfectly circular orbit, using the formulas for centripetal acceleration and the Newtonian gravitational force.

Denote the mass of the large planet by \(M \), the mass of the satellite by \(m\), and the distance between the centers of mass of the two objects by \(r\). Assume that the planet is sufficiently massive as to approximately remain fixed. Then if the satellite is in circular orbit, the planet attracts the satellite towards itself with a gravitational force \( F = \frac{ G M m } { r^{2} } \), which is equal to the centripetal force keeping the satellite in circular orbit:

\[ \frac{ G M m } {r^{2} } = \frac{ m v^{2} }{ r }. \]

Rearranging, one immediately finds the velocity (speed) of the satellite in circular orbit:

\[ \boxed{v = \sqrt{ \frac{ GM } { r } } }.\]

The direction of the velocity is always directed tangent to the circle of orbit. Note that the speed of the satellite is independent of the mass of the satellite, since the small \(m\) was cancelled from both sides of the above. This shows that orbital velocity is a characteristic of the planet, not necessarily the bodies orbiting it.

From the above equation, one can also derive the angular velocity \(\omega\) and the period of rotation \(T\) of the satellite. Using \(v = r \omega\) and \(T = \frac{2\pi}{\omega}\), the basic equations of angular kinematics, one finds:

\[\boxed{\omega = \sqrt{ \frac{ GM } { r^{3} } }, \qquad T = 2 \pi \sqrt{ \frac{ r^{3} } { GM } }}.\]

The period of Earth's orbit around the sun is 365 days. Given that the mass of the sun is about \(2 \times 10^{30} \text{ kg}\), compute the velocity to the nearest 1000 meters per second of the Earth as it orbits the sun. Approximate the orbit of the Earth around the sun as circular.

**Useful constant**: Newton's gravitational constant is \(6.67\times 10^{-11} \text{ N} \cdot \text{m}^2 / \text{kg}^2\).

## Angular Momentum and Central Forces

Recall that the angular momentum of an object is in general given by the equation:

\[\vec{L} = \vec{r} \times \vec{p},\]

where \(\vec{r}\) is the vector from the center of rotation and \(p\) is the linear momentum. In circular orbit, the position vector is always perpendicular to the momentum (which is tangential to the path), so the angular momentum reduces to \(L = rp = mvr\). Plugging in for the velocity \(v\) of rotation as derived above, the angular momentum of the satellite with respect to the planet is

\[ \boxed{L = m \sqrt{ GMr } }.\]

This is consistent with the alternative derivation using \(L = I \omega\) and the moment of inertia of a point mass \(I = mr^2\), since this yields:

\[L = mr^2 \sqrt{\frac{GM}{r^3}} = m\sqrt{GMr},\]

consistent with the above.

Because the force of Newtonian gravity is a central force, there is no torque on a satellite orbiting a planet: the force is always parallel to the position vector. Therefore, angular momentum is conserved always in Newtonian gravitation, provided there are no externally acting forces. In circular orbit, since the radius is fixed, one can see that the conservation of angular momentum is equivalent to the fact that the angular velocity remains constant.

A satellite of mass \(m\) circularly orbiting a planet of mass \(M\) has an angular momentum per unit mass of \(a\). How long does the satellite take to complete a revolution around the planet?

## Energy of a Bound Satellite

The kinetic, potential, and total mechanical energies of an object in circular orbit can be computed using the usual formulae, with the orbital velocity derived above plugged in.

Recall that the kinetic energy of an object in general translational motion is:

\[K = \frac12 mv^2.\]

Plugging in for the velocity, one obtains the kinetic energy of the satellite:

\[ \boxed{K = \frac{ GM m }{2r } }\]

Because Newtonian gravity is an inverse-square force, the potential energy of the satellite is just:

\[U = -\frac{ G M m } { r } \]

The total energy of the satellite, the sum of the kinetic and potential energies, is therefore:

\[E = K + U = - \frac{ G M m } { 2r }.\]

The total energy is negative in circular orbit. This makes sense because circular orbits can be thought of as bound states, much like the electron in a hydrogen atom. It requires positive energy to send the satellite out to infinity and zero energy, so the satellite must start at negative energy.

A small cosmic object of mass \(m\) a distance \(r\) from a planet of mass \(M \gg m\) has total mechanical energy:

\[E = -\frac{GMm}{3r}.\]

Which of the following correctly describes the behavior of the small object?

**Cite as:**Characteristics of Circular Orbits.

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