The most important thing to understand about planetary orbits is that the orbital energy — the sum of kinetic and potential energy — is a constant. This means orbits in a vacuum require no energy source and can continue perpetually unless acted on by an outside force, consistent with Newton's First Law. It also means that an elliptical orbit, an orbit that changes its distance from the parent body as well as its velocity, represents a constant exchange between kinetic and potential energy, in a way consistent with the requirement for energy conservation.

First we will show and explain the equations for kinetic and potential orbital energy.

(1) $ \displaystyle e_k = \frac{mv^2}{2}$

Where:

• e_k = Kinetic energy, joules
• m = Mass of the moving body, kilograms
• v = velocity, m/s

(2) $ \displaystyle e_p = - G \frac{m_1 m_2}{r}$

Where:

• e_p = Potential orbital energy, joules
• G = Universal gravitational constant, described above
• m₁ and m₂ = Masses separated by distance r, kilograms
• r = Radius (distance) between m₁ and m₂, meters

How can energy be negative?

The potential gravitational energy equation (2) is derived from the gravitational force equation, by this operation:

(3) $ \displaystyle e_p = \int G \frac{m_1 m_2}{r^2} dr = - G \frac{m_1 m_2}{r} $

This shows that the negative sign for energy arises naturally in the mathematics, but it also follows from the fact that moving toward the parent body should decrease the potential gravitational energy acquired while moving away, and this relationship can only be expressed meaningfully if gravitational potential energy is seen as negative. This aspect of gravitational potential energy has important cosmological implications (see below).

The total orbital energy e_t of kinetic energy e_k and potential orbital energy e_p is:

(4) $ \displaystyle e_t = e_k + e_p = - G \frac{m_1 m_2}{r} + \frac{m_{2}v^2}{2} = \frac{m_{2} r v^{2} - 2 \, G m_{1} m_{2}}{2 \, r} $

Where:

• e_t = total orbital energy, joules
• e_k = kinetic energy, joules
• e_p = potential energy, joules
• v = Velocity of orbiting body, m/s
• m₁ = Mass of parent body, kilograms
• m₂ = Mass of orbiting body, kilograms
• r = Radius (distance) between m₁ and m₂, meters

Again, for an orbit with no forces apart from gravity, the total orbital energy e_t is a constant. For an elliptical orbit this means the ratio of kinetic to potential energy is constantly changing. As the orbiting body approaches the parent, kinetic energy increases (higher velocity) and potential energy declines. As the orbiting body moves away, the reverse.

The Zero-Energy Solution

Because kinetic and potential orbital energies have opposite signs, equation (4) has a zero-energy solution. If the kinetic energy in an orbit is modest, for example a circular orbit or any elliptical orbit, negative potential energy exceeds positive kinetic energy. But if the kinetic energy has a value consistent with something called "escape velocity" (see below) the total orbital energy is zero. This solution plays a part in cosmology, in particular the details of the Big Bang.

For more on this topic, click here.

Next, we will show some equations derived from those above.

(5) $ \displaystyle v_0 = \sqrt{\frac{GM}{r}} $

Where:

• v₀ = Orbital velocity required for a circular orbit, m/s
• G = Universal gravitational constant described above
• M = Mass of the parent body, kilograms
• r = Distance to the center of the parent body, meters

Notice about this equation that it doesn't include a mass term for the orbiting body. This means the equation is only reasonably accurate in a case where the orbiting body's mass is much less than that of the parent. Also, real-world natural orbits are rarely circular. An artificial example of a circular orbit is a geostationary satellite, which must have a circular orbit to fulfill its purpose (see below for more on this topic).

(6) $ \displaystyle v_e = \sqrt{\frac{2GM}{r}} $

Where:

• v_e = Minimum velocity required to escape from orbit, m/s
• G = Universal gravitational constant described above
• M = Mass of the parent body, kilograms
• r = Distance to the center of the parent body, meters

Escape velocity has a number of special properties. First, in the escape velocity profile, even though the body never returns, the escaping body's velocity constantly decreases, reaching zero at infinity (the asymptotic result). This means escape velocity has zero net energy (e_k + e_p = 0).

At any velocity less than escape velocity, the orbiting body has negative net energy (e_k + e_p < 0) so it will eventually return. At any velocity greater than escape velocity, the orbiting body escapes with positive net energy (e_k + e_p > 0), so it will have positive energy at an infinite distance.

Again, escape velocity plays a part in cosmological theory.

These two related equations give an orbital time for a radius argument and the reverse, for circular orbits.

Time for Radius:

(7) $ \displaystyle t_0 = \frac{2 \pi r^{(\frac{3}{2})}}{\sqrt{GM}} $

Where:

• t₀ = Orbital period time, seconds
• r = Radius from the center of the mass being orbited to the orbital height, meters
• G = Universal gravitational constant described above
• M = Mass of body being orbited, kilograms

Radius for Time:

(8) $ \displaystyle r = \frac{(2GM)^{\left(\frac{1}{3}\right)} t_{0}^{\left(\frac{2}{3}\right)}}{2 \, \pi^{\left(\frac{2}{3}\right)}} $

Where:

• r = Radius from the center of the mass being orbited to the orbital height, meters
• t₀ = Orbital period time, seconds
• G = Universal gravitational constant described above
• M = Mass of body being orbited, kilograms

These equations can be used to compute the required altitude for geostationary satellites, but remember these points:

• The radius arguments are with respect to the center of the orbited body, not the surface. To convert a radius result to an altitude above the surface, subtract the planet's radius from the result.
• The orbital period for a geostationary satellite must be expressed in sidereal time to account for the fact that an earthy 24-hour day is with respect to the ever-changing direction of the sun, but the true rotation rate of the earth is with respect to the background stars, and is therefore not quite equal to 24 hours (it is 23 hours, 56 minutes and 4.091 seconds, more or less).

It's possible to use kinetic equations to find the velocity required to get to a particular altitude, but in the general case, with sufficient altitude that the gravitational force changes, such a solution requires numerical methods. Let's say we want to know what the departure velocity should be for a ballistic flight to a given altitude h. Neglecting air resistance, we can write relatively simple differential equation terms to solve it:

• p(0) = r, the planet's radius
• p'(0) = v, the required velocity to achieve height h
• p''(t) = -GM/p(t)², the changing gravitational acceleration

As it turns out, if we assume that the gravitational acceleration changes enough to influence the outcome, this differential equation is insoluble in closed form — it must be processed numerically. But there is a much simpler way to solve such a problem — compute the departure energy, compute the energy at altitude h, write an equation that compares the two, and solve for the departure velocity v. We will solve two forms:

First, a ballistic flight to an altitude h with no velocity remaining:

(9) $ \displaystyle -\frac{GM}{r} + \frac{v^2}{2} = -\frac{GM}{r+h} $

The left-hand side of equation (9) is the sum of potential and kinetic energy at departure. The right-hand side is the potential energy at altitude h — the velocity is zero. By solving for v we get:

(10) $ \displaystyle v = \sqrt{\frac{\mbox{2GM} h}{h r + r^{2}}} $

And if we solve for h we get:

(11) $ \displaystyle h = -\frac{r^{2} v^{2}}{r v^{2} - 2 \, G M} $

Our second example is a ballistic flight that ends in a circular orbit, which means there is substantial velocity at altitude, and therefore kinetic as well as potential energy:

(12) $ \displaystyle -\frac{GM}{r} + \frac{v^2}{2} = \frac{-\mbox{GM}}{2 \, {\left(h + r\right)}} $

As before, the left side represents the potential and kinetic energy at departure, and the right side represents potential energy plus the kinetic energy of a circular orbit at altitude r+h. The solution for v:

(13) $ \displaystyle v = \sqrt{\frac{{\left(h + 2 \, r\right)} \mbox{GM}}{h^{2} + h r}} $

And the solution for h:

(14) $ \displaystyle h= -\frac{r v^{2} - G M - \sqrt{r^{2} v^{4} + 6 \, G M r v^{2} + G^{2} M^{2}}}{2 \, v^{2}} $

Here is a calculator for the above energy problem, selectable for the case with, and without, a circular orbit:

     

Mode:   No Orbit   Circular Orbit

Value	Entry	Control
G (gravitational constant, units m3 kg-1 s-2)
M (mass, units kilograms)
r (radius, units meters)
v (velocity, units meters/second)
h (height, units meters)

You may either enter a value for v and then click to solve for h, or the reverse.