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Orbital mechanics · Lesson 2

Describing an orbit

An ellipse describes where a spacecraft can go. Kepler's laws also tell us how it moves along that path and how long a complete orbit takes.

Circular speed was a special case. Here we keep gravity central but allow the radius and speed to vary.

Put the planet at a focus

An ellipse is an oval with two fixed points called foci. The sum of distances to the two foci is constant along the curve. In the two-body approximation, the central body lies at one focus, not at the ellipse's center. A circle is an ellipse whose foci coincide.

The semi-major axis a is half the ellipse's longest diameter. Eccentricity e measures its departure from circular shape. A circle has e = 0; bound ellipses have 0 ≤ e < 1. Periapsis is the closest point to the central body, and apoapsis is the farthest. Around Earth these are also called perigee and apogee.

Closest and farthest radii

r_{p} = a (1 - e) r_{a} = a (1 + e)

Recover the size

a = \frac{r_{p} + r_{a}}{2}

Recover the eccentricity

e = \frac{r_{a} - r_{p}}{r_{a} + r_{p}}

An orbit with rₚ = 7,000 km and rₐ = 13,000 km has a = 10,000 km and e = 0.30. Its closest altitude above a mean-radius Earth is 629 km. The semi-major axis is the average of these two extreme radii, not the average of every radius sampled uniformly in time.

Recover the geometry before calculating motion

The center of an ellipse is its point of symmetry. A focus is a different point, displaced along the long axis. Let the distance from the center to either focus be c. The eccentricity is c divided by a. Moving from the attracting focus to the nearer end of the long axis covers a minus c; moving to the farther end covers a plus c. Substituting c equal to a times e gives the two apsis formulas above.

Adding the closest and farthest radii cancels the focus displacement. The sum is twice the semi-major axis. Subtracting them instead gives twice the displacement. Dividing that difference by their sum therefore gives eccentricity. This derivation explains why the formulas use center-to-spacecraft radii rather than heights above the surface. The focus represents the attracting center, not the nearest patch of ground.

Consider a different Earth example with periapsis radius 8,000 km and apoapsis radius 12,000 km. Their sum is 20,000 km, so a is 10,000 km. Their difference is 4,000 km, so e is 0.20. The focus displacement is 2,000 km. The two altitudes are 1,629 km and 5,629 km after subtracting Earth's mean radius. Reconstructing the extreme radii from a and e returns 8,000 and 12,000 km, checking the calculation.

If you instead average those altitudes, you obtain 3,629 km. That number is not the semi-major axis to insert into Kepler's period law. Adding Earth's radius restores 10,000 km. Using the altitude average directly would describe a much smaller mathematical ellipse, largely inside Earth, rather than the stated orbit. Carrying units is necessary, but units alone cannot distinguish a radius from an altitude because both are lengths.

Keeping a fixed while increasing e moves the near end inward and the far end outward by equal amounts. It does not simply enlarge every distance. In the demonstration, the long ellipse has a closest radius of 0.40a and a farthest radius of 1.60a. Its full major-axis length is still 2a, the same as the circular reference's diameter. The focus stays at the attracting body while the center of the ellipse shifts relative to it.

This provides a physical feasibility check separate from the geometry. If a were 10,000 km and e were 0.60 around Earth, the nearest radius would be 4,000 km, below Earth's surface. The equation still defines an ellipse, but a spacecraft could not coast around that entire path outside Earth. The widget uses normalized distances and enlarged central markers; it does not claim that every displayed eccentricity is a viable orbit at a particular Earth altitude.

Finally, eccentricity is a shape ratio, not a distance or a percentage of time spent near the planet. Two ellipses can both have eccentricity 0.30 while having very different sizes and periods. Conversely, two paths with the same semi-major axis can differ in shape while sharing a period. Keep size, shape, and physical clearance separate when deciding what a pair of numbers actually specifies.

The craft moves faster near periapsis

Equal time does not mean equal distance

Eccentricity × 10030

Circular reference

Ellipse, e = 0.30

0% of one period. Closest radius: 0.70a. Farthest radius: 1.30a.

Speed Step 0 / 12

Both orbits have the same semi-major axis and central mass, so they have the same period. Each dot is one twelfth of a period apart. The two shaded sectors in the ellipse each sweep one twelfth of its total area. Distance scales match; central bodies are enlarged. Changing eccentricity resets playback.

Kepler's first law gives an ellipse with the attracting body at a focus. His second law says that the line from the focus to the spacecraft sweeps equal areas in equal times. Near periapsis that line is shorter, so it must turn through a larger angle in the same time to sweep the same area. The craft covers more distance there.

Do not animate an ellipse by moving through equal geometric angles at equal time steps. That would usually violate the second law. The demonstration solves Kepler's equation, which connects elapsed time to position. You do not need to solve that equation by hand to use the period and speed formulas.

Kepler's third law relates the period T to the semi-major axis:

Time for one orbit

T = 2 π \sqrt{\frac{a^{3}}{μ}}

a is the semi-major axis. With kilometers and seconds as the input units, T is in seconds.

Compare two orbits around the same body

\frac{T_{2}}{T_{1}} = {(\frac{a_{2}}{a_{1}})}^{\frac{3}{2}}

The power 3/2 means take the cube, then its square root. A circle of radius 7,000 km has a period of about 97.1 minutes. Doubling its radius multiplies the period by √8 ≈ 2.83, giving 274.8 minutes. A circle and an ellipse with the same a have the same period, even though their shapes and speeds differ.

For the 7,000 by 13,000 km ellipse, a = 10,000 km and T ≈ 165.9 minutes. Half the period is the time from periapsis to apoapsis. It is not generally the time needed to cover half the visible arc from an arbitrary starting point.

Use the dots as a clock rather than a ruler

Choose Circle in the eccentricity presets. The slider reads zero and both panels now show circles. Their dots are equally spaced because equal elapsed times correspond to equal angles and distances in circular motion. Step forward and compare the two craft. Their positions match throughout the cycle. This is the limiting case that an ellipse animation should recover when its eccentricity becomes zero.

Choose Moderate ellipse. The slider reads thirty, meaning eccentricity 0.30 rather than thirty. Changing this value resets playback. The two panels still have equal semi-major axes and equal periods. Each advance represents one twelfth of a period. The ellipse's dots spread out near the closest point and crowd together near the farthest point. That spacing records distance traveled in equal time, not changing time intervals.

At step six, both craft have completed half a period. The elliptical craft is at apoapsis, opposite its starting periapsis. At step twelve it has returned to the start. Now compare the two shaded sectors, one starting at periapsis and one at apoapsis. Each represents one twelfth of the ellipse's area and one twelfth of its period. Their outlines differ because their radii and turning angles differ. Equal area does not require the sectors to have the same shape.

For a small turning angle measured in radians, a narrow sector has area approximately one half times radius squared times angle. If the radius were twice as large, the angle would need to be one quarter as large to sweep the same small area. This is a local comparison; it does not replace the exact swept-area calculation over a large sector where radius changes substantially. It explains why uniform angle steps would give the wrong timing.

The program advances a clock-like angle uniformly, then solves Kepler's equation to locate the craft. That clock variable is called mean anomaly. It is not generally the actual angle of the spacecraft measured from the focus. The distinction matters because a visually smooth animation can still have incorrect timing. Here the equal-time samples and equal-area sectors let you inspect the intended physical relationship without having to implement the equation solver.

For the 10,000 km semi-major axis, the period formula first forms a cubed, giving one trillion cubic kilometers. Dividing by Earth's mu leaves approximately 2,508,778 square seconds. Taking its square root gives about 1,584 seconds; multiplying by two pi gives about 9,952 seconds, or 165.9 minutes. The intermediate square-root value is not yet the full period. Forgetting the factor of two pi would give a substantially shorter answer.

Period comparisons need the same central body. At fixed mu, increasing a by a factor of four increases a cubed by sixty-four and its square root by eight. The period therefore grows eightfold, not fourfold. If the central body's mu also changes, it no longer cancels in the ratio. A similarly sized orbit around a more massive central body has a shorter period, because its stronger gravity supports faster orbital motion.

A shape does not specify a complete orbit

You also need the plane of the orbit, the direction of its long axis within that plane, and the craft's position at a stated time. Inclination is the angle between the orbital plane and a chosen reference plane, often Earth's equator. An inclination of 0° is equatorial in the prograde direction; 90° is polar.

A top-down ellipse hides inclination. Two identical ellipses drawn on separate sheets could occupy different planes and never share the same spacecraft position except at possible crossings. Later we will separate changing an orbit's size from changing its plane.

These ellipses remain fixed only in the ideal model. Earth's oblateness and other bodies can rotate the plane and the line connecting periapsis to apoapsis. The model is useful because it isolates the main motion, not because every real satellite repeats an unchanging curve forever.

Add a plane, a direction, and a time reference

Imagine cutting the ellipse from a sheet of paper and placing the attracting focus at Earth's center. Tilting the sheet changes the orbital plane without changing a or e. Rotating the ellipse within that sheet changes where periapsis points. Moving the spacecraft marker around the unchanged curve changes its location at the chosen time. These are distinct freedoms that a size-and-shape description leaves unspecified.

Inclination fixes the tilt relative to a reference plane, but not every aspect of orientation. Two orbital planes can have the same inclination and be rotated differently around Earth's polar axis. Their intersections with the equatorial plane then point in different directions. A full orientation description identifies this intersection line as well as the direction of periapsis within the orbital plane. Later plane-change calculations must respect those actual intersections.

A time reference is often called an epoch. Saying that a craft is at periapsis at a specified epoch, together with the orbit and direction of travel, lets you advance its location using the elapsed fraction of the period. Saying merely that it is somewhere on the ellipse does not. If two craft share the same geometric orbit but occupy different locations, they need not meet just because enough time passes.

An alternative is to provide position and velocity vectors at the epoch. Those vectors specify where the craft is and how it is moving in a stated reference frame. Under the ideal two-body force law, they determine the subsequent trajectory. Orbital size, shape, orientation, and phase can then be derived. This is why navigation records require more than an image of a path or a single altitude measurement.

Some orientation labels lose their separate meaning in special cases. A perfect circle has no unique closest point, so an arbitrary periapsis direction cannot be physically distinguished from another. An exactly equatorial orbit likewise has no unique line where its plane crosses the equator, because the planes coincide. The motion remains well defined. It is the chosen description that needs care, not the physical orbit that becomes indeterminate.

A time average also differs from a geometric average. The arithmetic mean of periapsis and apoapsis radii gives a by definition. A craft spends more time in the more distant part of an eccentric orbit, so sampling its radius at equal time intervals weights that region more heavily. You cannot infer a by treating an average of such samples as though it were the average of the two extremes. The dot pattern provides a visible reason for the distinction.

When checking an orbit description, ask which prediction it supports. Two apsis radii can establish size and eccentricity. Adding mu supplies the ideal period. Predicting a location relative to Earth at a particular time also needs orientation, direction, and phase. Predicting a close approach to another craft requires its state too. Each additional question needs information that the previous, narrower calculation could legitimately leave out.

Exercises

Recover the ellipse from its extremes

An Earth orbit has closest radius 7,000 km and farthest radius 13,000 km. Find a, e, and both altitudes.

Hint

Use the sum for a and the difference divided by the sum for e. Subtract 6,371 km only when converting a radius to altitude.

Worked answer

Compare the two ends of the same ellipse

Periapsis
7,000 km · 8.604 km/s

Apoapsis
13,000 km · 4.633 km/s

Earth sits at a focus, not at the ellipse's center. Radii are measured from the focus. Points mark equal time intervals; the denser points near apoapsis show slower motion.

a = (7,000 + 13,000)/2 = 10,000 km. e = 6,000/20,000 = 0.30. Periapsis altitude is 7,000 − 6,371 = 629 km; apoapsis altitude is 6,629 km. These radii describe the ellipse, but its orientation and the craft's position still need specification.

Does twice the radius mean twice the period?

Compare circular Earth orbits of radii 7,000 km and 14,000 km. Calculate the period ratio and explain why it exceeds two.

Hint

Apply the period ratio before inserting any dimensional constants.

Worked answer

A larger orbit takes disproportionately longer

7,000 km radius: 97.14 minutes

14,000 km radius: 274.76 minutes

These are circular two-body orbits around Earth. Doubling radius multiplies the period by 2√2, about 2.83, rather than by two.

No. The period increases by more than a factor of two.

Period multiplier when radius doubles

\frac{T_{2}}{T_{1}} = \sqrt{2^{3}} = \sqrt{8} \approx 2.828

Starting from 97.1 minutes gives about 274.8 minutes. The path is larger and the circular speed is lower, so a simple doubling of travel time is insufficient.

Find the faulty animation

Two animations show the same ellipse. One advances the craft by equal angles about the focus each second. The other advances it through equal swept areas. Which obeys Kepler's second law, and why?

Hint

Equal time requires equal swept area, not equal angle or equal arc length.

Worked answer

Equal swept areas

The amber and green sectors cover equal time intervals. Their angular widths differ. The orbit uses e = 0.6; focus marker enlarged.

The equal-angle animation is wrong for an eccentric orbit. For a small angular interval, the swept area is approximately one half times r² times the angle in radians. Near apoapsis, a larger r would make that equal-angle slice sweep more area in the same time. The correct animation turns through less angle there and more near periapsis.

Sources and model limits

The calculations use Newtonian two-body motion and ideal instantaneous burns. They omit atmospheric drag, an uneven gravity field, other attracting bodies, and finite engine thrust unless the text explicitly discusses them. They are teaching calculations, not flight instructions.

NASA, Basics of Space Flight: Gravity & Mechanics. Ellipses, Kepler laws, gravity, and orbital motion.
OpenStax, University Physics: Kepler laws of planetary motion. A textbook treatment of periods, angular momentum, and transfer ellipses. Some optional derivations use calculus.

Earth examples use μ = 398,600.4418 km³/s² and a rounded mean radius of 6,371 km. Reported answers are rounded, not measurements. The Sun-centered example uses astronomical units and years with a one-year circular Earth orbit.

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