Gravitation, planetary motion, and orbits

A satellite keeps falling

Imagine a small satellite moving sideways above a planet. Gravity continually changes its velocity toward the planet. With a suitable initial position and velocity, its path can keep missing the surface. An orbit is motion under gravity; being in orbit does not mean gravity has disappeared.

We will first hold the much larger primary approximately fixed in an inertial frame. Position is measured from its center. This is a useful limit when its mass M is much greater than the orbiting mass m. Later we will return to what changes when both bodies move.

Units, standards, and physical constants prepares the units; Algebra, functions, graphs, and logarithms prepares substitution and square roots; Vectors and vector spaces prepares direction and perpendicular components. The force, energy and angular-momentum ideas needed here are introduced locally. No unpublished mechanics article is required.

An attractive force decreases with distance

Let r be the positive distance between two point masses. Newtonian gravity has magnitude F = GMm/r², where G is the gravitational constant. For a point outside a spherically symmetric primary, its mass can be treated as concentrated at the center. The distance is measured from that center, rather than from the surface.

Write μ = GM, the gravitational parameter, with units m³/s². For the position vector r from the fixed primary to the satellite, the force on the satellite is F = −μm r/|r|³. The minus sign makes the force attractive. The primary experiences the equal and opposite interaction force, although its acceleration is much smaller when M is much larger.

Keeping the masses fixed and doubling the separation makes the force one quarter as large. This rule concerns separation: doubling an altitude above a surface generally does not double the center-to-center distance. We will use the invented parameter μ = 20 m³/s² and satellite mass m = 1 kg for arithmetic, without identifying the model with a real planet.

First balance a circular orbit

For a circle of radius r traversed at constant speed v, the inward acceleration is v²/r. Gravity supplies that acceleration: μm/r² = mv²/r. There is one force in this model. “Centripetal” names its inward role and does not add a second force that balances gravity.

Cancel m to obtain v² = μ/r. At r = 5 m, v² = 4 m²/s², so the circular speed is 2 m/s. Its inward acceleration is 0.8 m/s². The speed is constant, but the velocity changes because its direction changes.

The period T is the time for one complete circuit. Circumference divided by speed gives T = 2πr/v = 5π s in this example. A different sideways speed at the same radius generally starts a different orbit; the circular formula cannot be applied to every point of an ellipse.

One ellipse, two different speeds

A nonradial bound orbit under the isolated inverse-square interaction is an ellipse, including a circle as a special case. Its semimajor axis a is half its longest diameter. Eccentricity e describes its shape: 0 ≤ e < 1, with e = 0 for a circle. In the fixed-primary limit, the primary is at a focus, which generally differs from the ellipse’s center.

Our original ellipse has a = 5 m and e = 0.6. Its semiminor axis is b = a√(1 − e²) = 4 m. Put the primary at (0, 0) m and the ellipse center at (−3, 0) m. The nearest point, periapsis, is (2, 0) m; the farthest, apoapsis, is (−8, 0) m. Their distances from the primary are a(1 − e) = 2 m and a(1 + e) = 8 m.

The diagram uses one scale for both spatial directions. A square marks periapsis and a triangle marks apoapsis; a diamond marks the primary, and a cross marks the ellipse center. Velocity arrows have their own common scale, point tangent to the path and show the counterclockwise speeds 4 m/s and 1 m/s. They compare two instants on one orbit, rather than showing the satellite in two places at once or recording elapsed times.

Angular momentum explains the speed change

For a satellite of fixed mass, angular momentum about the primary is L = r × mv. Gravity is parallel to the radius, so its torque r × F about this origin is zero. Thus L stays constant even though the satellite has a nonzero force and its linear momentum changes.

In the orbit plane, write h = |r × v| = |L|/m, the specific angular momentum, with units m²/s. At periapsis and apoapsis the radius and velocity are perpendicular, so h = rv there. Our values give (2 m)(4 m/s) = (8 m)(1 m/s) = 8 m²/s. Away from those two points, use the perpendicular velocity component v⊥: h = rv⊥. The full speed generally is not all perpendicular to the radius.

An imaginary radius from the primary sweeps out area at the rate dA/dt = h/2. Here that rate is 4 m²/s. A constant rate gives equal swept sector areas in equal time intervals: Kepler’s area law. It does not give equal distances traveled. A straight triangle between two finite-separated positions is only an approximation to the swept region bounded by the curved orbit.

Energy stays fixed while speed changes

Choose gravitational potential energy zero at infinite separation. For the fixed-primary model, U = −μm/r. Kinetic energy is K = mv²/2, and total mechanical energy is E = K + U. The negative potential energy reflects this reference choice; it is not a negative speed or a negative mass.

At periapsis, K = 8 J and U = −10 J. At apoapsis, K = 0.5 J and U = −2.5 J. Both sums are −2 J. The satellite slows while moving outward because kinetic energy decreases as potential energy increases. Its total energy remains fixed under the assumed conservative interaction.

For an ellipse in this model, E = −μm/(2a) and therefore v² = μ(2/r − 1/a). The circular example with radius 5 m has the same a and the same energy −2 J, but a different angular momentum. Energy alone does not fix the shape and direction of an orbit.

Negative E gives bound nonradial Kepler motion. With the same reference, E = 0 is the escape-energy threshold, and E > 0 permits unbound motion. At a given radius, the corresponding escape speed is √(2μ/r). Energy and speed alone do not show that a trajectory avoids a solid primary or an atmosphere. Radial collision paths are outside the nonradial ellipse classification used here.

The period follows the semimajor axis

Kepler’s period law for the fixed-primary problem is T² = (4π²/μ)a³. For circles, a equals r. For ellipses, a is half the sum of the nearest and farthest distances; it is not the instantaneous radius and is not generally the time average of that radius.

The area of our ellipse is πab = 20π m². Divide by its constant areal rate 4 m²/s to obtain T = 5π s. The circle of radius 5 m also takes 5π s. Their shapes, speeds and angular momenta differ, but the same semimajor axis and gravitational parameter give the same period.

For orbits about the same primary in this approximation, doubling a multiplies T² by 8 and T by 2√2. Comparing periods around different primaries also requires their different gravitational parameters. An orbit measurement can constrain that parameter only after the geometry, timing and model assumptions have been checked.

Return to two moving bodies

In the exact isolated Newtonian two-body problem, both bodies move about their shared center of mass. For their separation vector, the orbital formulas use the gravitational parameter G(M + m). The semimajor axis in that period law belongs to the relative orbit. When M is much greater than m, this parameter is approximately GM and the primary’s motion is small.

Real orbit predictions also account for the effects relevant to their purpose: other gravitating bodies, departures from spherical symmetry, atmospheric drag and thrust can change the simple trajectory. This lesson ignores relativistic corrections. A closed curve in the ideal model does not establish how accurately it predicts a measured orbit.

To use the model, name the reference frame, identify the center-to-center distance and initial velocity, choose the applicable forces and check the conservation laws. A spacecraft burn changes its orbital state; the new orbit must be calculated from the changed state. The toy ellipse here is a coherent comparison, rather than a launch plan or evidence about any particular planet.

Worked examples

Try it yourself

1. For unchanged masses, a separation grows from 3 m to 6 m. By what factor does the Newtonian force magnitude change?

   Show the answer and reasoning

   It becomes one quarter of its original magnitude.

   The separation doubles and the force scales as 1/r². An altitude change would first have to be converted to center-to-center distance.

2. A bound nonradial ellipse has nearest distance 3 m and farthest distance 9 m. Find a and e.

   Show the answer and reasoning

   a = 6 m and e = 0.5.

   Use a = (rp + ra)/2 = 6 m and e = (ra − rp)/(ra + rp) = 6/12. Both distances are from the occupied focus.

3. In the supplied ellipse, L has magnitude 8 kg m²/s for the 1 kg satellite. Find the speed at apoapsis, where r = 8 m.

   Show the answer and reasoning

   1 m/s.

   Radius and velocity are perpendicular at apoapsis, so v = L/(mr). That shortcut requires the perpendicularity stated here; at a general point it gives v⊥.

4. For the same μ = 20 m³/s², a separate ellipse has semimajor axis 10 m. What is its period relative to the supplied a = 5 m ellipse?

   Show the answer and reasoning

   The period is 2√2 times as long, or 10√2π s.

   T² scales as a³ at fixed μ, so the period ratio is √(2³). Apply it to the original period 5π s.

5. Between the two apsides of the supplied ellipse, U changes from −10 J to −2.5 J. Find ΔU and ΔK.

   Show the answer and reasoning

   ΔU = +7.5 J and ΔK = −7.5 J.

   Subtract the initial values from the final values. Mechanical-energy conservation makes ΔK = −ΔU; the final kinetic energy is 0.5 J.

6. Two equal masses M interact in isolation. For their relative orbit, is the gravitational parameter GM or 2GM, and is either body stationary at their center of mass?

   Show the answer and reasoning

   The relative parameter is 2GM. Neither body is stationary at their shared center of mass in a nonradial orbit.

   The exact parameter is G(M + M). Both bodies move about the shared center of mass; the fixed-primary approximation is unsuitable for equal masses.

What to carry forward

• Gravity changes velocity in an orbit; centripetal acceleration is the inward result of the relevant force.
• Read an ellipse from its occupied focus and retain the assumptions behind a fixed primary.
• A central force conserves angular momentum about its center; equal swept areas do not mean equal path lengths.
• With zero potential at infinity, kinetic and potential energy can change while their sum stays fixed.
• At fixed gravitational parameter, equal semimajor axes give equal Kepler periods, even for different eccentricities.
• The exact relative two-body problem uses both masses; real trajectory predictions require the relevant perturbations and observations.