Motion of Orbiting Satellites

Learning Goals

After this lesson, you should be able to:

• define the satellite–central-object system and state the massive-central-object approximation;
• draw the free-body diagram of an orbiting satellite;
• derive circular-orbit speed and period from Newton's laws;
• calculate kinetic, gravitational potential, and total mechanical energy in a circular orbit;
• identify what remains constant in circular and elliptical orbits;
• use angular-momentum conservation to compare periapsis and apoapsis speeds;
• interpret energy-versus-radius and period-versus-radius graphs;
• derive escape speed from the \(E=0\) condition;
• distinguish orbital radius from altitude; and
• explain apparent weightlessness as continuous free fall.

1. The System and Its Approximation

Consider a central object of mass \(M\) and a satellite of mass \(m\), interacting only through gravity. When \(m\ll M\), the central object's motion is negligible and the satellite can be modeled as orbiting a nearly fixed center.

If the masses are comparable, both objects move around their common center of mass. That more general two-body motion does not invalidate conservation laws, but the fixed-center picture must then be replaced.

3. Orbit Means Continuous Free Fall

Ignoring thrust, drag, and other bodies, the satellite's free-body diagram contains one force:

The direction is always toward the center of the central object. In a circular orbit this inward force changes the direction of the velocity but not its magnitude. Astronauts feel apparently weightless because the spacecraft and everything inside it accelerate together in free fall—not because gravity has disappeared.

4. Circular-Orbit Speed

For a circular orbit, gravity supplies the centripetal acceleration:

Cancel the satellite mass and solve:

A larger circular-orbit radius therefore corresponds to a smaller orbital speed. The satellite's mass does not affect that speed under the fixed-central-mass approximation.

5. Circular-Orbit Period and Kepler Scaling

The satellite travels circumference \(2\pi r\) during one period:

Combining this with the orbital-speed equation gives

Squaring produces the circular-orbit form of Kepler's third-law relationship:

For satellites orbiting the same central mass, a graph of \(T^2\) versus \(r^3\) is linear through the origin with slope \(4\pi^2/(GM)\).

6. Worked Example: A Low Circular Orbit

A satellite circles Earth at \(r=7.00\times10^6\,\mathrm m\) from Earth's center. Use \(GM_{\mathrm E}=3.986\times10^{14}\,\mathrm{m^3/s^2}\).

Answer: The orbital speed is \(7.55\,\mathrm{km/s}\) and the period is about \(97.2\) minutes.

7. Gravitational Potential Energy Uses Infinity as Zero

For the satellite–central-object system, define gravitational potential energy to approach zero as the separation becomes infinite:

The negative sign means a finite-radius gravitational system has less energy than the separated state at infinity. Moving outward makes \(U\) increase toward zero; moving inward makes \(U\) more negative.

Near a planet's surface, \(mgh\) can approximate a small change in potential energy, but \(-GMm/r\) is the required model when \(r\) changes significantly.

8. Energy of a Circular Orbit

From the circular-orbit force condition, \(v^2=GM/r\). Therefore,

Thus, for a circular orbit,

The negative total energy identifies the orbit as gravitationally bound.

9. Energy Values for the Example Orbit

For the \(r=7.00\times10^6\,\mathrm m\) orbit, it is useful to calculate energy per unit satellite mass:

These values verify \(U=-2K\) and \(E=-K\). A heavier satellite at the same radius has the same speed and period but proportionally larger \(K\), \(U\), and \(E\).

10. Circular and Elliptical Orbits Compared

In both cases, gravity is internal to the chosen two-object system and conservative. Its central direction also gives zero torque about the central object.

11. Why Angular Momentum Is Constant

Gravitational force points along the line from the satellite to the central object, so its lever arm about the central object's center is zero:

With zero external torque about that point,

In a circle, \(v_\perp=v\). At periapsis and apoapsis of an ellipse, the velocity is also perpendicular to the radius, making those locations especially convenient for comparisons.

12. Periapsis and Apoapsis Speed

At the nearest point \(r_p\) and farthest point \(r_a\), angular-momentum conservation gives

If \(r_p=8.0\times10^6\,\mathrm m\), \(r_a=1.20\times10^7\,\mathrm m\), and \(v_p=7.5\,\mathrm{km/s}\), then

The satellite moves faster near periapsis and slower near apoapsis while its total mechanical energy remains unchanged.

13. Energy Exchange Along an Ellipse

Between periapsis and apoapsis, separation increases, so potential energy increases toward zero:

Because \(E=K+U\) is constant,

On the inward trip, the reverse occurs: \(U\) decreases and \(K\) increases. Energy conservation and angular-momentum conservation describe the same speed trend from different perspectives.

14. Deriving Escape Speed

The exact escape speed is the initial speed that lets a satellite approach rest at infinite separation. At infinity, \(K_f=0\) and \(U_f=0\), so the system's total mechanical energy is zero:

Satellite mass cancels. Comparing with circular-orbit speed at the same radius gives

A greater launch speed gives positive total energy and a nonzero speed at infinity; a smaller speed does not escape under the ideal two-object model.

15. Worked Example: Escape from the Example Radius

At \(r=7.00\times10^6\,\mathrm m\),

Equivalently,

Answer: The exact escape speed at that radius is \(10.7\,\mathrm{km/s}\), assuming no atmosphere, propulsion after launch, or other celestial bodies.

16. Graph and Proportional Reasoning

“Energy increases” for a higher circular orbit means the negative value becomes less negative. Moving a satellite to a higher circular orbit requires adding energy even though its final speed is smaller.

17. From One Circular Orbit to a Higher One

A tangential thrust adds energy and angular momentum, placing the satellite on a transfer ellipse rather than instantly creating a larger circle. At the far side of that ellipse, another appropriately directed thrust can circularize the orbit.

This resolves a common paradox: a higher circular orbit has a lower final speed, yet energy must be supplied to reach it. The added energy raises the system's total mechanical energy from a more negative value to a less negative value; the path between circular orbits is not itself circular.

18. A Reliable Problem-Solving Workflow

1. Define the central object, satellite, system boundary, and whether \(m\ll M\).
2. Use center-to-center distance \(r\), not altitude alone.
3. Draw gravity toward the central object's center.
4. For a circular orbit, set \(GMm/r^2=mv^2/r\).
5. Use \(U=-GMm/r\) with zero potential at infinity.
6. For an isolated orbit, conserve \(E=K+U\) and angular momentum about the center.
7. At periapsis and apoapsis, use \(r_pv_p=r_av_a\).
8. For exact escape, set the total energy equal to zero.
9. Check signs, units, limiting behavior, and whether the result depends correctly on satellite mass.

19. Investigation: Infer a Central Mass from Orbit Data

Use a provided set of orbital radii and periods for several satellites around one central object, or generate an internally written synthetic data set.

1. Convert all radii to center-to-center SI units and periods to seconds.
2. Calculate \(r^3\) and \(T^2\) with appropriate uncertainties.
3. Plot \(T^2\) vertically against \(r^3\) horizontally.
4. Fit a line and test whether the intercept is consistent with zero.
5. Use the slope \(s=4\pi^2/(GM)\) to calculate the central mass.
6. Compare the result with the accepted value and calculate percent difference.
7. Discuss uncertainty in radius, period, orbital eccentricity, drag, and the fixed-central-object approximation.

This investigation uses the orbit itself as evidence for the central object's mass.

20. Common Traps and Key Takeaways

• An orbiting satellite is in free fall; gravity is not absent.
• Use orbital radius from the center, so \(r=R_{\mathrm{planet}}+h\).
• Gravity is the centripetal force; do not add a separate “centripetal force.”
• Circular speed decreases as radius increases, while period increases.
• Satellite mass cancels from speed, period, and escape speed, but not from energy or angular momentum.
• Use \(U=-GMm/r\), not \(+GMm/r\).
• A bound orbit has negative total mechanical energy.
• In an ellipse, total energy and angular momentum remain constant, while \(K\), \(U\), \(r\), and speed change.
• Exact escape corresponds to \(E=0\).
• A higher circular orbit has less speed but more total mechanical energy.