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4.4 · Relate kinetic, potential, and total energy in an orbit

Learn to relate kinetic, potential, and total energy in an orbit through clear examples and targeted practice.

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Gravitation and Orbital Motion

Relating kinetic, gravitational potential, and total energy

An orbiting satellite can speed up or slow down while its total mechanical energy stays constant. As it moves closer to the central body, gravitational potential energy becomes more negative, and kinetic energy increases by the corresponding amount. This lesson uses Newtonian gravitation and the work–energy principle to describe that exchange. We treat the satellite as a particle and the central body as fixed in an inertial reference frame. The model neglects other bodies, atmospheric drag, and any thrust, so gravity is the only force doing work.

What you will learn

  • Define the system and reference frame for a particle orbiting a central body.
  • Use gravitational potential energy with a zero level at infinity.
  • Relate kinetic energy, potential energy, and total mechanical energy along an orbit.
  • Use total energy to compare speeds at different orbital positions and identify bound and escape motion.

System, frame, and energy model

Choose the satellite as the system. An observer fixed to the central body uses an approximately inertial frame, with the origin at the body's centre. Let rr be the satellite's distance from that origin, measured in metres, and let the positive radial direction point outward. The satellite's velocity is measured relative to this frame. Its speed vv is the magnitude of that velocity; it is not the same as its position or the distance it has travelled along its path.
For a central body of mass MM, Newton's gravitational force points inward. Define the gravitational parameter as μ=GM\mu=GM, where GG is the universal gravitational constant. Gravitational potential energy is defined to be zero at infinite separation. At finite rr, it is negative: U=−GMm/rU=-GMm/r. The sign reflects that energy must be supplied to separate the satellite and central body to infinity.
The satellite's kinetic energy is K=mv2/2K=mv^2/2. When gravity is the only force doing work, the work–energy principle says that changes in kinetic energy equal the negative changes in potential energy. Thus the total mechanical energy E=K+UE=K+U is constant. Dividing by satellite mass gives the specific energies, which are energy per unit mass and have units of joules per kilogram.
E=K+U=12mv2−GMmrE=K+U=\frac{1}{2}mv^2-\frac{GMm}{r}
  • Use one central-body-fixed inertial frame for both position and velocity.
  • With potential energy zero at infinity, gravitational potential energy is negative.
  • With gravity as the only force doing work, K+UK+U stays constant.

How energy changes along an orbit

For a fixed satellite mass, conservation of energy can be written per unit mass as ϵ=v2/2−μ/r\epsilon=v^2/2-\mu/r, where ϵ=E/m\epsilon=E/m. This form is convenient because the satellite mass cancels. If the satellite moves from an initial radius r1r_1 and speed v1v_1 to a final radius r2r_2 and speed v2v_2, conservation gives a direct relation between the two speeds and radii.
The relation is useful even when the path is not circular. At a smaller radius, the term −μ/r-\mu/r is more negative, so the kinetic-energy term must be larger to keep the sum fixed. At a larger radius, kinetic energy is lower. In an elliptical orbit the satellite is therefore faster at its closest point and slower at its farthest point. Energy alone relates speed to radius; it does not specify the direction of the velocity.
A circular orbit has constant radius and constant speed. Newton's second law, with inward gravity providing the required inward acceleration, gives v2=μ/rv^2=\mu/r. Substitution into the energy relation shows that the kinetic energy is positive, the potential energy has twice its magnitude and is negative, and the total energy is negative. These are specific to a circular orbit; for a noncircular orbit, the energy relation still applies but the circular speed equation does not.
\epsilon=v22\frac{v^2}{2}-μr\frac{\mu}{r}=constant
  • Specific mechanical energy is constant along an ideal orbit.
  • Closer to the central body means greater speed when comparing points on the same orbit.
  • For a circular orbit, kinetic energy is half the magnitude of potential energy.

Bound motion, escape, and checks

With the zero of potential energy at infinity, a bound orbit has negative total energy: the satellite does not have enough energy to reach infinity with nonzero speed. The boundary case for just reaching infinity with zero speed has total energy zero. At a specified radius, setting the total energy to zero gives the escape speed. A speed below this value gives negative energy; a speed above it gives positive energy in this ideal model.
These energy statements are useful checks, not substitutes for choosing a suitable equation. Confirm that the potential-energy reference is stated, that the same central body and frame are used throughout, and that no neglected force does work. Check dimensions: μ/r\mu/r has units of square metres per square second, the same as v2v^2, so specific energy has units of joules per kilogram. Check signs as well: KK cannot be negative, while UU is negative under this reference choice.
A useful limiting check is large distance. As rr tends to infinity, UU approaches zero. For the escape boundary, the speed also tends to zero there, so total energy approaches zero. This matches the definition of escape speed and helps catch sign errors.
vesc=2μrv_{\mathrm{esc}}=\sqrt{\frac{2\mu}{r}}
  • Negative total energy indicates bound motion in this two-body, gravity-only model.
  • Escape speed is found by setting total energy to zero at the stated radius.
  • State the potential-energy reference before interpreting the sign of total energy.

Worked example

Energy in a circular orbit

A satellite is in a circular orbit of radius 7.00×106 m7.00\times10^6\ \mathrm{m} from Earth's centre. Use μ=3.986×1014 m3/s2\mu=3.986\times10^{14}\ \mathrm{m^3/s^2}. Find its speed and its kinetic, potential, and total specific energies.
Gravity in a circular orbit
Gravity in a circular orbitOutward radial direcTangential directionSatelliteGravity inward

At the shown location, inward gravity supplies the circular motion's inward acceleration.

  1. Set the system and knowns
    The system is the satellite, observed in an inertial frame centred on Earth. The radius is positive outward; gravity and the required centripetal acceleration point inward. The orbit is circular, so the radius and speed are constant. The unknowns are speed and the three specific energies.
  2. Find the circular speed
    For circular motion, Newton's second law equates inward gravitational acceleration to the circular-motion acceleration. The satellite mass cancels. Use the positive root because speed is a magnitude.
    v=μr=7.55×103 m/sv=\sqrt{\frac{\mu}{r}}=7.55\times10^3\ \mathrm{m/s}
  3. Compute the energy terms
    Divide each energy by satellite mass. The potential term is negative because zero potential energy was assigned at infinity. Add the two terms to get total specific energy.
    k=v22=28.5 MJ/kg,u=−μr=−56.9 MJ/kg,ϵ=k+u=−28.5 MJ/kgk=\frac{v^2}{2}=28.5\ \mathrm{MJ/kg},\quad u=-\frac{\mu}{r}=-56.9\ \mathrm{MJ/kg},\quad \epsilon=k+u=-28.5\ \mathrm{MJ/kg}
Answer: The speed is 7.55×103 m/s7.55\times10^3\ \mathrm{m/s}. The specific kinetic, potential, and total energies are respectively 28.5 MJ/kg28.5\ \mathrm{MJ/kg}, −56.9 MJ/kg-56.9\ \mathrm{MJ/kg}, and −28.5 MJ/kg-28.5\ \mathrm{MJ/kg}.
Check: The circular-orbit relation predicts u=−2ku=-2k, which the results satisfy. Also, μ/r\mu/r has units of m2/s2\mathrm{m^2/s^2}, equivalent to J/kg\mathrm{J/kg}.

Worked example

Comparing speeds at two orbital radii

A satellite follows an elliptical orbit about Earth, with closest radius 8.00×106 m8.00\times10^6\ \mathrm{m} and farthest radius 2.00×107 m2.00\times10^7\ \mathrm{m}. Take μ=3.986×1014 m3/s2\mu=3.986\times10^{14}\ \mathrm{m^3/s^2}. At the closest point its velocity is perpendicular to the radius. Find its speed there and at the farthest point using conservation of energy.
Gravity at closest approach
Gravity at closest approachOutward radial direcTangential directionSatelliteGravity inward

The energy comparison needs the two radii; gravity points toward the central body.

  1. Define states and energy
    The satellite is the system and the frame is inertial with its origin at Earth's centre. State 1 is closest approach, and state 2 is farthest distance. The positive radial direction is outward. With gravity as the only force doing work, specific energy is conserved.
  2. Use the closest-point condition
    At closest approach the velocity is perpendicular to the radius, but energy conservation only needs its speed. The given radii and the unknown closest-point speed determine the constant specific energy.
    ϵ=v122−μr1\epsilon=\frac{v_1^2}{2}-\frac{\mu}{r_1}
  3. Solve for both speeds
    At the farthest radius, set the expression for specific energy equal to its value at closest approach. Solving for each speed gives positive magnitudes. The smaller speed belongs to the larger radius, as expected for this orbit.
    v1=2μr2r1(r1+r2)=8.44×103 m/s,v2=2μr1r2(r1+r2)=3.38×103 m/sv_1=\sqrt{\frac{2\mu r_2}{r_1(r_1+r_2)}}=8.44\times10^3\ \mathrm{m/s},\quad v_2=\sqrt{\frac{2\mu r_1}{r_2(r_1+r_2)}}=3.38\times10^3\ \mathrm{m/s}
Answer: The speed at closest approach is 8.44×103 m/s8.44\times10^3\ \mathrm{m/s}, and at the farthest point it is 3.38×103 m/s3.38\times10^3\ \mathrm{m/s}.
Check: Substitution gives the same specific energy at both radii, about −14.2 MJ/kg-14.2\ \mathrm{MJ/kg}. It is negative, consistent with a bound orbit, and the speed is lower at the larger radius.

Worked example

Escape speed from a given radius

A probe is at radius 9.00×106 m9.00\times10^6\ \mathrm{m} from Earth's centre. With μ=3.986×1014 m3/s2\mu=3.986\times10^{14}\ \mathrm{m^3/s^2}, find the speed that makes its total mechanical energy zero. Ignore all forces except Earth's gravity.
Probe at escape threshold
Probe at escape thresholdOutward radial direcTangential directionProbeGravity inward

The energy threshold is zero when the probe reaches infinity with zero speed.

  1. Set the system and threshold
    The probe is the system, and the observer uses the Earth-centred inertial frame. At the initial state the radius is known and the speed is unknown. Escape threshold means the probe reaches infinite distance with zero final speed, so total energy is zero.
  2. Apply energy conservation
    The initial kinetic and potential specific energies must sum to zero. Solve for the speed, choosing the positive root because speed cannot be negative.
    v22−μr=0,v=2μr\frac{v^2}{2}-\frac{\mu}{r}=0,\quad v=\sqrt{\frac{2\mu}{r}}
  3. Substitute SI values
    Substitute the stated gravitational parameter and radius. The result is the threshold speed at this radius, not a constant speed throughout the probe's motion.
    v=2(3.986×1014 m3/s2)9.00×106 m=9.41×103 m/sv=\sqrt{\frac{2(3.986\times10^{14}\ \mathrm{m^3/s^2})}{9.00\times10^6\ \mathrm{m}}}=9.41\times10^3\ \mathrm{m/s}
Answer: The escape-threshold speed is 9.41×103 m/s9.41\times10^3\ \mathrm{m/s}.
Check: At the threshold, initial kinetic energy per unit mass equals the magnitude of the negative gravitational potential energy per unit mass. As the probe moves outward, its speed decreases toward zero while its potential energy approaches zero.

Common mistakes and how to avoid them

Taking gravitational potential energy as positive while also setting it to zero at infinity.
Correction: With zero at infinity, use U=−GMm/rU=-GMm/r. A different reference is possible, but the reference and signs must be consistent.
Using the circular-orbit speed relation at every point in an elliptical orbit.
Correction: The relation v2=μ/rv^2=\mu/r applies to a circular orbit. For a general orbit, use conservation of energy between the stated positions.
Assuming that total energy is the same as kinetic energy.
Correction: Total mechanical energy includes both kinetic and gravitational potential energy: E=K+UE=K+U.
Treating zero total energy as the energy of every escaping trajectory.
Correction: Zero is the escape threshold for a probe that reaches infinity with zero speed. A probe with greater energy can reach infinity with positive speed.

Lesson summary

  • For a satellite in a gravity-only model, E=K+UE=K+U is conserved.
  • With gravitational potential energy zero at infinity, U=−GMm/rU=-GMm/r and total energy is negative for a bound orbit.
  • As orbital radius decreases, speed increases along the same orbit.
  • For a circular orbit, v2=μ/rv^2=\mu/r, and the total specific energy is negative.
  • Escape speed follows by setting total energy to zero at the starting radius.

Check your understanding

Question 1

A satellite moves from a larger radius to a smaller radius on the same ideal orbit. What happens to its kinetic energy?
  1. It increases because gravitational potential energy becomes more negative.
  2. It decreases because potential energy becomes more negative.
  3. It stays constant because total energy is conserved.
  4. It becomes negative.
Show answer and explanation
It increases because gravitational potential energy becomes more negative.
Total energy stays constant, so the decrease in gravitational potential energy is balanced by an increase in kinetic energy.

Question 2

At a fixed radius, what does zero total energy represent in the ideal gravity-only model?
  1. A circular orbit at that radius.
  2. The escape threshold, reaching infinity with zero speed.
  3. A satellite at rest at that radius.
  4. A bound orbit with minimum speed.
Show answer and explanation
The escape threshold, reaching infinity with zero speed.
At infinity the potential energy tends to zero; zero total energy therefore corresponds to zero speed there.

Question 3

For a circular orbit, how does the specific potential energy compare with the specific kinetic energy?
  1. It is equal and positive.
  2. It is half as large and positive.
  3. It is negative and twice the kinetic energy's magnitude.
  4. It is zero.
Show answer and explanation
It is negative and twice the kinetic energy's magnitude.
Using v2=μ/rv^2=\mu/r gives k=μ/(2r)k=\mu/(2r) and u=−μ/r=−2ku=-\mu/r=-2k.

Key terms

Specific energy
Energy divided by mass, measured in joules per kilogram.
Gravitational potential energy
Energy associated with position in a gravitational field; here it is defined as zero at infinite separation.
Escape speed
The speed at a given radius that gives zero total energy, so the object reaches infinity with zero speed in the ideal model.
Bound orbit
An orbit with negative total mechanical energy in the stated gravity-only model.

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