Energy Changes in Orbital Motion

Expert reviewed • 04 March 2025 • 10 minute read

Gravitational Potential Energy in Orbit

Gravitational potential energy (U)(U) of an object in orbit is determined by its mass and its distance from the centre of the orbited body. It is given by the equation:

U=−GMmrU = -\frac{GMm}{r}

Where:

  • GG is the gravitational constant
  • MM is the mass of the central body
  • mm is the mass of the orbiting object
  • rr is the distance between the centres of the two bodies

In elliptical orbits, gravitational potential energy varies with orbital radius:

  • It increases (becomes less negative) as orbital radius increases
  • It decreases (becomes more negative) as orbital radius decreases

For circular orbits, gravitational potential energy remains constant due to the fixed orbital radius.

Kinetic Energy in Orbit

Objects in orbital motion also possess kinetic energy (K)(K), given by:

K=12mv2K = \frac{1}{2}mv^2

Where:

  • mm is the mass of the orbiting object
  • vv is its velocity

Circular Orbits

In a stable circular orbit, the orbital velocity remains constant and is given by:

vorbital=GMrv_{\text{orbital}} = \sqrt{\frac{GM}{r}}

Substituting this into the kinetic energy equation yields:

K=GMm2rK = \frac{GMm}{2r}

Elliptical Orbits

In elliptical orbits, the orbital velocity and thus the kinetic energy of an object changes with its position:

  • Velocity and kinetic energy increase as the object moves closer to the central body
  • Velocity and kinetic energy decrease as the object moves farther from the central body

This behaviour aligns with Kepler's Second Law of Planetary Motion, which states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

Total Mechanical Energy in Orbit

The total mechanical energy (E)(E) of an object in orbit is the sum of its kinetic and potential energies:

E = K + UE = K + U

Circular Orbits

For circular orbits, the total energy is:

E=GMm2r−GMmr=−GMm2rE = \frac{GMm}{2r} - \frac{GMm}{r} \\= -\frac{GMm}{2r}

This shows that the total energy is exactly half the gravitational potential energy for circular orbits.

Elliptical Orbits

For elliptical orbits, the total energy is given by:

E=−GMm2aE = -\frac{GMm}{2a}

Where a is the semi-major axis of the elliptical orbit.

Energy Changes Within an Orbit

In the absence of non-conservative forces, the total mechanical energy of an orbiting object remains constant throughout its orbit, adhering to the law of conservation of energy. However, there is a continuous exchange between potential and kinetic energy:

  • As orbital radius decreases:
    • Gravitational potential energy decreases (becomes more negative)
    • Kinetic energy increases
  • As orbital radius increases:
    • Gravitational potential energy increases (becomes less negative)
    • Kinetic energy decreases

Energy Changes Between Orbits

When an object moves between different orbits, its total mechanical energy changes:

  • Moving to an orbit with a larger radius or semi-major axis:
    • Total energy increases (becomes less negative)
    • Kinetic energy decreases
    • Gravitational potential energy increases (becomes less negative)
  • Moving to an orbit with a smaller radius or semi-major axis:
    • Total energy decreases (becomes more negative)
    • Kinetic energy increases
    • Gravitational potential energy decreases (becomes more negative)

Practice Question 1

Calculate the total energy of a satellite with mass 1000 kg orbiting Earth (M = 5.97 × 10^24 kg) in a circular orbit with radius 7000 km.

Given:

  • G=6.67×10−11 Nm2/kg2G = 6.67 × 10^{-11} \:Nm^2/kg^2
  • m=1000 kgm = 1000 \:kg
  • M=5.97×1024 kgM = 5.97 × 10^{24} \:kg
  • r=7,000,000 mr = 7,000,000 \:m

Solution

E=−GMm2rE=−(6.67×10−11)(5.97×1024)(1000)2(7,000,000)E ≈− 2.84 × 1010JE = -\frac{GMm}{2r}\\E = -\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(1000)}{2(7,000,000)}\\E ≈− 2.84 × 10^{10} J

The negative value indicates that the satellite is bound to Earth's orbit.

Practice Question 2

A spacecraft is in an elliptical orbit around Mars. At periapsis (closest approach), it has a speed of 4 km/s and is 4000 km from Mars' centre. At apoapsis (farthest point), it is 20,000 km from Mars' centre. Calculate its speed at apoapsis.

Given:

  • vperiapsis=4000m/sv_{\text{periapsis}} = 4000 m/s
  • rperiapsis=4,000,000mr_{\text{periapsis}} = 4,000,000 m
  • rapoapsis=20,000,000mr_{\text{apoapsis}} = 20,000,000 m

Solution

Use conservation of energy: Eperiapsis=EapoapsisE_{periapsis} = E_{apoapsis}

12mvp2−GMmrp=12mva2−GMmra\frac{1}{2}mv_p^2 - \frac{GMm}{r_p} = \frac{1}{2}mv_a^2 - \frac{GMm}{r_a}

Divide by mm and rearrange:

va2=vp2−2GM(1ra−1rp)v_a^2 = v_p^2 - 2GM(\frac{1}{r_a} - \frac{1}{r_p}) va=40002−2GM(120,000,000−14,000,000)v_a = \sqrt{4000^2 - 2GM(\frac{1}{20,000,000} - \frac{1}{4,000,000})}

Solving this (with GMGM for Mars ≈4.282×1013 m3/s2≈ 4.282 × 10^{13} \:m^3/s^2):

va ≈ 1789 m/sor≈1.79km/sv_a ≈ 1789 \:m/s \quad or\quad \approx1.79 km/s

This example illustrates how velocity (and thus kinetic energy) changes within an elliptical orbit, being higher at periapsis and lower at apoapsis.

Previous

Understanding Escape Velocity in Gravitational Fields

Next

Low Earth and Geostationary Orbits

Return to Module 5: Advanced Mechanics