Orbital Energy

Orbital energy describes the mechanical energy of an object moving in a gravitational field. For a satellite, planet, moon, or spacecraft, the two main forms of mechanical energy are:

• kinetic energy (Eₖ) because the object is moving
• gravitational potential energy (Eₚ) because of its position in the gravitational field

The total orbital energy is:

E_total = Eₖ + Eₚ

For an object of mass m orbiting a much larger central body of mass M:

Eₖ = 1/2 mv²

Eₚ = −GMm/r

Therefore:

E_total = 1/2 mv² − GMm/r

The sign and magnitude of this total energy tell us a great deal about the object's orbit.

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Kinetic Energy in Orbit

Any orbiting object has kinetic energy because it is moving.

Eₖ = 1/2 mv²

where:

• m = mass of the orbiting object
• v = orbital speed

Kinetic energy is always:

Eₖ ≥ 0

For an orbiting satellite with nonzero speed:

Eₖ > 0

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Worked Example: Satellite Kinetic Energy

A satellite has:

m = 1000 kg

v = 7500 m/s

Then:

Eₖ = 1/2(1000)(7500²)

Eₖ = 2.81 × 10¹⁰ J

Therefore, the satellite has approximately:

28.1 billion joules of kinetic energy.

Orbital speeds are extremely high, so even relatively small satellites can have enormous kinetic energies.

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Gravitational Potential Energy

For two masses separated by distance r:

Eₚ = −GMm/r

where:

• G = 6.67 × 10⁻¹¹ N·m²/kg²
• M = mass of the central body
• m = mass of the orbiting object
• r = distance between their centres

Unlike the near-surface equation:

ΔEₚ = mgΔh

the equation:

Eₚ = −GMm/r

is appropriate when gravitational field strength changes significantly with distance.

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Why Gravitational Potential Energy Is Negative

For orbital calculations, gravitational potential energy is conventionally defined as:

Eₚ = 0 at infinite separation

A satellite at any finite distance from a planet is gravitationally bound to it.

Energy must be supplied to move the satellite from that position all the way to infinity.

Therefore:

Eₚ < 0

The negative sign represents the fact that the two objects form a gravitationally bound system relative to the zero-at-infinity reference.

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Potential Energy Becomes Less Negative with Distance

Consider:

Eₚ = −GMm/r

As r increases, the magnitude of Eₚ decreases.

For example:

Moving farther away increases gravitational potential energy.

It becomes:

less negative

and approaches zero.

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Total Orbital Energy

Total mechanical energy is:

E_total = Eₖ + Eₚ

or:

E_total = 1/2 mv² − GMm/r

The value of E_total can be:

• negative
• zero
• positive

These three possibilities correspond to fundamentally different gravitational states.

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Negative Total Energy

If:

E_total < 0

the object is gravitationally bound.

It does not have enough mechanical energy to escape to infinity.

Bound gravitational orbits include:

• circular orbits
• elliptical orbits

Therefore:

bound orbit → negative total orbital energy

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Zero Total Energy

If:

E_total = 0

the object is exactly at the ideal escape threshold.

It can move infinitely far from the central body while its speed approaches zero.

This corresponds to the minimum escape condition.

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Positive Total Energy

If:

E_total > 0

the object is unbound.

It can escape to infinity while retaining kinetic energy.

Therefore:

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Energy in a Circular Orbit

Circular orbits have an especially useful relationship between kinetic, potential, and total energy.

For circular motion:

F_gravity = F_centripetal

Therefore:

GMm/r² = mv²/r

Cancel m and simplify:

v² = GM/r

Now substitute this into kinetic energy:

Eₖ = 1/2 mv²

Eₖ = 1/2 m(GM/r)

Therefore:

Eₖ = GMm/(2r)

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Potential Energy in a Circular Orbit

Gravitational potential energy is:

Eₚ = −GMm/r

Compare this with kinetic energy:

Eₖ = GMm/(2r)

Therefore:

Eₚ = −2Eₖ

This is an important circular-orbit relationship.

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Total Energy in a Circular Orbit

Total energy is:

E_total = Eₖ + Eₚ

Substitute:

E_total = GMm/(2r) − GMm/r

Therefore:

E_total = −GMm/(2r)

So for a circular orbit:

Eₖ = +GMm/(2r)

Eₚ = −GMm/r

E_total = −GMm/(2r)

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The 1 : −2 : −1 Relationship

For a circular orbit, the energies have the ratio:

Eₖ : Eₚ : E_total = 1 : −2 : −1

For example, if:

Eₖ = +4 × 10⁹ J

then:

Eₚ = −8 × 10⁹ J

and:

E_total = −4 × 10⁹ J

This provides a useful shortcut for circular-orbit calculations.

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Worked Circular-Orbit Example

A 500 kg satellite orbits Earth at:

r = 7.00 × 10⁶ m

Use:

G = 6.67 × 10⁻¹¹ N·m²/kg²

M_Earth = 5.97 × 10²⁴ kg

First calculate:

GM ≈ 3.98 × 10¹⁴ m³/s²

Kinetic Energy

Eₖ = GMm/(2r)

Eₖ = (3.98 × 10¹⁴)(500) / [2(7.00 × 10⁶)]

Eₖ ≈ 1.42 × 10¹⁰ J

Potential Energy

Eₚ = −GMm/r

Eₚ ≈ −2.84 × 10¹⁰ J

Total Energy

E_total = −GMm/(2r)

E_total ≈ −1.42 × 10¹⁰ J

Check:

Eₖ + Eₚ

= 1.42 × 10¹⁰ − 2.84 × 10¹⁰

= −1.42 × 10¹⁰ J

The negative total energy confirms that the satellite is gravitationally bound.

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Why Bound Orbits Have Negative Energy

Imagine a satellite in orbit around Earth.

To remove it completely from Earth's gravitational influence, energy must be supplied.

At infinity:

Eₚ → 0

For the minimum escape condition:

Eₖ → 0

Therefore:

E_total → 0

If the satellite's current total energy is negative, additional energy must be supplied to raise its energy to zero.

This is why negative energy indicates a bound gravitational system.

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How Strongly Is an Object Bound?

The magnitude of the negative total energy indicates how strongly the object is bound.

For example:

E_total = −10¹¹ J

represents a more tightly bound orbit than:

E_total = −10⁹ J

assuming the same orbiting mass and central body.

The more negative the energy:

the more energy must be added to escape.

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Orbital Energy and Radius

For a circular orbit:

E_total = −GMm/(2r)

Therefore:

E_total ∝ −1/r

As orbital radius increases, total energy becomes:

less negative

and approaches zero.

So:

higher orbit → greater total energy

This is true even though the satellite's orbital speed becomes lower.

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Comparing Circular Orbits

Suppose Satellite A orbits at radius r.

Its total energy is:

E_A = −GMm/(2r)

Satellite B orbits at:

2r

Then:

E_B = −GMm/(4r)

Therefore:

E_B = 1/2 E_A

in magnitude.

If:

E_A = −100 MJ

then:

E_B = −50 MJ

The higher orbit has greater total energy because −50 MJ is greater than −100 MJ.

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Higher Orbit but Lower Speed

This can initially seem contradictory.

A satellite in a higher circular orbit has:

• lower orbital speed
• lower kinetic energy
• greater gravitational potential energy
• greater total mechanical energy

The increase in gravitational potential energy is larger than the decrease in kinetic energy.

Therefore, energy must be added to move a satellite into a higher orbit.

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Energy Changes Between Circular Orbits

Suppose a satellite moves from:

r₁ → r₂

where:

r₂ > r₁

Initial total energy:

E₁ = −GMm/(2r₁)

Final total energy:

E₂ = −GMm/(2r₂)

The required change is:

ΔE = E₂ − E₁

Because the higher orbit has less-negative energy:

ΔE > 0

Energy must be supplied.

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Worked Example: Raising an Orbit

Suppose a satellite has:

E₁ = −8.0 × 10⁹ J

in its original circular orbit.

Its new circular orbit has:

E₂ = −5.0 × 10⁹ J

Then:

ΔE = E₂ − E₁

= −5.0 × 10⁹ − (−8.0 × 10⁹)

= +3.0 × 10⁹ J

The spacecraft must gain:

3.0 × 10⁹ J

of orbital energy.

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Lowering an Orbit

Suppose instead:

E₁ = −5.0 × 10⁹ J

E₂ = −8.0 × 10⁹ J

Then:

ΔE = −8.0 × 10⁹ − (−5.0 × 10⁹)

= −3.0 × 10⁹ J

The satellite must lose:

3.0 × 10⁹ J

of orbital energy.

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Energy in Elliptical Orbits

In an elliptical orbit:

• orbital radius changes
• speed changes
• kinetic energy changes
• gravitational potential energy changes

However, if gravity is the only force doing work:

total orbital energy remains constant.

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At Periapsis

At the closest point:

• r is smallest
• Eₚ is most negative
• speed is greatest
• Eₖ is greatest

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At Apoapsis

At the farthest point:

• r is largest
• Eₚ is less negative
• speed is smallest
• Eₖ is smallest

Yet:

E_total remains constant

throughout the orbit.

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Energy Transformation in an Elliptical Orbit

As the satellite moves from apoapsis toward periapsis:

gravitational potential energy → kinetic energy

As it moves from periapsis toward apoapsis:

kinetic energy → gravitational potential energy

This is similar to an object falling and rising near Earth's surface, but the gravitational field changes substantially throughout the orbit.

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Total Energy of an Elliptical Orbit

For an elliptical orbit:

E_total = −GMm/(2a)

where:

a = semi-major axis

The semi-major axis describes the overall size of the ellipse.

This equation is extremely important because total orbital energy depends on the semi-major axis, not the satellite's instantaneous distance from the central body.

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Circular Orbit as a Special Case

For a circular orbit:

a = r

Therefore:

E_total = −GMm/(2r)

So the circular-orbit equation is simply a special case of the more general elliptical-orbit equation.

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Same Semi-Major Axis, Same Total Energy

Two elliptical orbits with different shapes can have the same semi-major axis.

If they orbit the same central body and involve the same satellite mass, they have the same total orbital energy.

Their kinetic and potential energies at particular locations may differ, but their total orbital energy is identical.

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Specific Orbital Energy

It is often useful to calculate energy per kilogram of spacecraft.

Specific orbital energy is:

ε = E_total/m

For an object at distance r with speed v:

ε = v²/2 − GM/r

For a circular orbit:

ε = −GM/(2r)

For an elliptical orbit:

ε = −GM/(2a)

The unit is:

J/kg

which is equivalent to:

m²/s²

Using specific energy removes the spacecraft mass from the calculation.

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The Vis-Viva Equation

Orbital energy can also be used to determine speed anywhere in an elliptical orbit.

Start with:

E_total = 1/2 mv² − GMm/r

For an elliptical orbit:

E_total = −GMm/(2a)

Set them equal:

1/2 mv² − GMm/r = −GMm/(2a)

Cancel m and rearrange:

v² = GM(2/r − 1/a)

Therefore:

v = √[GM(2/r − 1/a)]

This is called the vis-viva equation.

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What Vis-Viva Tells Us

For a particular orbit, a is constant.

As r decreases:

v increases

As r increases:

v decreases

This mathematically explains why objects move:

• fastest at periapsis
• slowest at apoapsis

For a circular orbit:

a = r

so:

v² = GM(2/r − 1/r)

= GM/r

giving:

v = √(GM/r)

as expected.

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Orbital Transfers

A spacecraft changes orbit by changing its orbital energy.

Rocket engines do this by changing the spacecraft's velocity.

Since:

Eₖ = 1/2 mv²

an engine burn can alter kinetic energy and therefore total orbital energy.

A spacecraft can:

• add orbital energy to move toward a higher orbit
• remove orbital energy to move toward a lower orbit

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Raising an Orbit

Suppose a spacecraft is travelling in a circular orbit.

It fires its engine forward, in the direction of travel.

Immediately:

• speed increases
• kinetic energy increases
• total orbital energy increases

The spacecraft does not instantly jump to a higher circular orbit.

Instead, it enters an elliptical transfer orbit.

Its original location becomes the low point of the new ellipse.

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Climbing Through the Transfer Orbit

After the engine shuts off, the spacecraft moves outward.

As it climbs:

• r increases
• gravitational potential energy increases
• kinetic energy decreases
• speed decreases

Total energy remains approximately constant during the unpowered coast.

The spacecraft eventually reaches the high point of the transfer orbit.

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Circularizing the Higher Orbit

At the high point, another forward engine burn increases the spacecraft's speed.

This raises the opposite side of the orbit.

If the burn is correct, the elliptical trajectory becomes a new higher circular orbit.

This produces an interesting result:

energy was added, yet the final orbital speed is lower than in the original low orbit.

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Why This Is Not a Contradiction

The final higher orbit has:

• lower kinetic energy
• greater gravitational potential energy
• greater total energy

The increase in potential energy is larger than the decrease in kinetic energy.

Therefore:

higher orbit = greater total orbital energy

even though:

higher circular orbit = lower orbital speed

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Hohmann Transfer

A Hohmann transfer is an efficient way of moving between two circular orbits in the same plane.

It uses an elliptical transfer orbit.

A simplified transfer to a higher orbit involves:

1. initial circular orbit
2. forward engine burn
3. elliptical transfer orbit
4. coast toward apoapsis
5. second forward burn
6. final higher circular orbit

Both burns add orbital energy.

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Energy During a Hohmann Transfer

Suppose:

r₁ < r₂

The initial circular orbit has:

E₁ = −GMm/(2r₁)

The transfer ellipse has semi-major axis:

a = (r₁ + r₂)/2

Therefore:

E_transfer = −GMm/(2a)

The final circular orbit has:

E₂ = −GMm/(2r₂)

The energy ordering is:

E₁ < E_transfer < E₂

Remember that these energies are negative.

For example:

−100 MJ < −70 MJ < −50 MJ

The spacecraft becomes progressively less tightly bound.

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Energy Diagram for Orbit Raising

A useful conceptual sequence is:

low circular orbit

E_total = very negative

↓

engine adds energy

↓

transfer ellipse

E_total = less negative

↓

engine adds more energy

↓

high circular orbit

E_total = even less negative

The spacecraft is still bound as long as:

E_total < 0

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Lowering an Orbit

To lower an orbit, the spacecraft can fire its engine opposite its direction of motion.

Immediately:

• speed decreases
• kinetic energy decreases
• total orbital energy decreases
• energy becomes more negative

The spacecraft enters an ellipse with a lower point on the opposite side.

A second burn can then circularize the lower orbit.

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Deorbiting

To deliberately return a spacecraft to Earth, engineers can perform a deorbit burn.

The spacecraft fires its engine to reduce orbital energy.

Its new orbit intersects Earth's atmosphere.

Atmospheric drag then removes additional mechanical energy.

The spacecraft descends and re-enters.

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Atmospheric Drag and Orbital Energy

Satellites in low Earth orbit encounter traces of atmosphere.

Drag does negative work on the satellite.

Therefore:

orbital energy decreases

and becomes more negative.

The satellite gradually moves into a lower orbit.

This process is called:

orbital decay

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Why a Decaying Satellite Can Speed Up

This seems strange.

Drag slows the satellite, yet lower circular orbits have higher orbital speeds.

The sequence is important.

Drag first removes mechanical energy.

The orbit becomes elliptical and drops lower.

As the satellite falls:

• gravitational potential energy decreases
• some potential energy becomes kinetic energy

The satellite can eventually move faster in the lower orbit even though its total orbital energy has decreased.

This is an important example of why speed alone does not determine orbital energy.

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Escape as an Orbital-Energy Problem

A bound circular orbit has:

E_total < 0

To escape:

E_total must reach at least 0

Therefore, engines must add enough energy to move the spacecraft from negative orbital energy to zero or positive energy.

At the exact escape threshold:

E_total = 0

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Energy Needed to Escape from a Circular Orbit

For a circular orbit:

E_initial = −GMm/(2r)

At the minimum escape condition:

E_final = 0

Therefore:

ΔE = E_final − E_initial

ΔE = 0 − [−GMm/(2r)]

So:

ΔE = GMm/(2r)

Interestingly, this equals the satellite's existing circular-orbit kinetic energy.

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Orbital Speed and Escape Speed

For a circular orbit:

v_orbit = √(GM/r)

Escape speed at the same radius is:

v_escape = √(2GM/r)

Therefore:

v_escape = √2 v_orbit

An orbiting spacecraft already possesses substantial kinetic energy, so it does not need to acquire the full escape energy from rest at that location.

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Worked Escape Example

Suppose a spacecraft has circular orbital speed:

7.5 km/s

Then:

v_escape = √2(7.5)

≈ 10.6 km/s

Increasing its speed in the appropriate direction can raise its total energy from negative toward zero.

At:

10.6 km/s

at that same radius, the spacecraft would be at the ideal local escape threshold.

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Bound and Unbound Trajectories

Orbital energy allows trajectories to be classified.

Circular Orbit

E_total < 0

Bound.

Elliptical Orbit

E_total < 0

Bound.

Escape Threshold

E_total = 0

Ideal limiting escape trajectory.

Hyperbolic Trajectory

E_total > 0

Unbound.

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Orbital Energy and Gravity Assists

Gravity assists can also change a spacecraft's orbital energy relative to the Sun.

During a planetary flyby, the planet bends the spacecraft's velocity vector.

Because the planet itself is moving around the Sun, the spacecraft can:

• gain heliocentric orbital energy
• lose heliocentric orbital energy

This can move the spacecraft onto a different solar orbit without requiring the equivalent energy change entirely from onboard propulsion.

Thus orbital energy provides another way of understanding gravitational slingshots.

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Real Mission Applications

Orbital-energy calculations are important for:

• satellite deployment
• orbit raising
• geostationary transfers
• lunar missions
• interplanetary missions
• planetary orbit insertion
• rendezvous
• deorbiting
• gravity assists
• asteroid and comet missions

Mission planners continually analyze how much orbital energy must be added or removed.

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Orbit Insertion

A spacecraft arriving at another planet may initially have:

E_total > 0

relative to that planet.

It is therefore unbound and would fly past.

To enter orbit, the spacecraft must lose energy.

An engine burn can reduce kinetic energy until:

E_total < 0

The spacecraft then becomes gravitationally bound.

This manoeuvre is called:

orbit insertion.

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Aerobraking and Energy

If a planet has an atmosphere, spacecraft may sometimes use atmospheric drag to remove orbital energy.

This is called aerobraking.

Instead of using large amounts of propellant, repeated atmospheric passes can gradually lower the spacecraft's orbital energy.

The lost mechanical energy is transferred mainly into thermal energy.

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Comparing Two Orbits

Suppose two identical satellites orbit the same planet.

Satellite A:

r = R

Satellite B:

r = 2R

For circular orbits:

Kinetic Energy

Eₖ ∝ 1/r

Therefore:

Eₖ,B = 1/2 Eₖ,A

Potential Energy

Eₚ ∝ −1/r

Therefore, Satellite B's potential energy has half the negative magnitude.

Total Energy

E_total ∝ −1/r

Therefore, Satellite B's total energy also has half the negative magnitude.

Satellite B is less tightly bound.

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Energy Graphs

Orbital energy can be represented graphically as a function of distance.

Gravitational potential energy:

Eₚ = −GMm/r

approaches:

0

as:

r → ∞

For a circular orbit:

Eₖ = +GMm/(2r)

and:

E_total = −GMm/(2r)

At any circular-orbit radius:

• Eₚ is negative
• Eₖ is positive
• |Eₚ| = 2Eₖ
• E_total is negative

As radius increases, all three approach zero.

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Reading an Orbital-Energy Diagram

Suppose an energy diagram shows:

Eₚ = −80 MJ

For a circular orbit:

Eₖ = +40 MJ

and:

E_total = −40 MJ

If energy is added until:

E_total = −10 MJ

the spacecraft remains bound.

If energy is added until:

E_total = 0

it reaches the escape threshold.

If:

E_total = +10 MJ

the spacecraft is unbound.

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Orbital Energy Is Relative to the Central Body

Orbital energy depends on the gravitational system being considered.

A spacecraft may be:

• unbound from Earth
• but still bound to the Sun

For example, a spacecraft that escapes Earth usually enters a new orbit around the Sun.

Therefore, statements such as "the spacecraft has escaped" must specify:

escaped from what?

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Leaving Earth Versus Leaving the Solar System

Escaping Earth requires enough energy to become unbound from Earth.

Escaping the Solar System requires enough heliocentric energy to become unbound from the Sun.

These are different energy conditions.

A spacecraft can have:

E_Earth ≥ 0

while still having:

E_Sun < 0

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Energy and Mission Planning

When planning an orbital manoeuvre, engineers can ask:

What is the initial orbital energy?

What is the desired final orbital energy?

How much energy must be added or removed?

What velocity change will produce that energy change?

Where in the orbit should the engine burn occur?

Because kinetic energy depends on v², the location of a burn can strongly affect the resulting orbit.

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Why Burn Location Matters

A spacecraft's speed changes around an elliptical orbit.

An engine burn performed where the spacecraft is already moving rapidly can produce a particularly useful change in orbital energy.

This is one reason orbital manoeuvres are carefully timed rather than performed at arbitrary locations.

Mission design involves not only:

how much thrust

but also:

where and when the thrust is applied.

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A General Problem-Solving Strategy

For an orbital-energy problem:

Identify the system

Determine:

• central mass M
• spacecraft mass m
• orbital radius r
• orbital speed v

Calculate kinetic energy if needed

Eₖ = 1/2 mv²

Calculate gravitational potential energy

Eₚ = −GMm/r

Calculate total energy

E_total = Eₖ + Eₚ

For circular orbits, use the shortcuts:

Eₖ = GMm/(2r)

Eₚ = −GMm/r

E_total = −GMm/(2r)

Interpret the sign

• negative → bound
• zero → escape threshold
• positive → unbound

For an orbital transfer

Calculate:

ΔE = E_final − E_initial

Then determine whether energy must be:

• added
• removed

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Common Mistakes

Forgetting that gravitational potential energy is negative

For orbital problems:

Eₚ = −GMm/r

Using mgh for large orbital changes

Use −GMm/r when gravitational field strength changes significantly.

Thinking negative energy means the satellite has no energy

Negative total energy means the satellite is gravitationally bound relative to the chosen zero.

Thinking higher orbits have less total energy

Higher circular orbits have greater, less-negative total energy.

Thinking higher orbital energy means higher circular speed

Higher circular orbits have greater total energy but lower orbital speed.

Confusing kinetic energy with total energy

Total orbital energy includes both kinetic and gravitational potential energy.

Thinking an elliptical orbit has changing total energy

Eₖ and Eₚ change, but their sum remains constant when only gravity does work.

Using r as altitude

Orbital radius is measured from the centre of the central body.

Forgetting the factor of 2

For circular orbits:

E_total = −GMm/(2r)

not −GMm/r.

Thinking zero total energy means the spacecraft is stationary

At a finite distance, an object at the escape threshold is moving. Its speed approaches zero only as r approaches infinity.

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Key Terms

Orbital energy — mechanical energy associated with an object's motion and position in a gravitational system.

Kinetic energy — energy due to motion, calculated using Eₖ = 1/2mv².

Gravitational potential energy — energy associated with the positions of masses in a gravitational field, Eₚ = −GMm/r.

Total orbital energy — sum of kinetic and gravitational potential energy.

Bound orbit — orbit with negative total mechanical energy.

Unbound trajectory — trajectory with positive total mechanical energy relative to the gravitational source.

Escape threshold — condition where total mechanical energy equals zero.

Specific orbital energy — orbital energy per unit spacecraft mass.

Semi-major axis — half the longest diameter of an ellipse; determines the total energy of a bound Keplerian orbit.

Periapsis — closest point of an orbit to the central body.

Apoapsis — farthest point of an orbit from the central body.

Vis-viva equation — equation relating orbital speed, distance and semi-major axis.

Orbital transfer — manoeuvre that changes a spacecraft from one orbit to another.

Hohmann transfer — efficient two-burn transfer between approximately circular coplanar orbits.

Orbit insertion — manoeuvre that reduces spacecraft energy sufficiently for gravitational capture.

Deorbit burn — manoeuvre that decreases orbital energy so a spacecraft's trajectory enters an atmosphere or intersects the central body.

Orbital decay — gradual loss of orbital energy, often due to atmospheric drag.

Aerobraking — use of atmospheric drag to remove spacecraft orbital energy.

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Key Takeaways

• Orbital energy consists primarily of kinetic energy and gravitational potential energy.
• E_total = Eₖ + Eₚ.
• Kinetic energy is Eₖ = 1/2mv².
• Gravitational potential energy is Eₚ = −GMm/r.
• Gravitational potential energy is negative because zero is defined at infinite separation.
• Bound circular and elliptical orbits have negative total energy.
• Zero total energy represents the ideal escape threshold.
• Positive total energy represents an unbound trajectory.
• For a circular orbit, Eₖ = GMm/(2r).
• For a circular orbit, Eₚ = −GMm/r.
• For a circular orbit, E_total = −GMm/(2r).
• Therefore, for circular orbits, Eₚ = −2Eₖ.
• Higher circular orbits have lower speeds but greater total orbital energy.
• Moving to a higher orbit requires adding energy.
• Moving to a lower orbit requires removing energy.
• In an elliptical orbit, kinetic and potential energy continually transform into one another.
• Total mechanical energy remains constant during an unpowered ideal orbit.
• Objects move fastest near periapsis and slowest near apoapsis.
• For an elliptical orbit, E_total = −GMm/(2a).
• The vis-viva equation relates orbital speed to position within an elliptical orbit.
• Rocket burns change kinetic energy and therefore change orbital energy.
• Orbital transfers commonly use temporary elliptical trajectories.
• Reaching escape conditions means raising total orbital energy to zero or above.
• Gravity assists can change a spacecraft's orbital energy relative to the Sun.
• Orbital-energy analysis is fundamental to satellite operations, interplanetary travel, orbit insertion, aerobraking and spacecraft return.

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Check Your Understanding

1. Define orbital energy and identify its two main components.
2. Explain why gravitational potential energy is negative for an orbiting satellite.
3. A 1200 kg satellite moves at 7.4 × 10³ m/s. Calculate its kinetic energy.
4. A satellite has Eₖ = +3.0 × 10¹⁰ J and Eₚ = −6.0 × 10¹⁰ J. Calculate its total energy and determine whether it is gravitationally bound.
5. Derive Eₖ = GMm/(2r) for a circular orbit using the circular orbital-speed relationship.
6. A circular-orbit satellite has kinetic energy +8.0 × 10⁹ J. Determine its gravitational potential energy and total orbital energy.
7. Explain why a bound orbit must have negative total mechanical energy.
8. Two identical satellites orbit the same planet at radii r and 2r. Compare their kinetic, potential and total energies.
9. Explain why a satellite in a higher circular orbit can have less kinetic energy but more total energy than a satellite in a lower orbit.
10. A spacecraft moves from an orbit with total energy −9.0 × 10⁹ J to one with total energy −5.0 × 10⁹ J. Calculate the change in orbital energy and state whether energy was added or removed.
11. Describe the changes in kinetic and gravitational potential energy as a satellite moves from apoapsis to periapsis in an elliptical orbit.
12. Explain the significance of the semi-major axis in the equation E_total = −GMm/(2a).
13. Describe the two main engine burns used to move a spacecraft from a lower circular orbit to a higher circular orbit using a Hohmann transfer.
14. A spacecraft approaching a planet has positive total orbital energy relative to that planet. Explain what must happen for the spacecraft to enter a bound orbit.
15. A student says, "A spacecraft in a higher circular orbit has less speed, so it must have less total energy." Evaluate this statement using kinetic energy, gravitational potential energy and total orbital energy.