Orbital Energy, Escape Speed, and Bound versus Unbound Motion

(Topic)

Orbital Energy, Escape Speed, and Bound versus Unbound Motion

M07-04 introduced the gravitational potential energy

GM m U(r) = − ------- r

(1)

for a test mass

m

(2)

moving outside a spherical source mass

M.

(3)

If no nongravitational work is done, the total mechanical energy is conserved:

|--------------------------------| | 1- 2 GM--m-- | |E = 2mv − r = constant.| ---------------------------------

(4)

The sign of this conserved energy immediately classifies gravitational motion.

With the usual reference

U (∞ ) = 0,

(5)

|------| |E < 0 | --------

(6)

corresponds to bound motion,

|------| -E-=-0--

(7)

is the escape threshold, and

|------| -E->-0--

(8)

corresponds to unbound motion.

Figure 1. Gravitational motion can be interpreted using the potential U = −GMm∕r and a horizontal total-energy level. The kinetic energy at any radius is the vertical difference K = E −U.

1 Specific orbital energy

The total energy scales with the orbiting test mass.

Divide by

m

(9)

to define the specific orbital energy:

|--------| | E | |𝜀 = -- .| -----m---

(10)

Then

|------------| | v2 μ | |𝜀 = ---− --,| -----2----r---

(11)

where

|--------| μ-=--GM---

(12)

is the gravitational parameter of the central body in the one-body approximation.

The units of

𝜀

(13)

are

|--------------| |J∕kg = m2 ∕s2.| ----------------

(14)

Specific energy is especially useful because it removes the test mass from the orbital dynamics.

2 Energy conservation between two radii

At two locations on the same trajectory,

1 2 μ 1 2 μ --v1 − --= -v2 − -- . 2 r1 2 r2

(15)

Therefore

|------------(--------)---| |2 2 1- -1 | v2 = v1 + 2μ r2 − r1 . | --------------------------

(16)

This equation is valid whenever gravity is the only force doing work.

It does not require circular motion.

3 Example 1: falling inward

Suppose an object moves from

r1 = 2R

(17)

r2 = R

(18)

and has initial speed

v . 1

(19)

Then

v₂²	= v₁² + 2μ	(20)
	= v₁² + .	(21)

Thus

|----∘----------| | μ | v2 = v21 +--. | ------------R----

(22)

The speed increases because gravitational potential energy becomes kinetic energy.

4 Circular orbit energy

For a circular orbit of radius

r,

(23)

M07-01 gave

|-2-----μ--| |vcirc = --.| --------r--|

(24)

The specific kinetic energy is

K--= 1v2circ = -μ-. m 2 2r

(25)

The specific potential energy is

U- = − μ-. m r

(26)

Therefore the total specific energy is

𝜀_circ	= −	(27)
	= −.	(28)

The total orbital energy is

|---------GM--m--| Ecirc = − ------.| ------------2r----

(29)

Figure 2. In a circular orbit, the kinetic energy is +GMm∕(2r), the potential energy is −GMm∕r, and the total energy is −GMm∕(2r).

5 A useful circular-orbit relation

For a circular orbit,

|----------| K = − 1-U.| -------2----

(30)

Therefore

|--------------------------| | 1- | |E = K + U = 2U = − K. | ---------------------------

(31)

These relations apply to the inverse-square circular orbit.

They should not be assumed for an arbitrary instantaneous point on a noncircular trajectory.

6 Bound motion

Suppose

𝜀 < 0.

(32)

At infinity,

− μ∕r → 0.

(33)

Energy conservation would require

1- 2 2 v∞ = 𝜀 < 0,

(34)

which is impossible for a real speed.

Therefore a negative-energy trajectory cannot reach infinity.

It is gravitationally bound:

|------------------------------| -𝜀-<-0---=-⇒----bound--motion.-|

(35)

In the ideal inverse-square two-body problem, noncollision bound orbits are ellipses, with the circle as a special case.

The conic-orbit derivation will appear later in the M07 sequence.

7 Escape threshold

At the threshold of escape, the body reaches infinity with zero remaining speed.

Thus

v∞ = 0

(36)

and

|------| -𝜀 =-0.-

(37)

At radius

r,

(38)

0 = 1v2 − μ. 2 esc r

(39)

Therefore

|------------| | ∘ --- | |vesc = 2μ-.| ----------r---

(40)

This result was introduced in M07-04. Here its role as the energy boundary between bound and unbound motion is central.

8 Unbound motion

𝜀 > 0,

(41)

the object can reach infinity and retain nonzero speed.

At infinity,

1- 2 𝜀 = 2 v∞.

(42)

Define the hyperbolic excess speed

|-----√----| v∞--=---2𝜀.-

(43)

Then

|--------| | v2∞- | -𝜀 =--2-.-

(44)

The name comes from the hyperbolic orbit that appears in the ideal inverse-square scattering problem.

Figure 3. The zero of total energy separates bound from unbound gravitational motion. Negative energy is bound, zero energy is the escape threshold, and positive energy leaves a nonzero speed at infinity.

9 Local speed for an unbound trajectory

For positive specific energy,

1 μ 1 -v2 − --= -v2∞. 2 r 2

(45)

Hence

|----------2-μ-| |v2 = v2∞ + ---.| ------------r---

(46)

Since

v2 = 2μ ∕r, esc

(47)

|--------------| v2 = v2∞ + v2esc.| ----------------

(48)

This is a very useful relation for unbound trajectories.

10 Example 2: unbound arrival speed

Suppose a body approaches a planet with

v∞ = 3.00 km ∕s.

(49)

At a radius where the local escape speed is

vesc = 10.0km ∕s,

(50)

its speed is

v	= km∕s	(51)
	= 10.44 km∕s.	(52)

Gravity increases the speed above the asymptotic value.

11 Escape speed versus radius

The escape speed decreases with radius:

|--------∘------| | 2μ- | vesc(r) = r . | -----------------

(53)

Thus

|-------−1∕2-| -vesc-∝-r----.-

(54)

Doubling the radius reduces escape speed by

√ -- 1∕ 2.

(55)

Quadrupling the radius reduces escape speed by

1∕2.

(56)

Figure 4. Escape speed decreases as r^−1∕2. The curve is normalized to the escape speed at a reference radius r₀.

12 Energy classification at one location

At a given radius

r,

(57)

compare the instantaneous speed with local escape speed.

v < vesc,

(58)

then

𝜀 < 0.

(59)

v = vesc,

(60)

then

𝜀 = 0.

(61)

v > vesc,

(62)

then

𝜀 > 0.

(63)

Thus

|----------------------------------| | | | v < vesc =⇒ bound energy, | | v = vesc =⇒ escape threshold, | | v > vesc =⇒ unbound energy. | -----------------------------------

(64)

The trajectory shape still depends on the velocity direction and angular momentum.

Energy alone classifies bound versus unbound behavior.

13 Speed-radius curves for fixed energy

Solve the specific-energy equation for speed:

|-------∘------------| | ( μ) | |v(r) = 2 𝜀 + r- .| ----------------------

(65)

For

𝜀 = 0,

(66)

this reduces to escape speed.

For

𝜀 > 0,

(67)

the speed approaches

√--- 2 𝜀

(68)

at infinity.

For

𝜀 < 0,

(69)

the expression cannot remain real to arbitrarily large

r.

(70)

That is another way to see the existence of a finite outer turning radius for purely radial bound motion.

Figure 5. Speed as a function of radius for representative negative, zero, and positive specific energies. The zero-energy curve is the escape-speed curve.

14 Radial motion and turning points

Now consider purely radial motion.

At an outer turning point,

v = 0.

(71)

Then

--μ-- 𝜀 = − rmax .

(72)

Therefore a radial trajectory with

𝜀 < 0

(73)

has

|---------μ--| |rmax = − --.| ----------𝜀--|

(74)

If the initial radius and speed are

r0, v0,

(75)

then

1 2 μ 𝜀 = --v0 − --. 2 r0

(76)

Thus

|----------------| |-1--- -1 -v20 | |rmax = r0 − 2 μ.| ------------------

(77)

15 Example 3: radial launch below escape speed

Let

v0 = 0.80vesc(r0).

(78)

Since

v2 = 2μ∕r0, esc

(79)

𝜀	= (0.80)² −	(80)
	= −0.36.	(81)

Therefore

|----------------------| | -r0- | -rmax-=-0.36-≈--2.78r0.-|

(82)

The object rises to a finite radius and then falls back in the ideal radial problem.

Figure 6. A purely radial launch below escape speed has negative energy and reaches a finite outer turning radius where its speed becomes zero.

16 Why radial turning points are a special case

For nonzero angular momentum, a turning point in radius does not imply that the total speed is zero.

At periapsis or apoapsis, the radial velocity is zero but the tangential velocity is generally nonzero.

Therefore the simple relation

𝜀 = − μ ∕rmax

(83)

applies only to purely radial motion.

M07-06 will introduce the effective potential that incorporates angular momentum and treats general radial turning points correctly.

17 Circular orbit versus escape at the same radius

At radius

r,

(84)

∘ -- μ vcirc = -- r

(85)

and

∘ --- v = 2μ-. esc r

(86)

Therefore

|------√-------| |v = 2v .| --esc-------circ--

(87)

If an object is initially in a circular orbit and receives an instantaneous prograde speed change at that same point, the ideal speed increment required to reach the escape threshold is

|-------√------------| -Δv--=-(--2-−-1)vcirc.|

(88)

This statement assumes the impulse changes only the velocity instantaneously and neglects other bodies and nongravitational forces.

Figure 7. At a fixed radius, escape speed is √2-- times circular speed. Raising a circular-orbit speed to the local escape value moves the total energy to the E = 0 threshold.

18 Example 4: circular orbit energy near Earth

Using

14 3 2 μE = 3.9860 × 10 m ∕s

(89)

r = 7.00 × 106 m,

(90)

the circular speed is

v_circ	=	(91)
	≈ 7.55 km∕s.	(92)

The escape speed at the same radius is

|------------------| -vesc-≈-10.67-km-∕s.|

(93)

The specific circular energy is

𝜀_circ	= −	(94)
	≈−2.85 × 10⁷ J∕kg.	(95)

19 Energy of an object falling from rest at infinity

Suppose an object starts at infinity with negligible speed.

Then

𝜀 = 0.

(96)

At radius

r,

(97)

1-v2 − μ-= 0. 2 r

(98)

Therefore

|--------------------| | ∘ --- | |v = 2μ-= v (r).| --------r------esc----

(99)

Thus the speed acquired by falling from rest at infinity equals the local escape speed.

This follows from time-reversal symmetry of the conservative trajectory.

20 Bound does not mean circular

A negative-energy orbit need not be circular.

The object may move through a range of radii with both radial and tangential velocity.

Energy determines whether infinity is energetically accessible.

Angular momentum determines additional geometric features of the trajectory.

The full orbit therefore requires both:

energy,
angular momentum.

21 Unbound does not mean purely radial

Likewise, positive energy does not imply radial escape.

An unbound trajectory may pass around the central mass and depart in a different direction.

In the ideal inverse-square problem, positive-energy noncollision trajectories are hyperbolas.

The energy sign gives the broad classification, while angular momentum fixes the impact geometry and periapsis distance.

22 Exact two-body energy

So far we have used the one-body approximation in which

M

(100)

is treated as fixed.

For two finite masses

m1

(101)

and

m2,

(102)

define the relative position

r = r2 − r1

(103)

and relative speed

vrel = |˙r|.

(104)

Define the reduced mass

|---------------| μ = -m1m2---. | | r m1 + m2 | -----------------

(105)

Then the relative mechanical energy is

|-------------------------| | 1 Gm m | Erel = -μrv2rel − ---1--2. | -------2------------r-----

(106)

Figure 8. The exact two-body problem separates into center of mass motion plus relative motion. The relative coordinate behaves like a particle of reduced mass moving in an inverse-square gravitational interaction.

23 Exact specific relative energy

Divide

Erel

(107)

μr.

(108)

Since

Gm1m2---= G (m1 + m2 ), μr

(109)

the specific relative energy is

|--------------------------| | 1-2 G-(m1-+-m2-) | 𝜀rel = 2vrel − r .| ----------------------------

(110)

Thus the exact two-body gravitational parameter is

|------------------| |μ = G (m1 + m2 ). | -------------------

(111)

When

m2 ≪ m1,

(112)

this reduces to

μ ≈ Gm1.

(113)

24 Exact two-body escape speed

The relative escape threshold satisfies

𝜀rel = 0.

(114)

Therefore

|---------∘----------------| |v = 2G-(m1-+--m2). | --rel,esc-----------r--------|

(115)

The familiar

∘ -------- 2GM ∕r

(116)

formula is the large-central-mass approximation to this exact relative result.

25 Energy sign and conic classification

For the ideal Newtonian inverse-square two-body problem,

|----------------------------------------| |𝜀 < 0 =⇒ elliptic bound orbit, | |𝜀 = 0 =⇒ parabolic escape orbit, | | | -𝜀->--0--=⇒---hyperbolic-unbound--orbit.--

(117)

A circle is a special ellipse.

This classification will be derived geometrically from the orbit equation later.

At this stage, the energy argument already establishes the bound versus unbound distinction.

26 Common mistakes

Forgetting the negative gravitational potential-energy term in the orbital energy.
Confusing total energy E with specific energy 𝜀 = E∕m.
Using GM∕r² where the energy formula requires GM∕r.
Assuming negative energy means circular motion.
Assuming positive energy means radial motion.
Forgetting that the zero-energy case is the escape threshold.
Treating escape speed as a constant independent of radius.
Forgetting that v_esc = v_circ only compares speeds at the same radius.
Using a radial turning-point formula for an orbit with nonzero angular momentum.
Assuming the total speed is zero at apoapsis.
Forgetting that v_∞ is the speed remaining at infinite separation.
Using v² = v_∞² + v_esc² for a negative-energy trajectory.
Confusing the gravitational parameter μ with the reduced mass μ_r.
Using μ = GM for an exact two-body problem when both masses are dynamically important.
Forgetting that energy classification alone does not determine the complete orbit geometry.

27 Practice exercises

Derive the specific orbital-energy equation from total mechanical energy.
Derive the two-radius speed relation from energy conservation.
Derive the circular-orbit total and specific energies.
Show that a negative-energy trajectory cannot reach infinity.
Derive escape speed by imposing 𝜀 = 0.
Show that positive specific energy equals v_∞²∕2.
Derive v² = v_∞² + v_esc².
Show that escape speed scales as r^−1∕2.
For purely radial bound motion, derive r_max = −μ∕𝜀.
Reproduce the v₀ = 0.80v_esc radial-turning-point example.
Derive the speed increment ( − 1)v_circ from circular orbit to local escape threshold.
Reproduce the circular speed, escape speed, and specific energy at r = 7.00 × 10⁶ m around Earth.
Show that falling from rest at infinity produces the local escape speed.
Derive the exact two-body relative energy using reduced mass.
Derive the exact relative escape speed .

28 Summary

For motion in a fixed spherical gravitational field,

|--------------------| | 1 2 GM m | E = 2-mv − --r---.| ----------------------

(118)

The specific orbital energy is

|-----2------| | v-- μ- | -𝜀 =-2--−-r-.-

(119)

For a circular orbit,

|------------| | -μ- | -𝜀circ =-−-2r-.

(120)

The energy sign classifies the motion:

|------------------------------------------------------------| -𝜀 <-0 bound,-----𝜀-=-0-escape-threshold,-----𝜀 >-0-unbound.--

(121)

The local escape speed is

|------------| | ∘ --- | |vesc = 2μ-.| ----------r---

(122)

For unbound motion,

|--------------| | 2 2 2 μ | |v = v∞ + -r-.| ----------------

(123)

For exact relative two-body motion,

|--------------------------| 𝜀 = 1v2 − G-(m1-+-m2-).| -rel---2-rel--------r--------

(124)

Energy determines whether a gravitational trajectory is bound or unbound. Angular momentum, introduced into the radial energy equation in M07-06, determines the allowed radial motion and turning points in the general case.

References

[1] PhysicsLibrary, M07-04, Gravitational Potential Energy and Gravitational Potential.

[2] PhysicsLibrary, M07-01, Newton’s Law of Universal Gravitation.

[3] PhysicsLibrary, M04-08, Conservation of Angular Momentum and Central Force Motion.

[4] J. R. Taylor, Classical Mechanics, University Science Books, 2005.

[5] D. Kleppner and R. Kolenkow, An Introduction to Mechanics, 2nd ed., Cambridge University Press, 2014.

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Orbital Energy, Escape Speed, and Bound versus Unbound Motion