Orbital vs Escape Velocity: What Is the Difference?

Orbital velocity is the speed that holds a circular orbit at a given radius. Escape velocity is the speed that gets away from that same radius forever. Escape is exactly the square root of two times orbital, about 41 percent faster, which takes twice the kinetic energy.

AP Physics: Unit 6 (topics 6.6 Motion of Orbiting Satellites, 2.9 Circular Motion). Escape velocity is required content in AP Physics 1 at EK 6.6.A.3 and in AP Physics C: Mechanics at EK 6.6.A.4, with identical wording in both: the escape velocity of a satellite is the satellite's velocity such that the mechanical energy of the system consisting of the satellite and the central object is equal to zero. Both attach the derived equation v_esc = square root of 2GM over r, and both add the same two sub-statements, that a satellite reaching escape velocity moves away until its speed reaches zero at an infinite distance, and that the escape velocity can be derived using conservation of energy laws. Orbital velocity is not named in any of the four CEDs; what is required is EK 2.9.B.1 in AP Physics 1, under LO 2.9.B, describe circular orbits using Kepler's third law, which states that for a satellite in circular orbit the centripetal acceleration is caused only by gravitational attraction and gives the derived equation T squared = 4 pi squared R cubed over GM. AP Physics C: Mechanics carries the same at Topic 2.10, LO 2.10.B, and both courses attach a boundary statement that students are not expected to know Kepler's first or second laws of planetary motion. AP Physics C: Mechanics additionally requires the circular-orbit energy budget at EK 6.6.A.3, with derived equations K = minus one half U and E_total = one half U = minus GMm over 2r. AP Physics 1 Unit 6 is weighted at 5 to 8 percent over about 5 to 10 class periods with suggested skills 1.C, 2.A, 2.C and 3.C for Topic 6.6; AP Physics C: Mechanics Unit 6 is weighted at 10 to 15 percent over about 13 to 19 class periods with suggested skills 1.C, 2.A and 3.C. Neither the escape velocity nor the orbital velocity expression is printed on any of the four equation sheets, although the gravitational force law, the gravitational potential energy and the centripetal acceleration all are.

Two speeds, one radius, one factor

Orbital velocity is the speed that keeps you going round. Escape velocity is the speed that gets you away. Quote them at the same radius around the same central body and they are locked to each other:

v_{\text{orb}} = \sqrt{\frac{GM}{r}}, \qquad v_{\text{esc}} = \sqrt{\frac{2GM}{r}} = \sqrt{2}\; v_{\text{orb}}

That factor of $\sqrt{2} = 1.414$ is the whole comparison, and it holds at every radius around every body. It is not a rule of thumb and not an approximation; it falls out of the two derivations in the next section and it is worth deriving rather than remembering, because the derivation is what an AP free-response question will ask for.

The AP CEDs are asymmetric about these two. Escape velocity is named and defined, in AP Physics 1 at EK 6.6.A.3 and in AP Physics C: Mechanics at EK 6.6.A.4, with identical wording: the escape velocity of a satellite is the satellite's velocity such that the mechanical energy of the system consisting of the satellite and the central object is equal to zero. Both courses attach the same derived equation $v_{\text{esc}} = \sqrt{2GM/r}$ .

Orbital velocity is never named in any of the four Course and Exam Descriptions. What is there is everything needed to produce it. AP Physics 1 EK 2.9.B.1 says that for a satellite in circular orbit around a central body, the satellite's centripetal acceleration is caused only by gravitational attraction, and that the period and radius of the circular orbit are related to the mass of the central body, with the derived equation $T^2 = \dfrac{4\pi^2}{GM}R^3$ . That first sentence is the orbital speed derivation in words. The C: Mechanics version sits at EK 2.10.B.1 with the same content.

So both are derived quantities in the CED's own labelling, and neither appears on any equation sheet. Section seven checks that claim against all four booklets.

Orbital vs escape velocity, side by side

Question you are asking	Orbital velocity	Escape velocity
What it achieves	A closed circular path at radius $r$	Reaching infinite distance with speed zero
Formula	$v_{\text{orb}} = \sqrt{\dfrac{GM}{r}}$	$v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$
Named in the CED	No, in none of the four	Yes, EK 6.6.A.3 and EK 6.6.A.4
What it is derived from	Newton's second law with centripetal acceleration	Conservation of energy with total energy zero
Total mechanical energy of the system	$-\dfrac{GMm}{2r}$ , negative	Exactly zero
Kinetic energy in terms of $U$	$K = -\tfrac{1}{2}U$ , EK 6.6.A.3 in C: Mechanics	$K = -U$
Depends on the satellite's mass	No	No
Depends on the direction of travel	Yes, it must be perpendicular to the radius	No, only the speed matters
Depends on radius as	$1/\sqrt{r}$	$1/\sqrt{r}$
Ratio to the other at the same $r$	$1/\sqrt{2} = 0.707$ of escape	$\sqrt{2} = 1.414$ of orbital
Kinetic energy ratio at the same $r$	Half	Twice
Printed on any AP equation sheet	No	No

Four rows are worth unpacking.

The total energy row is the real definition of both. Escape velocity is defined by the energy being zero, which is exactly what EK 6.6.A.3 says. Circular orbital velocity produces a total energy of $-GMm/2r$ , which is negative, and a negative total energy is what bound means. The comparison between the two speeds is a comparison between a bound system and one that is exactly on the boundary.

The mass row applies to both, and for the same reason. Every term in both derivations carries one factor of the satellite's mass, so it divides out. A bolt and a space station at the same radius need the same speed to orbit and the same speed to escape.

The direction row is the difference people miss. An orbital velocity is a genuine velocity: get the magnitude right and the direction wrong and you get an ellipse or a crash, not a circle. Escape velocity is really a speed. Its derivation uses only $K = \tfrac{1}{2}mv^2$ , which contains no direction, so any direction that does not run into the central body will do.

The radius row shows why the ratio is constant. Both fall off as $1/\sqrt{r}$ , so their quotient cannot depend on $r$ . Move out and both speeds drop, in lockstep, keeping the factor of $\sqrt{2}$ exactly.

Where the root two comes from

Both derivations are short. Doing them side by side is the fastest way to see that the $\sqrt{2}$ is a factor of two in the energy, wearing a square root.

Orbital velocity, from the second law. Start from EK 2.9.B.1: for a satellite in circular orbit around a central body, the satellite's centripetal acceleration is caused only by gravitational attraction. So set the gravitational force equal to the mass times the centripetal acceleration $a_c = v^2/r$ :

\frac{GMm}{r^2} = \frac{mv^2}{r}

Cancel one $m$ and one power of $r$ :

v_{\text{orb}}^2 = \frac{GM}{r} \quad \Longrightarrow \quad v_{\text{orb}} = \sqrt{\frac{GM}{r}}

Escape velocity, from the energy. Start from EK 6.6.A.3's definition, that the mechanical energy of the satellite and central object system is zero, using $U_G = -\dfrac{GMm}{r}$ , which is printed on all four sheets:

K + U = 0 \quad \Longrightarrow \quad \frac{1}{2}mv_{\text{esc}}^2 - \frac{GMm}{r} = 0 \quad \Longrightarrow \quad v_{\text{esc}}^2 = \frac{2GM}{r}

Divide one squared speed by the other.

\frac{v_{\text{esc}}^2}{v_{\text{orb}}^2} = \frac{2GM/r}{GM/r} = 2 \quad \Longrightarrow \quad \frac{v_{\text{esc}}}{v_{\text{orb}}} = \sqrt{2}

Everything cancels except the $2$ , and that $2$ traces back to the $\tfrac{1}{2}$ in the kinetic energy meeting a potential energy with no fraction in front of it. So the root two is not about orbits at all. It is the factor by which a $\tfrac{1}{2}mv^2$ has to grow to match a $GMm/r$ instead of half of one.

One consequence is worth carrying as a check. Since $v_{\text{esc}}(r) = \sqrt{2GM/r}$ and $v_{\text{orb}}(r/2) = \sqrt{GM/(r/2)} = \sqrt{2GM/r}$ , the escape speed at any radius equals the circular orbital speed at half that radius. Worked example three verifies it with numbers.

AP Physics C: Mechanics describes exactly this procedure in its instructional approaches, suggesting that when deriving the escape speed of a rocket launched from a planet, students choose an equation or fundamental physics principle such as conservation of energy from the equation sheet and work forward from it. Note the word choice there: the CED framework says escape velocity and its own instructional text says escape speed.

The energy accounting the CED writes down

AP Physics C: Mechanics gives the circular-orbit energy budget as required content, and once you have it the escape comparison becomes arithmetic rather than a new derivation.

EK 6.6.A.3 in C: Mechanics: the total energy of a system consisting of a satellite orbiting a central object in a circular path can be written in terms of the gravitational potential energy of that system or the kinetic energy of the satellite. Its derived equations are

K = -\tfrac{1}{2}U \qquad\text{and}\qquad E_{\text{total}} = \tfrac{1}{2}U = -\frac{GMm}{2r}

Read those two lines carefully, because they encode the whole comparison.

$K = -\tfrac{1}{2}U$ says the kinetic energy of a circular orbit is half the magnitude of the potential energy. Escape requires $K = -U$ , the full magnitude. Twice as much kinetic energy, which is $\sqrt{2}$ times the speed.
$E_{\text{total}} = -\dfrac{GMm}{2r}$ is negative, and its magnitude is exactly the kinetic energy the satellite already has. So the energy you must add to escape from a circular orbit equals the kinetic energy the satellite is already carrying.

AP Physics 1 stops short of these two equations but sets up the same picture qualitatively. EK 6.6.A.2.i: in circular orbits, the system's total mechanical energy, the system's gravitational potential energy, and the satellite's angular momentum and kinetic energy are constant. EK 6.6.A.2.ii: in elliptical orbits, the system's total mechanical energy and the satellite's angular momentum are constant, but the system's gravitational potential energy and the satellite's kinetic energy can each change. EK 6.6.A.2.iii: the gravitational potential energy of a system consisting of a satellite and a massive central object is defined to be zero when the satellite is an infinite distance from the central object, with the relevant equation $U_G = -\dfrac{Gm_1m_2}{r}$ .

That last statement is the one that makes escape velocity definable at all. Zero potential energy at infinity is a choice of reference point, and it is the choice that turns having enough energy to escape into the clean condition total energy equals zero. With any other reference point the number would change and the physics would not. EK 6.6.A.3.i spells out what reaching that boundary looks like: when the only force exerted on a satellite is gravity from a central object, a satellite that reaches escape velocity will move away from the central body until its speed reaches zero at an infinite distance from the central body.

So an object at exactly escape velocity does not fly off at constant speed. It slows the whole way, and arrives nowhere at zero. Anything faster arrives somewhere with speed left over, which is worked example three.

The case that separates them: 41 percent more speed, 100 percent more energy

Take a satellite in a circular orbit and ask what it takes to leave. Numbers from worked examples one and two, for a $1200$ kg satellite at $r = 7.0 \times 10^6$ m around a body of mass $6.0 \times 10^{24}$ kg.

Quantity	In circular orbit	At escape velocity	Ratio
Speed	$7561$ m/s	$10\,693$ m/s	$1.414$
Kinetic energy	$3.43 \times 10^{10}$ J	$6.86 \times 10^{10}$ J	$2.00$
Potential energy	$-6.86 \times 10^{10}$ J	$-6.86 \times 10^{10}$ J	$1.00$
Total energy	$-3.43 \times 10^{10}$ J	$0$	Not applicable

The speed column rises by $41$ percent and the energy column rises by $100$ percent. That mismatch is the practical content of the comparison and it is why escaping is much harder than the speed figures suggest. Kinetic energy goes as $v^2$ , so a $1.414$ times speed increase is a $2.000$ times energy increase, and it is energy that a rocket has to supply.

The potential energy row does not move, because the satellite has not gone anywhere yet. The whole change is kinetic.

Read the total energy row as the definition it is. In orbit the total is $-3.43 \times 10^{10}$ J, and the object is bound: it cannot get arbitrarily far away, because $U$ approaches zero at infinity and $K$ cannot be negative, so a negative total forbids infinite separation. At escape velocity the total is exactly zero, the boundary case. The extra energy needed is $3.43 \times 10^{10}$ J, which is numerically the same as the orbital kinetic energy, exactly as $E_{\text{total}} = -K$ requires for a circular orbit.

There is a second, sharper separation between the two speeds, and it is about direction rather than energy. Give a satellite $7561$ m/s at that radius pointing straight down and it does not orbit, it falls. Give it $10\,693$ m/s pointing straight down and it still does not escape, because it hits the planet, but that is a collision rather than a failure of the physics. Give it $10\,693$ m/s in any direction with a clear path and it escapes, sideways, at an angle, straight up, all the same. Orbital velocity constrains a vector. Escape velocity constrains a scalar.

What the words velocity and escape actually mean here

Two pieces of vocabulary cause more trouble on this topic than the physics does.

Escape velocity is a speed, and the CED calls it a velocity anyway. EK 6.6.A.3 says the escape velocity of a satellite is the satellite's velocity such that the mechanical energy of the system is equal to zero, and the equation attached to it, $v_{\text{esc}} = \sqrt{2GM/r}$ , returns a positive number with no direction. The CED's own instructional approaches section uses escape speed, and so does the sample multiple-choice question in the C: Mechanics exam information. Both terms refer to the same quantity. Use whichever word the question uses, and remember that no direction is required when you compute it.

Escape does not mean leaving the field. The gravitational field never reaches zero at any finite distance, since $\lvert \vec{g} \rvert = GM/r^2$ is positive everywhere. What escape means is precisely EK 6.6.A.3.i: moving away from the central body until the speed reaches zero at an infinite distance. It is a statement about the limit of the motion, not about crossing a boundary into field-free space. A rocket at escape velocity is decelerating at every moment of its journey and never stops decelerating; it simply runs out of distance to be pulled back over.

Escape velocity is not the launch speed of anything real. It applies when gravity is the only force acting, which is EK 6.6.A.3.i's condition, so it describes a projectile given all its speed at once. A rocket under continuous thrust never needs to reach $11$ km/s at the ground, because it keeps adding energy on the way up. The formula is still the right way to ask whether an object already moving freely will come back.

Orbital velocity is not a single number for a body either. Both speeds carry a radius, and both fall off as $1/\sqrt{r}$ . Quoting an orbital speed without a radius is like quoting a rotational inertia without an axis: the question has no answer until the missing input arrives.

Where the confusion costs a mark

Each of these is a specific scoring error.

Using $v_{\text{esc}} = \sqrt{GM/r}$ or $v_{\text{orb}} = \sqrt{2GM/r}$ . The factor of $2$ is on the escape side, because escape needs the full $\lvert U \rvert$ and a circular orbit needs half of it. If you cannot remember which, rederive: escape comes from $K + U = 0$ .
Saying escape velocity is twice orbital velocity. It is $\sqrt{2}$ times, about $1.41$ . The factor of two is in the kinetic energy, not the speed.
Forgetting the minus sign on $U_G$ . With $U_G = -GMm/r$ the condition $K + U = 0$ gives $\tfrac{1}{2}mv^2 = GMm/r$ . Writing $U_G$ positive gives a total energy that is never zero and an escape condition that does not exist.
Measuring $r$ from the surface. Both formulas take $r$ from the center of the central body. Escape from the surface uses the body's radius, and escape from orbit uses the orbital radius, and they are different numbers.
Putting the satellite's mass in the answer. It cancels in both derivations. If $m$ survives to your final expression, a cancellation was missed.
Using the orbiting object's mass as $M$ . $M$ is the central body. This one produces answers wrong by many orders of magnitude and is worth a units check.
Treating escape velocity as directional. It comes from an energy equation, and kinetic energy has no direction. Any escape path that misses the central body works.
Saying an object at escape velocity travels at constant speed once it is far away. EK 6.6.A.3.i says its speed reaches zero at an infinite distance. It decelerates forever.
Applying $v_{\text{orb}} = \sqrt{GM/r}$ to an elliptical orbit. The derivation assumed circular motion with $a_c = v^2/r$ . EK 6.6.A.2.ii says that in elliptical orbits the gravitational potential energy and the kinetic energy each change, so there is no single orbital speed to quote.
Assuming the central body is unaffected. EK 6.6.A.1 grants this only when the satellite's mass is negligible in comparison with the central object's mass, in which case the motion of the central object itself is negligible.
Expecting either formula on the equation sheet. Neither is printed. The CED labels the escape velocity equation a derived equation, and orbital velocity it does not name at all.

What the CED requires, and what the sheets do not print

Both quantities live in Unit 6 for the energy treatment and Unit 2 for the circular-motion treatment, in the two mechanics courses.

AP Physics 1 Unit 6, Energy and Momentum of Rotating Systems: 5 to 8 percent of the multiple-choice section, about 5 to 10 class periods. Topic 6.6, Motion of Orbiting Satellites, carries a single learning objective, 6.6.A, describe the motions of a system consisting of two objects interacting only via gravitational forces, with EK 6.6.A.1 on the negligible-mass satellite, EK 6.6.A.2 with its three sub-statements on what is conserved in circular and elliptical orbits, and EK 6.6.A.3 with its two sub-statements on escape velocity. Suggested skills for Topic 6.6 are 1.C, 2.A, 2.C and 3.C.

AP Physics C: Mechanics Unit 6: 10 to 15 percent, about 13 to 19 class periods. Its Topic 6.6 carries the same LO 6.6.A and the same EK 6.6.A.1 and 6.6.A.2, then inserts the energy statement at EK 6.6.A.3 with $K = -\tfrac{1}{2}U$ and $E_{\text{total}} = \tfrac{1}{2}U = -GMm/2r$ , pushing escape velocity down to EK 6.6.A.4. Suggested skills for that topic are 1.C, 2.A and 3.C. So the essential knowledge numbers for escape velocity differ between the two courses, and quoting the wrong one is quoting the wrong course.

The circular-orbit half sits earlier, in Unit 2: AP Physics 1 Topic 2.9, Circular Motion, at LO 2.9.B, describe circular orbits using Kepler's third law, with EK 2.9.B.1 and the derived equation $T^2 = \dfrac{4\pi^2}{GM}R^3$ . AP Physics C: Mechanics carries the same at Topic 2.10, LO 2.10.B. Both courses attach a boundary statement saying students are not expected to know Kepler's first or second laws of planetary motion.

AP Physics 2 and AP Physics C: Electricity and Magnetism do not cover orbital motion, though both booklets print the gravitational force law and $U_G$ .

On the equation sheets, checked against the appendix pages of all four Course and Exam Descriptions:

Line	AP Physics 1	AP Physics 2	C: Mechanics	C: E and M
$\lvert \vec{F}_g \rvert = Gm_1m_2/r^2$	Printed	Printed	Printed	Printed
$U_G = -Gm_1m_2/r$	Printed	Printed	Printed	Printed
$a_c = v^2/r$	Printed	Printed	Printed	Printed
$v_{\text{esc}} = \sqrt{2GM/r}$	Not printed	Not printed	Not printed	Not printed
$v_{\text{orb}} = \sqrt{GM/r}$	Not printed	Not printed	Not printed	Not printed
$T^2 = 4\pi^2R^3/GM$	Not printed	Not printed	Not printed	Not printed

Everything the derivations need is printed and neither result is. That is the CED's design rather than an oversight, and it is why both quantities are labelled derived equations: the exam wants the derivation, and it has given you the ingredients on the sheet to build it from.

A CED sample multiple-choice question for C: Mechanics tests exactly this understanding. It launches a small rock from an atmosphereless planet at twice the escape speed and asks for its speed very far from the planet. The answer is $\sqrt{3}$ times the escape speed, and worked example three shows why in three lines.

For the CED framing, see Topic 6.6 in AP Physics 1 and Topic 6.6 in C: Mechanics. The conservation of energy guide has the energy routine, the centripetal force guide has the circular-motion side, and the circular motion and gravitation practice set has problems to work.

Both speeds at one radius, derived and then computed

A central body has mass $M = 6.0 \times 10^{24}$ kg. Find the circular orbital speed and the escape speed at a radius of $r = 7.0 \times 10^6$ m from its center, and confirm their ratio. Then find both again at the body's surface, $r = 6.0 \times 10^6$ m. Use $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ .

Compute $GM$ once and reuse it. $GM = (6.67 \times 10^{-11})(6.0 \times 10^{24}) = 4.002 \times 10^{14}\ \text{m}^3/\text{s}^2$ .
Orbital speed at $r = 7.0 \times 10^6$ m. From EK 2.9.B.1's statement that the centripetal acceleration is caused only by gravitational attraction, $GMm/r^2 = mv^2/r$ gives $v_{\text{orb}} = \sqrt{GM/r}$ . So $v_{\text{orb}}^2 = (4.002 \times 10^{14})/(7.0 \times 10^6) = 5.717 \times 10^{7}\ \text{m}^2/\text{s}^2$ , and $v_{\text{orb}} = 7561\ \text{m/s}$ , that is $7.56$ km/s.
Escape speed at the same radius. From EK 6.6.A.3's condition that the mechanical energy is zero, $v_{\text{esc}} = \sqrt{2GM/r}$ . So $v_{\text{esc}}^2 = 2(5.717 \times 10^{7}) = 1.1434 \times 10^{8}\ \text{m}^2/\text{s}^2$ , and $v_{\text{esc}} = 10\,693\ \text{m/s}$ , that is $10.7$ km/s.
Check the ratio. $10\,693 / 7561 = 1.4142$ , and $\sqrt{2} = 1.41421$ . The agreement is exact to the digits carried, as it must be, since the two expressions differ only by that factor of $2$ inside the root.
Now the surface, $r = 6.0 \times 10^6$ m. $v_{\text{orb}}^2 = (4.002 \times 10^{14})/(6.0 \times 10^6) = 6.670 \times 10^{7}$ , so $v_{\text{orb}} = 8167\ \text{m/s}$ . And $v_{\text{esc}} = \sqrt{2(6.670 \times 10^{7})} = \sqrt{1.334 \times 10^{8}} = 11\,550\ \text{m/s}$ .
Check the ratio again. $11\,550/8167 = 1.4142$ . Unchanged, because both speeds scale as $1/\sqrt{r}$ and the $r$ divides out of the ratio.
Check the radius dependence. Going from $6.0 \times 10^6$ m to $7.0 \times 10^6$ m multiplied $r$ by $7/6 = 1.167$ , so both speeds should fall by $\sqrt{1.167} = 1.080$ . Orbital: $8167/7561 = 1.080$ . Escape: $11\,550/10\,693 = 1.080$ . Both match.
Note what did not appear anywhere in this example: the mass of whatever is doing the orbiting or escaping. It cancelled in both derivations, so no value was needed.

At $r = 7.0 \times 10^6$ m the orbital speed is $7561$ m/s and the escape speed is $10\,693$ m/s, a ratio of $1.4142$ , which is $\sqrt{2}$ . At the surface, $r = 6.0 \times 10^6$ m, they are $8167$ m/s and $11\,550$ m/s, the same ratio. Both fall off as one over the square root of the radius, so the factor of $\sqrt{2}$ between them never changes.

The energy budget for leaving a circular orbit

A $1200$ kg satellite is in a circular orbit of radius $7.0 \times 10^6$ m around a body of mass $6.0 \times 10^{24}$ kg, moving at $7561$ m/s. Find its kinetic energy, the system's gravitational potential energy, and the total mechanical energy. Then find the kinetic energy it would need to escape, the extra energy required, and the extra speed. Use $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ .

Kinetic energy in orbit. $K = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(1200)(7561)^2$ . Using $v^2 = 5.717 \times 10^7$ from the previous example, $K = (600)(5.717 \times 10^7) = 3.430 \times 10^{10}\ \text{J}$ .
Potential energy, from the sheet. $U_G = -\dfrac{GMm}{r} = -\dfrac{(4.002 \times 10^{14})(1200)}{7.0 \times 10^6} = -\dfrac{4.802 \times 10^{17}}{7.0 \times 10^6} = -6.861 \times 10^{10}\ \text{J}$ . The negative sign is the convention EK 6.6.A.2.iii sets, with $U$ defined to be zero at infinite separation.
Check against the CED's derived equation. EK 6.6.A.3 in C: Mechanics gives $K = -\tfrac{1}{2}U$ for a circular orbit. Here $-\tfrac{1}{2}(-6.861 \times 10^{10}) = 3.430 \times 10^{10}\ \text{J}$ , which is the kinetic energy computed independently in step one. The two agree.
Total mechanical energy. $E = K + U = 3.430 \times 10^{10} - 6.861 \times 10^{10} = -3.430 \times 10^{10}\ \text{J}$ . Also equal to $\tfrac{1}{2}U = -GMm/2r$ , the CED's second derived equation, and it is negative, which is what bound means.
Kinetic energy needed to escape. Escape requires $K + U = 0$ at this radius, so $K_{\text{esc}} = -U = 6.861 \times 10^{10}\ \text{J}$ , exactly twice the orbital kinetic energy.
Extra energy required. $\Delta E = 6.861 \times 10^{10} - 3.430 \times 10^{10} = 3.430 \times 10^{10}\ \text{J}$ . That is the same number as the orbital kinetic energy, and it must be, because $E_{\text{total}} = -K$ for a circular orbit and escape means raising the total energy to zero.
Extra speed required. $v_{\text{esc}} - v_{\text{orb}} = 10\,693 - 7561 = 3132\ \text{m/s}$ , an increase of $3132/7561 = 41.4$ percent.
Line the two increases up. Speed up by $41.4$ percent, energy up by $100$ percent. The mismatch is the $v^2$ in the kinetic energy: $(1.414)^2 = 2.000$ . This is why a modest-sounding speed increase is a major engineering problem, and it is the practically important content of the factor $\sqrt{2}$ .

In orbit: $K = 3.430 \times 10^{10}$ J, $U = -6.861 \times 10^{10}$ J, total $E = -3.430 \times 10^{10}$ J. To escape from that radius the satellite needs $K = 6.861 \times 10^{10}$ J, twice as much, so it must be given an extra $3.430 \times 10^{10}$ J, which is numerically its whole orbital kinetic energy. The speed must rise by $3132$ m/s, only $41.4$ percent.

Launched at twice escape speed, and the CED sample question

A CED sample multiple-choice question for AP Physics C: Mechanics launches a small rock from the surface of a planet with no atmosphere at an initial speed of $2 v_{\text{esc}}$ , where $v_{\text{esc}}$ is the planet's escape speed, and asks for the rock's speed when it is very far from the surface. Work it symbolically. Then confirm, using the planet from worked example one, that the escape speed at a radius equals the circular orbital speed at half that radius, and that neither speed depends on the rock's mass.

Write the energy statement. With no atmosphere, gravity is the only force, so the mechanical energy of the rock and planet system is constant: $\tfrac{1}{2}mv_i^2 + U_i = \tfrac{1}{2}mv_f^2 + U_f$ .
Fix the two potential energies. At the surface, $U_i = -GMm/R$ . Very far away, $U_f \to 0$ by EK 6.6.A.2.iii, which defines the gravitational potential energy to be zero at infinite separation.
Express $U_i$ using the escape speed. By definition $\tfrac{1}{2}mv_{\text{esc}}^2 = GMm/R$ , so $GMm/R = \tfrac{1}{2}mv_{\text{esc}}^2$ and therefore $U_i = -\tfrac{1}{2}mv_{\text{esc}}^2$ . This substitution is what makes the problem come out in one line.
Initial kinetic energy. $v_i = 2v_{\text{esc}}$ , so $\tfrac{1}{2}mv_i^2 = \tfrac{1}{2}m(4v_{\text{esc}}^2) = 2mv_{\text{esc}}^2$ .
Total energy. $E = 2mv_{\text{esc}}^2 - \tfrac{1}{2}mv_{\text{esc}}^2 = \tfrac{3}{2}mv_{\text{esc}}^2$ .
Solve for the final speed. Far away, $E = \tfrac{1}{2}mv_f^2$ , so $\tfrac{1}{2}mv_f^2 = \tfrac{3}{2}mv_{\text{esc}}^2$ , giving $v_f^2 = 3v_{\text{esc}}^2$ and $v_f = \sqrt{3}\,v_{\text{esc}}$ . That is option (D) among the four the CED offers.
Read the result. Launching at twice the escape speed leaves only $\sqrt{3} = 1.73$ times the escape speed at infinity, not $2$ . The planet took away one escape speed's worth of kinetic energy on the way, and because energy goes as $v^2$ , subtracting $1$ from $4$ leaves $3$ , not $1$ .
Now the half-radius identity, with the planet from worked example one. $v_{\text{esc}}$ at $r = 7.0 \times 10^6$ m is $\sqrt{2(4.002 \times 10^{14})/(7.0 \times 10^6)} = \sqrt{1.1434 \times 10^8} = 10\,693\ \text{m/s}$ . The circular orbital speed at $r/2 = 3.5 \times 10^6$ m is $\sqrt{(4.002 \times 10^{14})/(3.5 \times 10^6)} = \sqrt{1.1434 \times 10^8} = 10\,693\ \text{m/s}$ . Identical, because $2GM/r$ and $GM/(r/2)$ are the same expression.
And the mass independence. The rock's mass $m$ appeared in every term of the energy statement and divided out before the final line. A $0.5$ kg rock and a $500$ kg rock launched at $2v_{\text{esc}}$ both end up at $\sqrt{3}\,v_{\text{esc}}$ .

The rock ends up at $\sqrt{3}\,v_{\text{esc}}$ , about $1.73$ times the escape speed, not twice it. For the planet of worked example one, the escape speed at $7.0 \times 10^6$ m and the circular orbital speed at $3.5 \times 10^6$ m are both $10\,693$ m/s, confirming that escape speed at any radius equals orbital speed at half that radius. Neither result depends on the rock's mass, which cancels out of the energy statement.

Frequently asked questions

What is the difference between orbital velocity and escape velocity?

Orbital velocity is the speed that holds a circular orbit at a given radius, equal to the square root of G times the central mass divided by the radius. Escape velocity is the speed at which an object's total mechanical energy with the central body is exactly zero, so it can reach an infinite distance, and it equals the square root of two G M over r. At the same radius, escape velocity is exactly the square root of two times orbital velocity, about 41 percent faster. The AP CEDs name and define escape velocity, at essential knowledge 6.6.A.3 in AP Physics 1 and 6.6.A.4 in AP Physics C: Mechanics, and never name orbital velocity at all, though they require everything needed to derive it.

Why is escape velocity the square root of two times orbital velocity?

Because circular orbit needs half as much kinetic energy as escape does. Setting the gravitational force equal to the mass times the centripetal acceleration gives an orbital speed squared of G M over r. Setting the total mechanical energy to zero, with potential energy minus G M m over r, gives an escape speed squared of two G M over r. Dividing one by the other leaves exactly two, so the speeds differ by the square root of two. The AP Physics C: Mechanics CED states the energy version directly at essential knowledge 6.6.A.3: for a circular orbit the kinetic energy is minus one half the potential energy, while escape requires the kinetic energy to equal the full magnitude of the potential energy.

Is escape velocity on the AP Physics equation sheet?

No. Neither the escape velocity formula nor the circular orbital speed formula is printed on any of the four AP Physics booklets, checked against the appendix pages of all four Course and Exam Descriptions. Both CEDs that cover the topic label escape velocity a derived equation rather than a relevant one, which is the CED's way of saying the exam expects you to produce it. What is printed, on all four sheets, is everything the derivation needs: the gravitational force law, the gravitational potential energy with its minus sign, and the centripetal acceleration.

Does escape velocity depend on the mass of the escaping object?

No. Every term in the energy statement carries one factor of the escaping object's mass, so it divides out before the final line, leaving the square root of two G M over r, where M is the central body's mass. A pebble and a spacecraft need the same speed to escape from the same radius. The same cancellation happens in the orbital velocity derivation, where one factor of the satellite's mass cancels from each side of the force equation. What both speeds do depend on is the central body's mass and the distance from its center.

Does escape velocity depend on direction?

No, which is why it is often called escape speed and why the AP Physics C: Mechanics CED uses that word in its own instructional text and sample questions even though the framework says velocity. The derivation uses only conservation of energy, and kinetic energy is one half m v squared, which contains no direction. So any launch direction that does not run into the central body will do. Orbital velocity is different: it is a genuine velocity, and a circular orbit needs the velocity to be perpendicular to the radius as well as the right magnitude. Get the direction wrong at orbital speed and you get an ellipse or a crash.

What happens if an object is launched faster than escape velocity?

It reaches infinity with speed left over, and the leftover is smaller than you might expect because energy goes as the square of speed. A CED sample multiple-choice question for AP Physics C: Mechanics makes exactly this point: a rock launched from an atmosphereless planet at twice the escape speed ends up moving at the square root of three times the escape speed when it is very far away, not twice it. The reasoning is that the initial kinetic energy is four escape-speed units while the gravitational potential energy costs one, leaving three. An object launched at exactly escape velocity is the boundary case, and essential knowledge 6.6.A.3.i says it moves away until its speed reaches zero at an infinite distance.

Why is the total energy of an orbiting satellite negative?

Because gravitational potential energy is defined to be zero at infinite separation and is negative everywhere else. Essential knowledge 6.6.A.2.iii states that convention, with the potential energy of a satellite and central object system equal to minus G times the masses divided by r. For a circular orbit the kinetic energy is only half the magnitude of that potential energy, so the sum is negative, and the AP Physics C: Mechanics CED writes it as minus G M m over two r at essential knowledge 6.6.A.3. A negative total is what bound means: since kinetic energy cannot be negative and potential energy approaches zero at infinity, an object with negative total energy can never reach an infinite distance. Escape velocity is the speed that raises that total to exactly zero.