AP Physics C: Mechanics · Topic 6.6

Topic 6.6: Motion of Orbiting Satellites

Q: Is the orbital speed formula on the AP Physics C equation sheet?

No, and it is not in the course and exam description either, not even as a labelled Derived Equation. You produce it by setting the gravitational force equal to the mass times the centripetal acceleration, both of which are printed on the sheet, and cancelling the satellite's mass. The result is the square root of the central mass times the gravitational constant divided by the orbital radius. Escape speed has a different status: it is printed in the framework under essential knowledge 6.6.A.4.ii with the Derived equation label, which the Required Equations page defines as marking results of derivations expected of students on the exam.

Q: Why is gravitational potential energy negative in AP Physics C?

Because of where the zero is set and which way the force points. Essential knowledge 6.6.A.2.iii defines gravitational potential energy to be zero when the satellite is an infinite distance from the central object. Integrating the printed change-in-potential-energy expression for the attractive inverse-square force, from infinity in to a distance r, then gives minus the gravitational constant times the two masses divided by that distance. Bringing an object in from infinity lowers its potential energy below the reference value, so every finite separation has a negative value. Differentiating that expression with the printed force-from-potential relation returns the inward force, which is a useful check.

Q: What is constant in a circular orbit and what changes in an elliptical one?

In a circular orbit, essential knowledge 6.6.A.2.i says the system's total mechanical energy, the system's gravitational potential energy, and the satellite's angular momentum and kinetic energy are all constant, because the distance never changes. In an elliptical orbit, 6.6.A.2.ii says only the total mechanical energy and the angular momentum are constant, while the gravitational potential energy and the kinetic energy can each change. Those two survive because gravity from the central body is a radial force, so its cross product with the position vector is zero and it exerts no torque about that body.

Q: Why do planets move faster when they are closer to the sun?

Two equivalent answers. By energy, the gravitational potential energy becomes more negative as the distance falls, and with the total mechanical energy fixed the kinetic energy must rise, so the speed rises. By angular momentum, gravity from the sun is a radial force, so it exerts no torque about the sun and the planet's angular momentum is constant. At the closest and farthest points the velocity is perpendicular to the radius, so the product of speed and distance is fixed there, and a smaller distance requires a larger speed. Essential knowledge 6.6.A.2.ii is the statement that both rest on.

Q: What is escape velocity in AP Physics C Mechanics?

Essential knowledge 6.6.A.4 defines it as the satellite's velocity such that the mechanical energy of the satellite and central-object system is equal to zero. Setting one half the mass times the speed squared equal to the gravitational potential energy's magnitude and solving gives the square root of twice the gravitational constant times the central mass divided by the distance, which 6.6.A.4.ii prints as a Derived equation. The satellite's own mass cancels, and because energy depends on speed rather than direction, the same speed escapes in any direction. It is exactly the square root of two times the circular orbital speed at the same radius.

Q: Is the escape velocity equation on the AP Physics C Mechanics equation sheet?

No. It appears in the framework under essential knowledge 6.6.A.4.ii carrying the Derived equation label, which the Required Equations page defines as marking the final results of derivations expected of students on the exam. Neither it nor the circular-orbit relations of 6.6.A.3 appear on the sheet. What the sheet does print is everything needed to produce them: the gravitational force, the gravitational potential energy with its minus sign, the kinetic energy, and the centripetal acceleration in both of its forms. The route to escape speed is to set the total mechanical energy to zero and solve.

Q: Does AP Physics C Mechanics test Kepler's laws?

The third law, yes; the first and second, not by name. Learning objective 2.10.B, in Unit 2 rather than Unit 6, asks students to describe circular orbits using Kepler's third law, and prints the relation between the square of the period and the cube of the radius as a Derived equation, so it is not on the sheet and you are expected to be able to produce it. A boundary statement under that same topic says that AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion. That does not remove elliptical orbits, which essential knowledge 6.6.A.2.ii covers directly, and it does not remove the physics behind the second law, which is conservation of angular momentum.

Unit 6: Energy and Momentum of Rotating Systems10-15% of the multiple-choice section

For a satellite in a circular orbit, gravity supplies the centripetal acceleration, which gives an orbital speed you derive rather than look up. Its kinetic energy is minus half its potential energy, and its total energy is half the potential energy. Escape speed makes the total energy zero.

AP Physics: Unit 6 (topics 6.6 Motion of Orbiting Satellites). Topic 6.6 of the current AP Physics C: Mechanics course and exam description, inside Unit 6, weighted 10 to 15% of the multiple-choice section. One learning objective, 6.6.A, and nine essential-knowledge statements: 6.6.A.1 on the negligible motion of the central object, 6.6.A.2 with sub-statements .i on circular orbits, .ii on elliptical orbits and .iii defining U_g to be zero at infinity with the relevant equation, 6.6.A.3 with two Derived equations K = -U/2 and E_total = U/2 = -GMm/2r for a circular path, and 6.6.A.4 defining escape velocity by zero mechanical energy with .i on the outcome and .ii giving the Derived equation v_esc = sqrt(2GM/r). Suggested skills 1.C, 2.A and 3.C, three in all, the fewest in Unit 6. No boundary statement under this topic. VERIFIED EQUATION STATUS: v = sqrt(GM/r) is not printed on the sheet AND does not appear anywhere in the C: Mechanics framework, not even as a Derived Equation; it is a pure derivation from the printed gravitational force law and the printed a_c = v^2/r = r omega^2. Escape speed IS a labelled Derived Equation (6.6.A.4.ii) but is not on the sheet. The circular-orbit energy relations are Derived Equations (6.6.A.3) and are not on the sheet. Kepler's third law, T^2 = 4 pi^2 R^3 / GM, is a Derived Equation of 2.10.B.1 in Unit 2, alongside a boundary statement saying AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion. Difference from the identically titled AP Physics 1 Topic 6.6, verified statement by statement against both CEDs: AP Physics 1 has eight statements and no counterpart to 6.6.A.3, and it numbers escape velocity 6.6.A.3 rather than 6.6.A.4. The C sheet additionally prints the potential-energy integral and F = -dU/dx, neither of which is on the AP Physics 1 sheet. Sample multiple-choice question 13, answer D, pairs skill 2.A with 6.6.A and 6.6.A.4.

What Topic 6.6 requires

Topic 6.6 carries one learning objective and nine essential-knowledge statements. 6.6.A: describe the motions of a system consisting of two objects or systems interacting only via gravitational forces.

Statement	What it says
6.6.A.1	In a system consisting only of a massive central object and an orbiting satellite whose mass is negligible in comparison, the motion of the central object itself is negligible
6.6.A.2	The motion of satellites in orbits is constrained by conservation laws
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
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
6.6.A.2.iii	Gravitational potential energy is defined to be zero when the satellite is an infinite distance from the central object. Relevant equation: $U_g = -G\frac{m_1 m_2}{r}$
6.6.A.3	The total energy of a system with a satellite in a circular orbit can be written in terms of the gravitational potential energy or the kinetic energy. Derived equations: $K = -\frac{1}{2}U$ and $E_{\text{total}} = \frac{1}{2}U = -\frac{GMm}{2r}$
6.6.A.4	The escape velocity of a satellite is the satellite's velocity such that the mechanical energy of the satellite and central-object system is equal to zero
6.6.A.4.i	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
6.6.A.4.ii	The escape velocity from a central body of mass $M$ can be derived using conservation of energy laws. Derived equation: $v_{\text{esc}} = \sqrt{\frac{2GM}{r}}$

Suggested skills: 1.C, 2.A and 3.C. Three, the fewest of any topic in Unit 6, which lists four apiece for 6.1 through 6.4 and five for 6.5. Topic 6.6 prints no boundary statement. Unit 6's only one sits under Topic 6.5 and concerns rolling friction.

The orbital speed formula is not in the CED at all, so derive it

This is the single most useful thing to know before an exam on this topic, and it cuts two ways.

$v = \sqrt{GM/r}$ is not printed on the equation sheet, and it is not a Derived Equation either. It does not appear anywhere in the AP Physics C: Mechanics course and exam description. Topic 6.6 states nine things and that is not one of them. Nor is it among the Derived equations of Topic 2.10, Circular Motion, which has exactly three: $v = \sqrt{gr}$ for the minimum speed at the top of a vertical loop in 2.10.A.2.i, $T = 2\pi r / v$ in 2.10.A.5.iii, and Kepler's third law in 2.10.B.1.

Escape speed, by contrast, is a labelled Derived Equation. Statement 6.6.A.4.ii prints $v_{\text{esc}} = \sqrt{2GM/r}$ under the heading Derived equation, and the framework's Required Equations page defines that label precisely: not all equations in the course framework appear on the equation sheet, and many are provided for reference and guidance "or to demonstrate the final results of derivations expected of students on the exam". Those are the Derived Equations. So the escape-speed result is one the exam expects you to be able to produce, and it is also not printed on the sheet.

The practical upshot: two of the most-quoted orbital results are things you write down yourself. Both derivations are short.

Orbital speed. For a circular orbit, gravity is the only force and it is entirely centripetal. Both halves are printed on the sheet: $\lvert \vec{F}_g \rvert = \dfrac{Gm_1m_2}{r^2}$ and $a_c = \dfrac{v^2}{r} = r\omega^2$ . Newton's second law for the satellite of mass $m$ :

\frac{GMm}{r^2} = \frac{mv^2}{r} \qquad \Longrightarrow \qquad v = \sqrt{\frac{GM}{r}}

The satellite's own mass cancels. That cancellation is a scoring point in its own right on a question that asks whether a heavier satellite orbits faster, and it is the reason the answer is no.

Escape speed. Set the total mechanical energy to zero, which is what 6.6.A.4 defines escape velocity to mean:

\tfrac{1}{2}mv_{\text{esc}}^2 - \frac{GMm}{r} = 0 \qquad \Longrightarrow \qquad v_{\text{esc}} = \sqrt{\frac{2GM}{r}}

Mass cancels again. Comparing the two, $v_{\text{esc}} = \sqrt{2}\,v_{\text{orbit}}$ at the same radius, which is a ratio worth carrying because it turns one number into the other in one step.

Deriving the potential energy rather than being handed it

The $U_g$ of statement 6.6.A.2.iii is printed on the sheet as $U_G = -G\frac{m_1m_2}{r}$ , and it is printed on the AP Physics 1 sheet too. What is different here is that the calculus-based sheet also prints the two lines that let you get to it and back from it, and the algebra-based sheet prints neither:

\Delta U = -\int_a^b \vec{F}_{\text{cf}}(r) \cdot d\vec{r} \qquad \qquad F_x = -\frac{dU(x)}{dx}

Run the first one for gravity. Take the reference point at infinity, which is exactly what 6.6.A.2.iii specifies, and integrate the attractive inverse-square force in from infinity to $r$ :

U(r) - U(\infty) = -\int_{\infty}^{r} \left(-\frac{GMm}{r'^2}\right) dr' = GMm\left[-\frac{1}{r'}\right]_{\infty}^{r} = -\frac{GMm}{r}

The minus sign inside the integrand is the force being attractive, that is, pointing toward decreasing $r$ . With $U(\infty) = 0$ by 6.6.A.2.iii's definition, the printed result falls out. That is why $U_G$ is negative, and it is a better answer to the exam question "why is gravitational potential energy negative" than "because the formula has a minus sign".

Run the second one backwards as a check: $F_r = -\dfrac{dU}{dr} = -\dfrac{d}{dr}\left(-\dfrac{GMm}{r}\right) = -\dfrac{GMm}{r^2}$ , negative meaning inward, which is the printed force law again. The two printed calculus lines are inverses of each other, and this is the cleanest place in the course to see it.

One consequence to keep straight. The sheet also prints $\Delta U_g = mg\Delta y$ , the flat-Earth approximation, which is what $-GMm/r$ becomes when $r$ barely changes and $g = GM/r^2$ is evaluated at the surface. On an orbital problem use the $-GMm/r$ form; reaching for $mg\Delta y$ across an orbital radius is a defect.

Circular and elliptical orbits hold different things constant

Statements 6.6.A.2.i and 6.6.A.2.ii are worth reading side by side, because the contrast between them is the whole of what conservation laws buy you here.

Quantity	Circular orbit (6.6.A.2.i)	Elliptical orbit (6.6.A.2.ii)
system's total mechanical energy	constant	constant
satellite's angular momentum	constant	constant
system's gravitational potential energy	constant	can change
satellite's kinetic energy	constant	can change

Everything is constant in a circular orbit because $r$ never changes, so $U$ cannot, and with $E$ fixed neither can $K$ . In an ellipse only the two conserved quantities survive, and $K$ and $U$ trade against each other.

The contrast is the answer to one of the unit's own essential questions, "Why do planets move faster when they travel closer to the sun?", and there are two ways to say it. The energy answer: as $r$ falls, $U = -GMm/r$ becomes more negative, and with $E$ fixed the kinetic energy must rise. The angular momentum answer, which is the one that uses this course's definitions: gravity from the central body is a radial force, so $\vec{r} \times \vec{F} = \vec{0}$ about that body and the net torque is exactly zero, which by 6.4.B.2 makes $\vec{L}$ constant. At the closest and farthest points of an ellipse the velocity is perpendicular to the radius, so $L = mvr$ there, and a smaller $r$ demands a larger $v$ .

The radial-force argument is worth spelling out on a free-response question because it is what makes the angular momentum constant without the orbit needing to be circular. It holds at every point of the ellipse, not just at the two apsides.

One trap to avoid, and it follows from the numbering. The two derived relations of 6.6.A.3, $K = -\frac{1}{2}U$ and $E_{\text{total}} = \frac{1}{2}U$ , are stated by the framework for a satellite orbiting a central object in a circular path. Do not carry them onto an ellipse. At an ellipse's perihelion the kinetic energy exceeds $-\frac{1}{2}U$ ; at aphelion it falls short. Worked example 3 computes both.

The statement AP Physics 1 does not have

Comparing the two frameworks statement by statement produces one clean difference, and it is not a difference of level.

AP Physics 1's Topic 6.6 has eight essential-knowledge statements: 6.6.A.1, then 6.6.A.2 with three sub-statements, then 6.6.A.3 on escape velocity with two sub-statements. AP Physics C: Mechanics has nine, and the extra one is 6.6.A.3, the circular-orbit energy relations, which has no counterpart in the algebra-based course at all. Escape velocity is then renumbered: it is 6.6.A.3 in AP Physics 1 and 6.6.A.4 here.

That renumbering is a practical trap. A revision resource that cites "6.6.A.3" for escape velocity is describing AP Physics 1. In this course 6.6.A.3 is:

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

both labelled Derived equations, both for a circular orbit. Deriving them takes two lines. Start from the circular-orbit condition of the previous section, $\frac{GMm}{r^2} = \frac{mv^2}{r}$ , so that $mv^2 = \frac{GMm}{r}$ and

K = \tfrac{1}{2}mv^2 = \frac{GMm}{2r} = -\tfrac{1}{2}U

since $U = -GMm/r$ . Then $E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r} = \frac{1}{2}U$ .

Three things follow, and each of them is a question this course can ask and the algebra-based one cannot pose from its own statements:

The total energy is negative, and that is what bound means. A satellite with $E < 0$ cannot reach infinity, because $U \to 0$ there and $K$ cannot be negative.
The kinetic energy is exactly half the size of the potential energy. Doubling the orbital radius halves $K$ and halves the size of $E$ .
Raising a satellite to a higher circular orbit requires energy input, and the satellite ends up slower. $E$ becomes less negative while $v = \sqrt{GM/r}$ falls. Both statements are true at once and the pairing is a standard multiple-choice trap.

Escape velocity, straight from the energy definition

Statement 6.6.A.4 defines escape velocity by an energy condition rather than by an outcome: it is "the satellite's velocity such that the mechanical energy of the satellite and central-object system is equal to zero". Statement 6.6.A.4.i then gives the outcome, with a condition attached that is easy to skip: 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.

Three consequences of the energy definition, all exam-relevant.

Direction does not matter. Mechanical energy depends on speed, not on the velocity's direction, so the same speed escapes whether the launch is straight up or at an angle, as long as nothing is in the way. The framework calls it a velocity; the equation is a speed. That is why sample multiple-choice question 13 can specify a planet with no atmosphere and give no launch direction.

The satellite's mass does not matter. It cancels, exactly as it does in the orbital-speed derivation.

Escaping means arriving with zero speed, not with the speed you launched at. A rock launched faster than escape speed keeps some speed at infinity, and the amount is set by energy conservation rather than by subtraction. Sample multiple-choice question 13 is exactly this: a small rock is launched from the surface of a planet with no atmosphere at $2v_{\text{esc}}$ , and the question asks its speed when it is very far from the planet. The published answer is choice D, $\sqrt{3}\,v_{\text{esc}}$ , and the route is one line:

\tfrac{1}{2}(2v_{\text{esc}})^2 - \frac{GM}{R} = \tfrac{1}{2}v_\infty^2, \quad \text{with} \quad \frac{GM}{R} = \tfrac{1}{2}v_{\text{esc}}^2

so $v_\infty^2 = 4v_{\text{esc}}^2 - v_{\text{esc}}^2 = 3v_{\text{esc}}^2$ . Speeds do not subtract; the squares do. That question is tagged with skill 2.A and aligned to 6.6.A and 6.6.A.4.

The same energy equation run below escape speed gives the maximum distance reached. Launch radially at $v_0 < v_{\text{esc}}$ from radius $r$ , set the kinetic energy to zero at $r_{\max}$ , and the algebra gives

r_{\max} = \frac{r}{1 - \left(v_0/v_{\text{esc}}\right)^2}

which correctly blows up as $v_0 \to v_{\text{esc}}$ . Worked example 2 evaluates it.

Kepler's third law lives in Unit 2, and so does the other boundary statement

Topic 6.6 says nothing about orbital periods, and people go looking for it here. It is in Unit 2.

Learning objective 2.10.B, inside Topic 2.10 Circular Motion, reads: describe circular orbits using Kepler's third law. Its single essential-knowledge statement, 2.10.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. Its Derived equation is

T^2 = \frac{4\pi^2}{GM}R^3

Derived, so not on the sheet, and expected of you. It follows in two lines from the orbital-speed result and the period, $T = 2\pi R/v$ , which is itself the Derived equation of 2.10.A.5.iii.

Topic 2.10 also carries a boundary statement, quoted whole: "AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion."

Be precise about what that removes and what it does not. It removes the statements of the first law (orbits are ellipses with the central body at a focus) and the second law (equal areas in equal times) as things you must know by name. It does not remove elliptical orbits from the course, because 6.6.A.2.ii is explicitly about them, and it does not remove the physical content of the second law, because that content is conservation of angular momentum, which Topics 6.3 and 6.4 require in full. A question can absolutely ask you to compare a comet's speed at two points of an ellipse. It just will not ask you to name the law.

So Unit 6's only boundary statement is the rolling-friction one under Topic 6.5, and the Kepler one belongs to Unit 2. Two boundary statements, two different units, and neither is under Topic 6.6.

What the equation sheet prints for Topic 6.6

Checked line by line against the printed Table of Information in the AP Physics C: Mechanics course and exam description:

Equation	On the C: Mechanics sheet
$\lvert \vec{F}_g \rvert = \frac{Gm_1m_2}{r^2}$	yes
$U_G = -\frac{Gm_1m_2}{r}$	yes
$a_c = \frac{v^2}{r} = r\omega^2$	yes, both forms
$K = \frac{1}{2}mv^2$	yes
$\Delta U = -\int_a^b \vec{F}_{\text{cf}}(r) \cdot d\vec{r}$	yes
$F_x = -\frac{dU(x)}{dx}$	yes
$\vec{L} = \vec{r} \times \vec{p} = I\vec{\omega}$	yes
$T = \frac{1}{f}$ and $T = \frac{2\pi}{\omega} = \frac{1}{f}$	yes
$\Delta U_g = mg\Delta y$	yes, but it is the local approximation
$G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$	yes, in the constants box
$v = \sqrt{GM/r}$	no, and it is not in the framework either
$v_{\text{esc}} = \sqrt{2GM/r}$	no, it is a Derived Equation in 6.6.A.4.ii
$K = -\frac{1}{2}U$ and $E_{\text{total}} = \frac{1}{2}U = -\frac{GMm}{2r}$	no, Derived Equations in 6.6.A.3
$T^2 = \frac{4\pi^2}{GM}R^3$	no, a Derived Equation in 2.10.B.1

So every headline result of this topic is absent from the sheet, and everything you need to produce them is present. Counting them individually: four results absent, in the last four rows, against ten printed lines that build them.

The constants box on the first page of the Table of Information prints $G = 6.67 \times 10^{-11}\ \mathrm{m^3/(kg \cdot s^2)} = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$ , along with $g = 9.8\ \mathrm{m/s^2}$ and $g = 9.8\ \mathrm{N/kg}$ for Earth's surface. No planetary masses or radii are printed, so a numerical orbital question supplies them.

The two calculus lines in the middle of the table are the ones an algebra-based student does not have, and they are what make the potential energy derivable rather than given.

How Topic 6.6 is tested, and if you are in AP Physics 1

Skill 2.A, deriving a symbolic expression by selecting and following a logical mathematical pathway, is suggested for this topic and carries 25 to 30% of the multiple-choice section, the largest single share of any skill. That is the right expectation to bring: almost every Topic 6.6 question is a derivation or a scaling argument, and the framework's own sample question on this objective is both.

Sample multiple-choice question 13 aligns to 6.6.A and essential knowledge 6.6.A.4, paired with skill 2.A, and is the launched-rock problem worked through above. Its published answer is choice D.

Skill 3.C carries 5 to 10% of the multiple-choice section and is also suggested here, and skill 1.C, qualitative graph sketches, points at the free-response section, where Science Practice 1 carries 20 to 35%.

AP Physics 1 has a Topic 6.6 with the same title, and on this topic the two frameworks overlap heavily: statements 6.6.A.1, 6.6.A.2, its three sub-statements and the whole escape-velocity treatment are the same content in both, and both sheets print the gravitational force, the gravitational potential energy and the centripetal acceleration.

The two real differences, stated plainly rather than inflated. First, this course adds essential knowledge 6.6.A.3, the circular-orbit energy relations, which the algebra-based framework does not contain at all. Second, this course's sheet prints the potential-energy integral and the force-from-potential derivative, so the potential energy is something you derive here and something you are handed there.

The [AP Physics 1 Topic 6.6 page](/ap-physics-1/unit-6-energy-and-momentum-of-rotating-systems/6-6-motion-of-orbiting-satellites) is for students in the algebra-based course. This page is for students in AP Physics C: Mechanics. If you are in Physics 1, the circular-orbit energy relations and the potential-energy integral on this page are beyond your framework, and the escape-velocity statement you want is numbered 6.6.A.3 rather than 6.6.A.4.

Where it connects: Topic 6.4 supplies the radial-force argument that makes a comet's angular momentum constant, and Topic 6.3 supplies the cross product it rests on. The centripetal force guide owns the circular-motion routine and the centripetal force calculator checks the arithmetic. The Unit 6 hub lists all six topics, and Unit 2 is where Kepler's third law and the circular-motion machinery live.

A circular orbit, with every result derived

An 850 kg satellite is in a circular orbit of radius $r = 5.00 \times 10^6$ m about a planet of mass $M = 6.42 \times 10^{23}$ kg. Use $G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$ from the sheet. Find (a) the orbital speed, (b) the kinetic, potential and total energies, checking them against essential knowledge 6.6.A.3, (c) the escape speed from that radius, and (d) the orbital period, checked against Kepler's third law.

Precompute the combination that appears everywhere: $GM = (6.67 \times 10^{-11})(6.42 \times 10^{23}) = 4.2821 \times 10^{13}\ \mathrm{m^3/s^2}$ .
(a) Neither $v = \sqrt{GM/r}$ nor anything equivalent appears in the framework or on the sheet, so derive it. Gravity supplies the centripetal acceleration: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ , and the satellite's mass cancels, leaving $v^2 = GM/r$ .
$v^2 = \dfrac{4.2821 \times 10^{13}}{5.00 \times 10^6} = 8.5643 \times 10^{6}\ \mathrm{m^2/s^2}$ , so $v = 2.926 \times 10^3\ \mathrm{m/s}$ , about 2.93 km/s.
(b) From the printed $U_G$ : $U = -\dfrac{GMm}{r} = -\dfrac{(4.2821 \times 10^{13})(850)}{5.00 \times 10^6} = -7.280 \times 10^{9}\ \mathrm{J}$ .
$K = \frac{1}{2}mv^2 = \frac{1}{2}(850)(8.5643 \times 10^6) = 3.640 \times 10^{9}\ \mathrm{J}$ .
Check against 6.6.A.3: $-\frac{1}{2}U = 3.640 \times 10^9$ J, matching $K$ digit for digit. The framework labels that a Derived equation, and this is the derivation.
$E_{\text{total}} = K + U = 3.640 \times 10^9 - 7.280 \times 10^9 = -3.640 \times 10^{9}\ \mathrm{J}$ , which is $\frac{1}{2}U$ and also $-\frac{GMm}{2r}$ , the other half of 6.6.A.3. The total is negative, which is what bound means.
(c) Statement 6.6.A.4 defines escape velocity by zero total mechanical energy: $\frac{1}{2}mv_{\text{esc}}^2 - \frac{GMm}{r} = 0$ , so $v_{\text{esc}} = \sqrt{2GM/r} = \sqrt{1.7128 \times 10^7} = 4.139 \times 10^3\ \mathrm{m/s}$ , about 4.14 km/s.
Ratio check: $4.139/2.926 = 1.4142 = \sqrt{2}$ , as it must be, since the escape expression differs from the orbital one only by the factor of 2 under the root.
(d) $T = \dfrac{2\pi r}{v} = \dfrac{2\pi(5.00 \times 10^6)}{2.926 \times 10^3} = 1.0735 \times 10^4\ \mathrm{s}$ , which is 2.98 hours.
Kepler check, using the Derived equation of 2.10.B.1: $T^2 = \dfrac{4\pi^2 r^3}{GM} = \dfrac{39.478(1.25 \times 10^{20})}{4.2821 \times 10^{13}} = 1.1524 \times 10^{8}\ \mathrm{s^2}$ , so $T = 1.0735 \times 10^4$ s. The two agree, as they must, since Kepler's third law is what you get by substituting the orbital speed into the period.
Note what never entered any of these: the 850 kg. It cancels from the speed, the escape speed and the period, and appears only in the energies.

(a) $v = 2.93$ km/s, derived from the force law and the centripetal acceleration. (b) $K = 3.640 \times 10^9$ J, $U = -7.280 \times 10^9$ J, $E = -3.640 \times 10^9$ J, confirming both derived relations of 6.6.A.3. (c) $v_{\text{esc}} = 4.14$ km/s, exactly $\sqrt{2}$ times the orbital speed. (d) $T = 1.07 \times 10^4$ s, or 2.98 hours, matching Kepler's third law.

Above escape speed, and below it

From the same distance $r = 5.00 \times 10^6$ m from the same planet, where the escape speed is $4.139 \times 10^3$ m/s, a probe is launched radially outward. Find (a) its speed very far away if it is launched at $1.5\,v_{\text{esc}}$ , (b) how far it gets if it is launched at $0.80\,v_{\text{esc}}$ , and (c) explain why the answer to (a) is not $0.5\,v_{\text{esc}}$ .

Set up once. With gravity the only force, mechanical energy is constant: $\frac{1}{2}mv^2 - \frac{GMm}{r} = \frac{1}{2}mv_f^2 - \frac{GMm}{r_f}$ . Divide through by $m$ , which is where the satellite's mass leaves the problem.
Rewrite the potential term using the escape-speed definition. Since $v_{\text{esc}}^2 = 2GM/r$ , we have $\dfrac{GM}{r} = \dfrac{1}{2}v_{\text{esc}}^2$ . Every step below is that substitution.
(a) Far away means $r_f \to \infty$ , so the final potential term vanishes, per 6.6.A.2.iii's choice of zero at infinity.
$\frac{1}{2}(1.5 v_{\text{esc}})^2 - \frac{1}{2}v_{\text{esc}}^2 = \frac{1}{2}v_\infty^2$ , so $v_\infty^2 = (2.25 - 1)v_{\text{esc}}^2 = 1.25\, v_{\text{esc}}^2$ and $v_\infty = 1.118\, v_{\text{esc}}$ .
Numerically, $v_\infty = 1.118(4.139 \times 10^3) = 4.627 \times 10^3\ \mathrm{m/s}$ , about 4.63 km/s.
(b) Below escape speed the probe stops at some finite $r_{\max}$ , where $v_f = 0$ : $\frac{1}{2}v_0^2 - \frac{1}{2}v_{\text{esc}}^2 = -\dfrac{GM}{r_{\max}} = -\dfrac{1}{2}v_{\text{esc}}^2\dfrac{r}{r_{\max}}$ .
Dividing by $\frac{1}{2}v_{\text{esc}}^2$ and rearranging gives $r_{\max} = \dfrac{r}{1 - (v_0/v_{\text{esc}})^2}$ .
With $v_0/v_{\text{esc}} = 0.80$ : $r_{\max} = \dfrac{5.00 \times 10^6}{1 - 0.64} = \dfrac{5.00 \times 10^6}{0.36} = 1.389 \times 10^{7}\ \mathrm{m}$ , about 2.78 times the launch radius.
Sanity check the limiting behaviour: as $v_0 \to v_{\text{esc}}$ the denominator goes to zero and $r_{\max}$ grows without bound, which is 6.6.A.4.i, the satellite reaching zero speed only at infinite distance.
(c) Speeds do not subtract because energy goes as the square of the speed. Launching at 1.5 times escape speed means launching with $2.25$ times the escape kinetic energy, of which $1$ unit is spent climbing out, leaving $1.25$ units, and the square root of that is $1.118$ , not $0.5$ . Sample multiple-choice question 13 in the framework is the same trap with $2v_{\text{esc}}$ : the answer is $\sqrt{3}\,v_{\text{esc}} \approx 1.73\,v_{\text{esc}}$ , not $v_{\text{esc}}$ .

(a) $v_\infty = \sqrt{1.25}\,v_{\text{esc}} = 1.118\,v_{\text{esc}} = 4.63$ km/s. (b) $r_{\max} = r/[1 - (v_0/v_{\text{esc}})^2] = 1.39 \times 10^7$ m, about 2.78 launch radii. (c) Because the energies subtract, not the speeds, so the surviving speed is the square root of the leftover energy.

A comet on an ellipse: two constants, two variables

A 500 kg comet orbits a star of mass $M = 2.00 \times 10^{30}$ kg. At its closest approach it is $r_p = 1.00 \times 10^{11}$ m from the star, moving at $v_p = 4.50 \times 10^4$ m/s perpendicular to the radius. Use $G = 6.67 \times 10^{-11}$ . Find (a) that this orbit is bound but not circular, (b) the comet's farthest distance and its speed there, (c) the kinetic and potential energies at both ends, and (d) check whether the circular relation of 6.6.A.3 applies.

$GM = (6.67 \times 10^{-11})(2.00 \times 10^{30}) = 1.334 \times 10^{20}\ \mathrm{m^3/s^2}$ .
(a) The circular speed at that radius would be $\sqrt{GM/r_p} = \sqrt{1.334 \times 10^9} = 3.652 \times 10^4$ m/s and the escape speed would be $\sqrt{2GM/r_p} = 5.165 \times 10^4$ m/s. At $4.50 \times 10^4$ m/s the comet is faster than circular and slower than escape, so the orbit is a bound ellipse with this point as its perihelion.
Confirm with the energy: per unit mass, $\varepsilon = \frac{1}{2}v_p^2 - \frac{GM}{r_p} = 1.0125 \times 10^9 - 1.334 \times 10^9 = -3.215 \times 10^{8}\ \mathrm{J/kg}$ . Negative, so bound.
(b) Two conservation laws, per 6.6.A.2.ii. Gravity is radial, so $\vec{r} \times \vec{F} = \vec{0}$ about the star and the angular momentum is constant. At perihelion and aphelion the velocity is perpendicular to the radius, so there $L = mvr$ with no sine factor.
Per unit mass, $\ell = r_p v_p = (1.00 \times 10^{11})(4.50 \times 10^4) = 4.50 \times 10^{15}\ \mathrm{m^2/s}$ , and at aphelion $v_a = \ell/r_a$ .
Substitute into the energy equation at aphelion: $\dfrac{\ell^2}{2r_a^2} - \dfrac{GM}{r_a} = \varepsilon$ . Multiply by $r_a^2$ and rearrange: $\lvert \varepsilon \rvert r_a^2 - GM\, r_a + \dfrac{\ell^2}{2} = 0$ .
$(3.215 \times 10^8) r_a^2 - (1.334 \times 10^{20}) r_a + 1.0125 \times 10^{31} = 0$ . The two roots are $1.000 \times 10^{11}$ m and $3.149 \times 10^{11}$ m. The smaller root reproducing the perihelion exactly is the check that the algebra is right; the larger is the aphelion.
$r_a = 3.149 \times 10^{11}$ m and $v_a = \dfrac{4.50 \times 10^{15}}{3.149 \times 10^{11}} = 1.429 \times 10^{4}\ \mathrm{m/s}$ .
Ratio check from angular momentum alone: $v_a/v_p = r_p/r_a = 1.000/3.149 = 0.3175$ , and $(4.50 \times 10^4)(0.3175) = 1.429 \times 10^4$ m/s. The comet is 3.15 times as far out and moving 3.15 times slower, which is the unit's essential question answered with numbers.
(c) With $m = 500$ kg. At perihelion: $K = \frac{1}{2}(500)(2.025 \times 10^9) = 5.063 \times 10^{11}$ J and $U = -\frac{GMm}{r_p} = -6.670 \times 10^{11}$ J, summing to $E = -1.608 \times 10^{11}$ J.
At aphelion: $K = \frac{1}{2}(500)(2.042 \times 10^8) = 5.104 \times 10^{10}$ J and $U = -2.118 \times 10^{11}$ J, summing to $-1.608 \times 10^{11}$ J. The total matches, as 6.6.A.2.ii requires, while $K$ fell by a factor of 9.9 and $U$ rose.
(d) At perihelion, $-\frac{1}{2}U = 3.335 \times 10^{11}$ J against an actual $K$ of $5.063 \times 10^{11}$ J, so the comet has more kinetic energy than the circular relation would give. At aphelion, $-\frac{1}{2}U = 1.059 \times 10^{11}$ J against an actual $K$ of $5.104 \times 10^{10}$ J, so it has less. The framework states 6.6.A.3 for a circular path, and this is why: on an ellipse it fails in both directions.

(a) Bound and elliptical: the speed lies between the circular $3.65 \times 10^4$ m/s and the escape $5.17 \times 10^4$ m/s, and the energy per unit mass is $-3.215 \times 10^8$ J/kg. (b) $r_a = 3.149 \times 10^{11}$ m and $v_a = 1.429 \times 10^4$ m/s, with $v_a/v_p = r_p/r_a$ exactly. (c) $K$ falls from $5.06 \times 10^{11}$ J to $5.10 \times 10^{10}$ J while $U$ rises, and the total holds at $-1.608 \times 10^{11}$ J. (d) It does not apply: $K$ exceeds $-U/2$ at perihelion and falls short at aphelion.

Frequently asked questions

Is the orbital speed formula on the AP Physics C equation sheet?

No, and it is not in the course and exam description either, not even as a labelled Derived Equation. You produce it by setting the gravitational force equal to the mass times the centripetal acceleration, both of which are printed on the sheet, and cancelling the satellite's mass. The result is the square root of the central mass times the gravitational constant divided by the orbital radius. Escape speed has a different status: it is printed in the framework under essential knowledge 6.6.A.4.ii with the Derived equation label, which the Required Equations page defines as marking results of derivations expected of students on the exam.

Why is gravitational potential energy negative in AP Physics C?

Because of where the zero is set and which way the force points. Essential knowledge 6.6.A.2.iii defines gravitational potential energy to be zero when the satellite is an infinite distance from the central object. Integrating the printed change-in-potential-energy expression for the attractive inverse-square force, from infinity in to a distance r, then gives minus the gravitational constant times the two masses divided by that distance. Bringing an object in from infinity lowers its potential energy below the reference value, so every finite separation has a negative value. Differentiating that expression with the printed force-from-potential relation returns the inward force, which is a useful check.

What is constant in a circular orbit and what changes in an elliptical one?

In a circular orbit, essential knowledge 6.6.A.2.i says the system's total mechanical energy, the system's gravitational potential energy, and the satellite's angular momentum and kinetic energy are all constant, because the distance never changes. In an elliptical orbit, 6.6.A.2.ii says only the total mechanical energy and the angular momentum are constant, while the gravitational potential energy and the kinetic energy can each change. Those two survive because gravity from the central body is a radial force, so its cross product with the position vector is zero and it exerts no torque about that body.

Why do planets move faster when they are closer to the sun?

Two equivalent answers. By energy, the gravitational potential energy becomes more negative as the distance falls, and with the total mechanical energy fixed the kinetic energy must rise, so the speed rises. By angular momentum, gravity from the sun is a radial force, so it exerts no torque about the sun and the planet's angular momentum is constant. At the closest and farthest points the velocity is perpendicular to the radius, so the product of speed and distance is fixed there, and a smaller distance requires a larger speed. Essential knowledge 6.6.A.2.ii is the statement that both rest on.

What is escape velocity in AP Physics C Mechanics?

Essential knowledge 6.6.A.4 defines it as the satellite's velocity such that the mechanical energy of the satellite and central-object system is equal to zero. Setting one half the mass times the speed squared equal to the gravitational potential energy's magnitude and solving gives the square root of twice the gravitational constant times the central mass divided by the distance, which 6.6.A.4.ii prints as a Derived equation. The satellite's own mass cancels, and because energy depends on speed rather than direction, the same speed escapes in any direction. It is exactly the square root of two times the circular orbital speed at the same radius.

Is the escape velocity equation on the AP Physics C Mechanics equation sheet?

No. It appears in the framework under essential knowledge 6.6.A.4.ii carrying the Derived equation label, which the Required Equations page defines as marking the final results of derivations expected of students on the exam. Neither it nor the circular-orbit relations of 6.6.A.3 appear on the sheet. What the sheet does print is everything needed to produce them: the gravitational force, the gravitational potential energy with its minus sign, the kinetic energy, and the centripetal acceleration in both of its forms. The route to escape speed is to set the total mechanical energy to zero and solve.

Does AP Physics C Mechanics test Kepler's laws?

The third law, yes; the first and second, not by name. Learning objective 2.10.B, in Unit 2 rather than Unit 6, asks students to describe circular orbits using Kepler's third law, and prints the relation between the square of the period and the cube of the radius as a Derived equation, so it is not on the sheet and you are expected to be able to produce it. A boundary statement under that same topic says that AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion. That does not remove elliptical orbits, which essential knowledge 6.6.A.2.ii covers directly, and it does not remove the physics behind the second law, which is conservation of angular momentum.