AP Physics 1 · Topic 6.6

Topic 6.6: Motion of Orbiting Satellites

Q: Is the escape velocity equation on the AP Physics 1 equation sheet?

No. The AP Physics 1 equation sheet does not print v_{\text{esc}} = \sqrt{2GM/r}, and the CED labels it a derived equation rather than a relevant one. What the sheet does print is everything you need to rebuild it in two lines: U_G = -Gm_1m_2/r, K = \frac{1}{2}mv^2, and |\vec{F}_g| = Gm_1m_2/r^2, along with the constant G = 6.67 \times 10^{-11} in the Table of Information. The orbital speed v = \sqrt{GM/r} and Kepler's third law T^2 = 4\pi^2 R^3/(GM) are likewise not printed and are meant to be derived.

Unit 6: Energy and Momentum of Rotating Systems5-8% of the multiple-choice section

Gravity is the only force on an orbiting satellite, so it supplies the centripetal acceleration. Total mechanical energy and angular momentum stay constant in any orbit. In a circular orbit the kinetic and gravitational potential energies are each constant too, because the radius never changes.

AP Physics: Unit 6 (topics 6.6 Motion of Orbiting Satellites). Topic 6.6 carries a single CED learning objective, 6.6.A, describe the motions of a system consisting of two objects interacting only via gravitational forces, with essential knowledge statements 6.6.A.1 through 6.6.A.3 and their sub-statements. The topic has no boundary statements of its own; the boundary excluding Kepler's first and second laws of planetary motion belongs to Topic 2.9 in Unit 2, which is also where Kepler's third law is taught. The CED's suggested skills here are 1.C, 2.A, 2.C, and 3.C. Unit 6 is weighted 5 to 8 percent of the multiple-choice section across about 8 to 14 class periods.

What Topic 6.6 requires

Topic 6.6 has exactly one learning objective, 6.6.A: describe the motions of a system consisting of two objects interacting only via gravitational forces. Three essential knowledge statements sit under it, two of which carry sub-statements. Here is the whole required content, item by item.

6.6.A.1 In a system consisting only of a massive central object and an orbiting satellite with mass that is negligible in comparison to the central object's mass, 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 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. The relevant equation is $U_g = -G\dfrac{m_1 m_2}{r}$ .
6.6.A.3 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.3.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.3.ii The escape velocity of a satellite from a central body of mass $M$ can be derived using conservation of energy laws, giving the derived equation $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$ .

Two things about that list are worth noticing before any physics. Topic 6.6 carries no boundary statements of its own. Several other topics in this course do, including both of the neighbouring rotation topics, but the two CED pages for Topic 6.6 contain none. The boundary that students most expect to find here, the one excluding Kepler's first and second laws, belongs to Topic 2.9 in Unit 2.

And the topic is built almost entirely out of conservation laws, not out of new equations. Statement 6.6.A.2 says so directly. Nothing here asks you to learn a new force law: the gravitational force came from Topic 2.6 and the circular-motion machinery came from Topic 2.9. What Topic 6.6 adds is a checklist of which quantities hold still in which kind of orbit.

The CED's suggested skills for the topic are 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Skill 1.C appears on no other topic in Unit 6, which is a hint about the graph-sketching questions this topic attracts. Unit 6 is weighted 5 to 8 percent of the multiple-choice section over about 8 to 14 class periods.

The two-body idealization, and what it lets you ignore

Statement 6.6.A.1 does one job: it lets you treat the central object as fixed. Strictly, two objects interacting gravitationally both orbit their common center of mass, because Newton's third law gives each the same size force. The satellite is far lighter, so it takes almost the whole acceleration, and the central object's own motion is negligible.

That is a conclusion you can reproduce rather than memorize, from $\vec{a}_{\text{sys}} = \vec{F}_{\text{net}}/m_{\text{sys}}$ on the equation sheet. Equal forces, very unequal masses, very unequal accelerations. The Earth does move in response to a satellite; it moves so little that no problem in this course accounts for it.

Two practical consequences. First, the $r$ in every equation on this page is measured from the center of the central object, not from its surface. A satellite 400 km up is not at $r = 400$ km; it is at Earth's radius plus 400 km. This is the single most common setup error in orbital problems. Second, the system whose energy you track is the satellite plus the central object together. Gravitational potential energy is a property of the pair, which is why $U_g$ carries both masses.

Gravity supplies the centripetal acceleration

For a circular orbit, statement 2.9.B.1 in Unit 2 puts it plainly: the satellite's centripetal acceleration is caused only by gravitational attraction. Nothing else is pulling. Set the sheet's gravitational force equal to the mass times the sheet's centripetal acceleration:

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

The satellite's own mass $m$ cancels, which is the answer to a whole family of questions. Two satellites at the same orbital radius have the same speed and the same period whatever they weigh, which is why an astronaut on a spacewalk stays alongside the station instead of drifting behind it.

That orbital-speed expression is not printed on the equation sheet. What is printed is $|\vec{F}_g| = G\frac{m_1m_2}{r^2}$ and $a_c = v^2/r$ , and you are expected to combine them, which is exactly suggested skill 2.A. The same is true of the orbital period: the sheet prints $T = 1/f$ , but not $T = 2\pi r/v$ , which the CED lists as a derived equation under 2.9.A.5.iii.

Note also that the AP Physics 1 Table of Information gives you $G = 6.67 \times 10^{-11}$ in its constants list, and nothing else you need here. Earth's mass, Earth's radius, and the Sun's mass are not on it. A question that needs them will supply them in the stem.

Kepler's third law, $T^2 = \frac{4\pi^2}{GM}R^3$ , comes out of the same substitution and it is squarely in the course, but its learning objective is 2.9.B, not anything in Topic 6.6. Topic 2.9 also carries the boundary statement that AP Physics 1 does not expect students to know Kepler's first or second laws of planetary motion.

What holds still in a circular orbit

Statement 6.6.A.2.i is a four-item list and it repays reading as a list, because the elliptical case in the next section keeps only two of the four. In circular orbits, all of the following are constant:

the system's total mechanical energy,
the system's gravitational potential energy,
the satellite's angular momentum,
the satellite's kinetic energy.

Every one of those follows from a single geometric fact: $r$ never changes, and neither does the speed. $U_g = -GMm/r$ depends only on $r$ , so it holds still. Speed is constant, so $K = \frac{1}{2}mv^2$ holds still. Their sum holds still. And $L = rmv\sin\theta$ from the sheet has $r$ , $m$ , and $v$ all fixed, with the velocity always perpendicular to the radius so $\sin\theta = 1$ .

There is a cleaner way to say why the energies do not move: gravity does no work on a satellite in a circular orbit. The sheet's $W = Fd\cos\theta$ has $\theta = 90$ degrees between the inward force and the tangential velocity, and $\cos 90^\circ = 0$ . A force can change the direction of motion forever without changing the speed, which is the whole content of uniform circular motion.

Two more results are worth deriving once, since skill 2.A invites them. Substituting $v^2 = GM/r$ into the kinetic energy gives $K = \frac{GMm}{2r}$ , so the total mechanical energy of a circular orbit is

E = K + U_g = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}

which is exactly half of $U_g$ , and negative. Neither expression is printed on the sheet; both come out of things that are. The negative sign is the point: a bound orbit has negative total mechanical energy, and that is the fact escape velocity is built on.

What changes in an elliptical orbit

Statement 6.6.A.2.ii is the counterpart, and the contrast is the thing to hold onto. 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.

So two survive and two do not. The reason is the same geometric fact read backwards: $r$ now varies, so $U_g = -GMm/r$ varies, and the speed varies with it so that the sum stays put. Energy sloshes between the potential and kinetic accounts exactly as it does for a ball thrown upward.

Angular momentum survives because the gravitational force always points straight at the central object, so its lever arm about that point is zero and it exerts no torque there. No external torque means no change in angular momentum, which is Topic 6.4 applied to an orbit.

That single conserved quantity answers one of Unit 6's own essential questions, which the CED prints as "Why do planets move faster when they travel closer to the sun?" At the closest and farthest points of the orbit the velocity is perpendicular to the radius, so $\sin\theta = 1$ and the sheet's $L = rmv\sin\theta$ reduces to $L = rmv$ . Holding $L$ fixed then forces $v$ up when $r$ comes down:

r_1 v_1 = r_2 v_2 \qquad \text{(closest and farthest points only)}

That restriction matters. At a general point of an ellipse the velocity is not perpendicular to the radius and the $\sin\theta$ has to stay in. The two apsides are where it drops out, which is why exam problems ask about those two points.

One scope note, because it looks like a contradiction and is not. Kepler's first and second laws are excluded by a boundary statement in Topic 2.9, so you are not expected to state that orbits are ellipses with the central body at a focus, or to reason with equal areas swept in equal times. Topic 6.6 nevertheless requires you to describe what is conserved in an elliptical orbit, and to explain the speeding-up qualitatively. The route the CED gives you is conservation of angular momentum, not Kepler.

Why gravitational potential energy is negative

Statement 6.6.A.2.iii fixes the zero: the gravitational potential energy of a satellite and a massive central object is defined to be zero when the satellite is an infinite distance from the central object. The sheet prints the consequence:

U_G = -\frac{Gm_1m_2}{r}

Students find that minus sign alarming, and it is worth being clear about what it does and does not mean. Potential energy has no absolute value, only differences, so a zero has to be chosen. Choosing infinity as the zero makes every finite separation a lower energy than that reference, and lower than zero means negative. Bringing two masses closer makes $U_G$ more negative, so $\Delta U_G$ is negative, which is correct: gravity does positive work as things fall together.

Compare the near-Earth version on the same sheet, $\Delta U_g = mg\Delta y$ , which comes from Topic 3.3. That form assumes a constant field and is only written as a change, with the zero wherever you find it convenient. The two are not rivals: $mg\Delta y$ is what $-GMm/r$ becomes over a height change small enough that $r$ barely moves.

This is the natural home of suggested skill 1.C, the qualitative graph sketch. A sketch of $U_G$ against $r$ starts steeply negative near the central body, rises as $r$ grows, and flattens toward zero from below, never crossing the axis. On the same axes, the total mechanical energy of a given bound orbit is a horizontal line below zero, and the vertical gap between that line and the $U_G$ curve is the kinetic energy. Where the two meet, $K = 0$ and the satellite is at its farthest point. That one picture answers most of what this topic can ask qualitatively.

Escape velocity, straight from the energy definition

Statement 6.6.A.3 defines escape velocity by energy rather than by motion: 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. Not "fast enough to get away" in the loose sense. Exactly the speed that makes the total zero.

The definition is the derivation, which is what 6.6.A.3.ii means by saying escape velocity can be derived using conservation of energy laws. Set the total to zero at the starting radius:

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

Three things fall out of it immediately.

The satellite's mass cancels, so escape velocity is a property of the central body and the starting distance alone. A pebble and a rocket need the same speed.

**Only $v^2$ appears**, so the direction does not matter. Any direction that avoids hitting the central body will do, which is why "escape speed" is the more honest name even though the CED says velocity.

Zero total energy is the borderline case. Statement 6.6.A.3.i spells out the motion it corresponds to: when the only force 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 it. Negative total energy means the satellite is bound and turns around at some finite $r$ . Positive total energy means it arrives at infinity still moving.

One relation is worth carrying: comparing $v_{\text{esc}} = \sqrt{2GM/r}$ with the circular orbital speed $v = \sqrt{GM/r}$ at the same radius gives $v_{\text{esc}} = \sqrt{2}\,v_{\text{orbit}}$ , about 1.41 times as fast. Neither $v_{\text{esc}}$ nor $v_{\text{orbit}}$ is printed on the equation sheet; the CED calls $v_{\text{esc}}$ a derived equation, and you should be able to rebuild it from $U_G = -Gm_1m_2/r$ and $K = \frac{1}{2}mv^2$ , both of which are printed.

How Topic 6.6 is tested, and what goes wrong

Take the four suggested skills as the question menu. Skill 1.C asks for qualitative sketches: $U_G$ against $r$ , kinetic energy against $r$ for an elliptical orbit, or speed against time over one orbit. Skill 2.A asks you to derive $v = \sqrt{GM/r}$ or $v_{\text{esc}} = \sqrt{2GM/r}$ from printed equations. Skill 2.C asks for comparisons between two orbital radii or between two points on one ellipse, usually as a factor of change. Skill 3.C asks you to justify a claim, and Unit 6's exam-preparation note warns that naming a conservation law is not by itself a sufficient justification: you have to walk from the principle to the claim.

Factor-of-change reasoning is worth practising directly, because $r$ enters the various quantities with different powers. Move a satellite to a circular orbit of four times the radius and: the gravitational force falls by 16, $U_G$ rises (becomes less negative) by a factor of 4, the orbital speed halves, the kinetic energy falls by 4, and the period grows by $4^{3/2} = 8$ .

The errors that cost points:

**Measuring $r$ from the surface.** It is measured from the center of the central body. Add the planet's radius to the altitude before anything else.
**Dropping the minus sign in $U_G$ .** A positive $-GMm/r$ turns a bound orbit into an unbound one and every energy conclusion inverts.
**Using $\Delta U_g = mg\Delta y$ over orbital distances.** That form assumes a field that does not weaken with height, which is false the moment $r$ changes appreciably.
**Applying $L = rmv$ at a general point of an ellipse.** It only reduces that far at the closest and farthest points, where the velocity is perpendicular to the radius. Elsewhere the $\sin\theta$ stays.
Saying kinetic energy is constant in an elliptical orbit. Statement 6.6.A.2.ii says the opposite: only the total mechanical energy and the angular momentum are constant there.
Thinking the satellite's mass matters. It cancels out of orbital speed, orbital period, and escape velocity alike. It does not cancel out of forces or energies.
Reaching for Kepler's first or second law. A Topic 2.9 boundary statement puts both outside the course. Use conservation of angular momentum instead.

For the circular-motion machinery underneath all of this, work through the centripetal force guide, and check a numerical setup with the centripetal force calculator.

Speed and period of a satellite in low Earth orbit

A satellite orbits Earth in a circular path 400 km above the surface. Earth's mass is $5.97 \times 10^{24}$ kg and its radius is $6.37 \times 10^6$ m. Using $G = 6.67 \times 10^{-11} \, \mathrm{N \cdot m^2/kg^2}$ , find the satellite's orbital speed and its period.

Get the orbital radius right first. It is measured from Earth's center: $r = 6.37 \times 10^6 + 4.00 \times 10^5 = 6.77 \times 10^6 \, \mathrm{m}$ .
Set gravity equal to the mass times the centripetal acceleration, per 2.9.B.1: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ . The satellite's mass cancels, leaving $v = \sqrt{GM/r}$ , so no satellite mass is needed.
Compute the product first: $GM = (6.67 \times 10^{-11})(5.97 \times 10^{24}) = 3.982 \times 10^{14} \, \mathrm{m^3/s^2}$ .
Divide: $GM/r = (3.982 \times 10^{14})/(6.77 \times 10^6) = 5.882 \times 10^{7} \, \mathrm{m^2/s^2}$ .
Take the root: $v = \sqrt{5.882 \times 10^{7}} = 7.67 \times 10^3 \, \mathrm{m/s}$ , about 7.67 km/s.
Period from the circumference: $T = 2\pi r / v = 2\pi(6.77 \times 10^6)/(7.669 \times 10^3) = (4.254 \times 10^7)/(7.669 \times 10^3) = 5.55 \times 10^3 \, \mathrm{s}$ .
Convert for a reality check: $5546 \, \mathrm{s} / 60 = 92.4$ minutes, which matches the roughly 90 minute orbit of a real low-Earth-orbit satellite.
Cross-check with Kepler's third law from Topic 2.9: $T^2 = 4\pi^2 r^3/(GM) = 4\pi^2 (6.77 \times 10^6)^3/(3.982 \times 10^{14}) = 3.076 \times 10^7 \, \mathrm{s^2}$ , and $\sqrt{3.076 \times 10^7} = 5.55 \times 10^3 \, \mathrm{s}$ . Same answer by a different route.

The orbital speed is $\mathbf{7.67 \times 10^3}$ m/s and the period is $\mathbf{5.55 \times 10^3}$ s, about 92.4 minutes. Neither result used the satellite's mass, because it cancels: every object in a 400 km circular orbit travels at this speed regardless of what it weighs.

Escape speed, and how it compares with orbital speed

Using the same Earth data, find the escape speed from Earth's surface, and show how the escape speed at the 400 km orbital radius compares with the circular orbital speed there.

Start from the definition in 6.6.A.3, not from a memorized formula: escape velocity is the speed that makes the system's total mechanical energy zero.
Write the total at the starting radius, with $U_G = -GMm/r$ from the equation sheet: $\frac{1}{2}mv^2 - \dfrac{GMm}{r} = 0$ .
The satellite's mass cancels from both terms, giving $v_{\text{esc}} = \sqrt{2GM/r}$ , the derived equation the CED lists in 6.6.A.3.ii.
At the surface, $r = 6.37 \times 10^6 \, \mathrm{m}$ : $2GM/r = 2(3.982 \times 10^{14})/(6.37 \times 10^6) = 1.250 \times 10^{8} \, \mathrm{m^2/s^2}$ .
Take the root: $v_{\text{esc}} = 1.12 \times 10^4 \, \mathrm{m/s}$ , about 11.2 km/s.
At the orbital radius $6.77 \times 10^6 \, \mathrm{m}$ instead: $v_{\text{esc}} = \sqrt{2(3.982 \times 10^{14})/(6.77 \times 10^6)} = \sqrt{1.176 \times 10^8} = 1.08 \times 10^4 \, \mathrm{m/s}$ .
Compare with the orbital speed found above: $(1.0846 \times 10^4)/(7.669 \times 10^3) = 1.414$ , which is $\sqrt{2}$ exactly, as dividing $\sqrt{2GM/r}$ by $\sqrt{GM/r}$ requires.
Interpret the sign, per 6.6.A.3.i. At exactly $v_{\text{esc}}$ the total mechanical energy is zero and the satellite's speed reaches zero only at an infinite distance. Below it the total is negative and the satellite is bound; above it the total is positive and the satellite arrives at infinity still moving.

Escape speed from Earth's surface is $\mathbf{1.12 \times 10^4}$ m/s, and at any radius it is exactly $\sqrt{2}$ times the circular orbital speed at that same radius. Both are independent of the escaping object's mass, and both fall as the starting radius grows.

A comet at its closest and farthest points

A comet orbits a star on a long ellipse. At its closest approach it is $1.0 \times 10^{11}$ m from the star and moving at $5.0 \times 10^4$ m/s. Its farthest point is $5.0 \times 10^{11}$ m from the star. Find its speed there, the factor by which its kinetic energy changes, and the change in the system's gravitational potential energy per kilogram of comet.

Choose the conserved quantity. Gravity always points at the star, so it exerts no torque about the star and the comet's angular momentum about it is constant, which is what 6.6.A.2.ii requires for an elliptical orbit.
At the closest and farthest points, and only there, the velocity is perpendicular to the radius, so the sheet's $L = rmv\sin\theta$ has $\sin\theta = 1$ and reduces to $L = rmv$ .
Set the two equal: $r_1 v_1 = r_2 v_2$ , so $v_2 = v_1 (r_1/r_2) = (5.0 \times 10^4)(1.0 \times 10^{11})/(5.0 \times 10^{11}) = (5.0 \times 10^4)(0.20) = 1.0 \times 10^4 \, \mathrm{m/s}$ .
Kinetic energy per kilogram at the closest point: $\frac{1}{2}v_1^2 = \frac{1}{2}(5.0 \times 10^4)^2 = 1.25 \times 10^9 \, \mathrm{J/kg}$ .
At the farthest point: $\frac{1}{2}v_2^2 = \frac{1}{2}(1.0 \times 10^4)^2 = 5.0 \times 10^7 \, \mathrm{J/kg}$ . The ratio is $(v_1/v_2)^2 = 5^2 = 25$ , so the kinetic energy falls to one twenty-fifth.
The total mechanical energy is constant (6.6.A.2.ii), so whatever kinetic energy is lost has gone into gravitational potential energy: $\Delta U_g = -\Delta K = 1.25 \times 10^9 - 5.0 \times 10^7 = 1.20 \times 10^9 \, \mathrm{J/kg}$ , an increase.
Consistency check on the numbers. Since $\Delta U_g = GMm(1/r_1 - 1/r_2)$ , the data implies $GM = (1.20 \times 10^9)/(1.0 \times 10^{-11} - 2.0 \times 10^{-12}) = 1.5 \times 10^{20}$ , so $M = 1.5 \times 10^{20}/(6.67 \times 10^{-11}) = 2.2 \times 10^{30}$ kg. That is a plausible star, a little heavier than the Sun, so the three given quantities describe a real orbit rather than an impossible one.

The comet moves at $\mathbf{1.0 \times 10^4}$ m/s at its farthest point, its kinetic energy there is one twenty-fifth of its value at closest approach, and the system's gravitational potential energy rises by $\mathbf{1.20 \times 10^9}$ J per kilogram. Angular momentum and total mechanical energy stayed constant throughout; kinetic and potential energy did not, which is exactly the split statement 6.6.A.2.ii describes.

Frequently asked questions

What does AP Physics 1 Topic 6.6 cover?

Topic 6.6, Motion of Orbiting Satellites, closes Unit 6 of AP Physics 1 and carries a single learning objective, 6.6.A: describe the motions of a system consisting of two objects interacting only via gravitational forces. Its essential knowledge statements say that a satellite far lighter than its central body leaves that body's motion negligible, that orbital motion is constrained by conservation laws, that gravitational potential energy is defined as zero at infinite separation so that $U_g = -Gm_1m_2/r$ , and that escape velocity is the speed making the system's total mechanical energy zero, giving $v_{\text{esc}} = \sqrt{2GM/r}$ . The topic has no boundary statements of its own.

What quantities are constant in a circular orbit?

Four, according to essential knowledge 6.6.A.2.i: the system's total mechanical energy, the system's gravitational potential energy, the satellite's angular momentum, and the satellite's kinetic energy. All four follow from the orbital radius and the speed both being fixed. A shorter way to see why no energy changes is that gravity points at the center while the velocity is tangential, so the angle between them is 90 degrees and $W = Fd\cos\theta$ gives zero work. In an elliptical orbit only two of the four survive, the total mechanical energy and the angular momentum.

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

Because angular momentum is conserved. The gravitational force always points straight at the sun, so it has no lever arm about the sun and exerts no torque there, which means the planet's angular momentum about the sun cannot change. At the closest and farthest points of the orbit the velocity is perpendicular to the radius, so $L = rmv\sin\theta$ reduces to $L = rmv$ , and holding that product fixed forces the speed up when the distance comes down. AP Physics 1 reaches this result through conservation of angular momentum rather than through Kepler's second law, which a Topic 2.9 boundary statement places outside the course.

What is escape velocity in AP Physics 1?

Essential knowledge 6.6.A.3 defines it by energy: 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. Setting $\frac{1}{2}mv^2 - GMm/r = 0$ and solving gives the CED's derived equation $v_{\text{esc}} = \sqrt{2GM/r}$ . Statement 6.6.A.3.i describes what that motion looks like: with gravity the only force acting, a satellite at escape velocity moves away until its speed reaches zero at an infinite distance. The satellite's own mass cancels, and because only $v^2$ appears, the direction does not matter either.

Is the escape velocity equation on the AP Physics 1 equation sheet?

No. The AP Physics 1 equation sheet does not print $v_{\text{esc}} = \sqrt{2GM/r}$ , and the CED labels it a derived equation rather than a relevant one. What the sheet does print is everything you need to rebuild it in two lines: $U_G = -Gm_1m_2/r$ , $K = \frac{1}{2}mv^2$ , and $|\vec{F}_g| = Gm_1m_2/r^2$ , along with the constant $G = 6.67 \times 10^{-11}$ in the Table of Information. The orbital speed $v = \sqrt{GM/r}$ and Kepler's third law $T^2 = 4\pi^2 R^3/(GM)$ are likewise not printed and are meant to be derived.

Does AP Physics 1 test elliptical orbits?

Yes, but only through conservation laws. Essential knowledge 6.6.A.2.ii requires you to know that in elliptical orbits the system's total mechanical energy and the satellite's angular momentum are constant, while the system's gravitational potential energy and the satellite's kinetic energy can each change. What is excluded sits in a Topic 2.9 boundary statement: AP Physics 1 does not expect students to know Kepler's first or second laws of planetary motion, so the geometry of ellipses and foci and the equal-areas rule are outside the course. You can be asked why a comet speeds up as it approaches; you answer with angular momentum, not with Kepler.

Does a satellite's mass affect its orbital speed or period?

No. Setting the gravitational force equal to the mass times the centripetal acceleration, $GMm/r^2 = mv^2/r$ , cancels the satellite's mass and leaves $v = \sqrt{GM/r}$ , which depends only on the central body's mass and the orbital radius. The period follows from the same expression and is likewise independent of the satellite's mass, as is escape velocity. This is why every object in the same orbit travels together, and why an astronaut floats alongside the station rather than falling behind. The satellite's mass does still matter for the force on it and for its kinetic and potential energies.