Unit 6 · Topic 6.6 · about 30 minutes · 7 parts · 13 scored checks, 2 of them multiple choice

AP Physics 1: Algebra-Based 6.6: Motion of Orbiting Satellites

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

A graph of energy in gigajoules against orbit radius in millions of meters, for a 500 kilogram satellite in circular orbits around Earth, from Earth's surface at 6.4 million meters out to 40 million meters. The satellite's kinetic energy is positive and falls toward zero as the radius grows. The gravitational potential energy of the satellite-Earth system is negative and rises toward zero. The total mechanical energy is negative, halfway between, always the kinetic energy with its sign flipped. At a radius of 10 million meters, where the satellite moves at 6.33 kilometers per second, the kinetic energy is 10.0 gigajoules, the total is minus 10.0 and the potential energy is minus 20.0.

Part 1 Think first Who pulls harder?

Commit to a guess first. Being wrong now makes the right idea stick better.

Predict first

A 500 kg weather satellite orbits Earth, and Earth's mass is about $10^{22}$ times the satellite's. Earth's gravitational pull keeps bending the satellite's path into a circle. How does the satellite's gravitational pull on Earth compare with Earth's pull on the satellite?

In short. A satellite orbiting a far more massive central object moves the way the conservation laws allow: the central object stays put, and the system's mechanical energy and the satellite's angular momentum decide the rest.

In a circular orbit, Newton's second law gives $v = \sqrt{\frac{G M}{r}}$ , and the mechanical energy, the gravitational potential energy $U_{g} = - \frac{G M m}{r}$ , the angular momentum and the kinetic energy all stay constant.
In an elliptical orbit, the mechanical energy and the angular momentum stay constant while $U_{g}$ and $K$ trade off, and at the closest and farthest points $r m v$ is the same.
Escape speed makes the mechanical energy zero, $v_{esc} = \sqrt{\frac{2 G M}{r}}$ , and a satellite moving at it slows toward zero speed only at an infinite distance.

Part 2 Learn it Four short ideas, each checked as you go

2a. One body barely moves

Gravity between a satellite and the planet it orbits is one interaction, and it pulls on both. They are a third law pair, as in 2.3 Newton’s Third Law, so the two forces are equal in size, $G \frac{M m}{r^{2}}$ from 2.6 Gravitational Force, and opposite in direction. What each force does is very different.

Take a 500 kg satellite $1.0 \times 10^{7}$ m from Earth's center. Earth's mass, $6.0 \times 10^{24}$ kg, is $1.2 \times 10^{22}$ times the satellite's. The pull on the satellite gives it an acceleration of $\frac{G M}{r^{2}} = 4.0$ m/s². The equal pull on Earth gives Earth an acceleration of about $3 \times 10^{- 22}$ m/s², smaller by that same factor.

So when the central object's mass is far greater than the satellite's, the central object's own motion is negligible. Treat it as fixed: all of the system's motion, and all of its kinetic energy, belongs to the satellite.

Take the satellite and the central object together as the system. The only forces are their gravitational pulls on each other, and those are internal, so nothing outside does work on the system. That is why conservation laws decide how a satellite moves: energy, as in 3.4 Conservation of Energy, and angular momentum, which only an angular impulse can change, as in 6.3 Angular Momentum and Angular Impulse.

Quick check · Decide and justify

A 500 kg satellite moves in a circular orbit around Earth, whose mass is $6.0 \times 10^{24}$ kg, and the only forces are the gravitational forces the two exert on each other. Should the kinetic energy of the satellite-Earth system include a term for the motion of Earth that the satellite's pull causes?

Which reason supports your answer?

2b. Circular orbits: nothing changes

In a circular orbit of radius $r$ , gravity alone gives the satellite its centripetal acceleration, as in 2.9 Circular Motion. Newton's second law toward the center gives

G \frac{M m}{r^{2}} = m \frac{v^{2}}{r} \Rightarrow v = \sqrt{\frac{G M}{r}}

and the satellite's mass cancels. Now the energies, with $U_{g} = 0$ at infinite separation as in 3.3 Potential Energy:

K = \frac{1}{2} m v^{2} = \frac{G M m}{2 r} U_{g} = - \frac{G M m}{r} E = K + U_{g} = - \frac{G M m}{2 r}

The velocity is always perpendicular to the line from the center, so the satellite's angular momentum about the center is $L = r m v$ , from 6.3 Angular Momentum and Angular Impulse.

With $r$ fixed, $v$ is fixed, so $K$ , $U_{g}$ , $E$ and $L$ all stay constant all the way round. Gravity points at the center, at right angles to the velocity, and does no work on the satellite.

The diagram follows a 500 kg satellite. At $1.0 \times 10^{7}$ m from Earth's center it moves at $6.33 \times 10^{3}$ m/s, with $K = 10.0$ GJ, $U_{g} = - 20.0$ GJ and $E = - 10.0$ GJ, where 1 GJ is $10^{9}$ J. The negative total means the satellite is bound: it does not have the energy to reach infinite separation.

Quick check · Calculate

A 500 kg spacecraft moves in a circular orbit $4.0 \times 10^{6}$ m from the center of Mars, whose mass is $6.4 \times 10^{23}$ kg. Use $G = 6.67 \times 10^{- 11} N \cdot m^{2} / {kg}^{2}$ , and put $U_{g} = 0$ at infinite separation. What is the total mechanical energy of the spacecraft-Mars system? Give your answer in gigajoules (1 GJ is $10^{9}$ J), to the nearest 0.01 GJ, with its sign.

Your answer GJ

2c. Elliptical orbits: two stay, two trade

In an elliptical orbit the satellite's distance from the central object changes, and so does its speed. Gravity is still the only force, and it still points at the central object's center. About that center it exerts no torque, because the force lies along the line from the center. With no angular impulse, the satellite's angular momentum about the center stays the same, as in 6.3 Angular Momentum and Angular Impulse. And with only gravity acting inside the system, the mechanical energy stays the same, as in 3.4 Conservation of Energy.

What changes is the split. As the satellite falls inward, $U_{g}$ becomes more negative and $K$ grows by the same amount. As it climbs back out, $U_{g}$ rises and $K$ shrinks.

At the closest and farthest points the velocity is perpendicular to the line from the center, so there $L = r m v$ :

r_{1} m v_{1} = r_{2} m v_{2} \Rightarrow r_{1} v_{1} = r_{2} v_{2}

If the farthest point is three times as far out as the closest point, the satellite moves a third as fast there. That is why a satellite speeds up as it swings in close. At every point in between, the velocity is not perpendicular to the line from the center, so use $r m v \sin θ$ .

Quick check · Sort it

Each card names one quantity for a satellite orbiting a far more massive planet, with only gravity acting between them. Tap a card, then whether that quantity stays the same or changes as the satellite goes around its orbit.

Stays the same

Changes

2d. Escape

A bound satellite has a negative mechanical energy. Give it more speed and the total rises. The speed that makes the total exactly zero is the size of the escape velocity, the escape speed:

\frac{1}{2} m v^{2} - \frac{G M m}{r} = 0 \Rightarrow v_{esc} = \sqrt{\frac{2 G M}{r}}

The satellite's mass cancels again, and the escape speed is $\sqrt{2}$ times the circular orbit speed at the same distance. From Earth's surface, $6.4 \times 10^{6}$ m from its center, it is about 11.2 km/s, if you ignore the air.

Follow a probe launched at exactly that speed, with gravity the only force. The total stays zero, so $K = - U_{g}$ at every distance. As $r$ grows, $U_{g}$ rises toward zero and $K$ falls toward zero with it. The probe keeps slowing and keeps moving away, and its speed reaches zero only at an infinite distance.

Launch it slower and the total is negative, so the probe can never reach an infinite distance. Launched straight up, it stops for an instant at its greatest height and falls back. Launched at any other angle, it never stops: it follows a closed path around the planet's center, still moving sideways at its farthest point, and a path that starts at the surface comes back to the surface. Launch it faster and the total is positive: it still has speed left however far it goes.

Quick check · Sketch the graph

A probe coasts straight away from a planet with its engines off, and only the planet's gravity acts on it. At 20,000 km ( $2.0 \times 10^{7}$ m) from the planet's center it moves at 10 km/s, exactly the escape speed at that distance. Sketch the probe's speed against its distance from the planet's center, from 20,000 km out to 100,000 km. The first point is fixed at 10 km/s.

Drag a point up or down, or tap it and use the arrow keys. The shape is checked, and the graph must pass through the values the question gives.

Part 3 Watch it done Far side of a lunar orbit

Watch each step, then finish the problem yourself.

A spacecraft moves in an elliptical orbit around the Moon, whose mass is $7.35 \times 10^{22}$ kg, with only the Moon's gravity acting on it. At its closest point it is $2.0 \times 10^{6}$ m from the Moon's center and moves at 1,870 m/s. Its farthest point is $5.0 \times 10^{6}$ m from the Moon's center. Find its speed at the farthest point.

Step 1 · Decide what stays constant
The spacecraft and the Moon make the system, and the Moon's motion is negligible. Gravity on the spacecraft points straight at the Moon's center, along the line from that center, so it exerts no torque about it. With no angular impulse, the spacecraft's angular momentum about the Moon's center is the same at every point of the orbit.
Step 2 · Write the angular momentum at the two points
At the closest and farthest points the velocity is perpendicular to the line from the Moon's center, so $\sin θ = 1$ and $L = r m v$ . Setting the two equal, $r_{1} m v_{1} = r_{2} m v_{2}$ , the mass cancels and $v_{2} = \frac{r_{1}}{r_{2}} v_{1}$ .
Why this step?
Why is it safe to drop the $\sin θ$ at these two points, but not at a point halfway between them?
At the closest and farthest points the spacecraft is moving neither toward the Moon nor away from it, so its velocity is at right angles to the line from the center and $\sin θ = 1$ . In between, its distance is changing, so part of its velocity runs along that line and the angle is not 90°. There $L = r m v \sin θ$ , with $\sin θ$ less than 1.
Your step · Calculate

Finish it. A spacecraft moves in an elliptical orbit around the Moon, with only the Moon's gravity acting on it. At its closest point it is $2.0 \times 10^{6}$ m from the Moon's center and moves at 1,870 m/s. Its farthest point is $5.0 \times 10^{6}$ m from the Moon's center. What is its speed at the farthest point?

Your answer m/s

Check it with energy, per kilogram of spacecraft. At the closest point, $\frac{1}{2} v_{1}^{2} - \frac{G M}{r_{1}} = \frac{1}{2} (1870)^{2} - \frac{(6.67 \times 10^{- 11}) (7.35 \times 10^{22})}{2.0 \times 10^{6}} = - 7.03 \times 10^{5}$ J/kg. At the farthest point, $\frac{1}{2} (748)^{2} - \frac{G M}{5.0 \times 10^{6}} = - 7.01 \times 10^{5}$ J/kg. They agree to within the rounding of 1,870 m/s: the mechanical energy is the same at both ends, as it must be.

Part 4 Try it Leaving from orbit

A new situation: the probe does not start from the ground. It starts from a space station that is already in orbit around Earth.

Try it · Calculate

A space station moves in a circular orbit $6.7 \times 10^{6}$ m from Earth's center. Earth's mass is $6.0 \times 10^{24}$ kg; use $G = 6.67 \times 10^{- 11} N \cdot m^{2} / {kg}^{2}$ . A small probe is launched from the station, and from then on only Earth's gravity acts on it. What is the least speed, measured relative to Earth's center, the probe needs at that distance to escape from Earth? Give your answer to the nearest 10 m/s.

Your answer m/s

Part 5 Explain it Circle, ellipse, escape

Nothing you type leaves this page. Check your answer against the model and each scoring point.

Free response is 50% of the exam score. This one is shaped like Question 1, Mathematical Routines.

Free-response practice · Question 1 style

A satellite of mass $m$ orbits a planet of mass $M$ , where $M$ is far greater than $m$ . Only the gravitational forces between them act, the planet's motion is negligible, and $U_{g} = 0$ at infinite separation.

(a) The satellite moves in a circular orbit of radius $R$ . Starting from Newton's second law, derive an expression for its speed $v_{0}$ in terms of $G$ , $M$ and $R$ .

(b) Derive an expression for the total mechanical energy of the satellite-planet system in that circular orbit, in terms of $G$ , $M$ , $m$ and $R$ .

(c) A short engine burn raises the satellite's speed to $v_{1}$ without changing its distance $R$ or turning its velocity, which stays perpendicular to the line from the planet's center. The satellite now follows an elliptical orbit whose closest distance is $R$ , at the burn point, and whose farthest distance is $3 R$ . Using conservation of angular momentum, derive an expression for its speed $v_{2}$ at the farthest point in terms of $v_{1}$ . Then, using conservation of energy, derive an expression for $v_{1}$ in terms of $G$ , $M$ and $R$ .

(d) Derive an expression for the escape speed $v_{esc}$ at a distance $R$ from the planet's center.

(e) $M = 6.0 \times 10^{24}$ kg and $R = 7.0 \times 10^{6}$ m, with $G = 6.67 \times 10^{- 11} N \cdot m^{2} / {kg}^{2}$ . Calculate $v_{0}$ , $v_{1}$ , $v_{2}$ and $v_{esc}$ . Use your values to explain why the satellite in (c) stays bound to the planet.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) Gravity is the only force and it points at the planet's center, so it alone gives the centripetal acceleration: $G \frac{M m}{R^{2}} = m \frac{v_{0}^{2}}{R}$ , so $v_{0} = \sqrt{\frac{G M}{R}}$ .

(b) $K = \frac{1}{2} m v_{0}^{2} = \frac{G M m}{2 R}$ and $U_{g} = - \frac{G M m}{R}$ , so $E = K + U_{g} = - \frac{G M m}{2 R}$ .

(c) Gravity points along the line to the planet's center, so it exerts no torque about that center, and the satellite's angular momentum about it is constant. At the closest and farthest points the velocity is perpendicular to that line, so $L = r m v$ there: $R m v_{1} = (3 R) m v_{2}$ , and $v_{2} = \frac{v_{1}}{3}$ . Only gravity acts, inside the system, so the mechanical energy is the same at both points:

\frac{1}{2} m v_{1}^{2} - \frac{G M m}{R} = \frac{1}{2} m {(\frac{v_{1}}{3})}^{2} - \frac{G M m}{3 R}

\frac{4}{9} v_{1}^{2} = \frac{2}{3} \frac{G M}{R} \Rightarrow v_{1} = \sqrt{\frac{3 G M}{2 R}}

(d) At escape speed the mechanical energy is zero: $\frac{1}{2} m v_{esc}^{2} - \frac{G M m}{R} = 0$ , so $v_{esc} = \sqrt{\frac{2 G M}{R}}$ .

(e) $\frac{G M}{R} = \frac{(6.67 \times 10^{- 11}) (6.0 \times 10^{24})}{7.0 \times 10^{6}} = 5.72 \times 10^{7}$ m²/s². Then $v_{0} = 7.56 \times 10^{3}$ m/s, $v_{1} = \sqrt{1.5} v_{0} = 9.26 \times 10^{3}$ m/s, $v_{2} = \frac{v_{1}}{3} = 3.09 \times 10^{3}$ m/s and $v_{esc} = \sqrt{2} v_{0} = 1.07 \times 10^{4}$ m/s. The burn speed $v_{1}$ is below $v_{esc}$ , so the mechanical energy is still negative, $\frac{1}{2} m v_{1}^{2} - \frac{G M m}{R} = - \frac{G M m}{4 R}$ , and the satellite cannot reach infinite separation. It stays bound, swinging out to $3 R$ and back.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Sets the gravitational force equal to $m \frac{v_{0}^{2}}{R}$ to derive $v_{0} = \sqrt{\frac{G M}{R}}$ , then adds $K = \frac{G M m}{2 R}$ and $U_{g} = - \frac{G M m}{R}$ to get $E = - \frac{G M m}{2 R}$ .
Uses $L = r m v$ at the closest and farthest points, justified by the velocity being perpendicular to the radius and by gravity exerting no torque about the center, to get $v_{2} = \frac{v_{1}}{3}$ .
Equates the mechanical energy at the closest and farthest points, with the minus signs on $U_{g}$ , and derives $v_{1} = \sqrt{\frac{3 G M}{2 R}}$ .
Sets the mechanical energy to zero and derives $v_{esc} = \sqrt{\frac{2 G M}{R}}$ .
Calculates $v_{0} = 7.56 \times 10^{3}$ m/s, $v_{1} = 9.26 \times 10^{3}$ m/s, $v_{2} = 3.09 \times 10^{3}$ m/s and $v_{esc} = 1.07 \times 10^{4}$ m/s, with units, and explains that $v_{1} < v_{esc}$ leaves the mechanical energy negative, so the satellite stays bound.

Part 6 Check yourself One orbit, three points

Both case questions use the satellite table.

A 1,000 kg satellite moves in an elliptical orbit around Earth, and only Earth's gravity acts on it. Point A is the closest point of its orbit and point C the farthest. The table gives the satellite's distance from Earth's center at three points, the gravitational potential energy of the satellite-Earth system there, with zero at infinite separation, and the satellite's kinetic energy. One kinetic energy has not been filled in. Energies are in gigajoules, where 1 GJ is $10^{9}$ J.

The satellite at three points of its orbit
Point	Distance from Earth's center (m)	Gravitational potential energy (GJ)	Kinetic energy (GJ)
A	$8.0 \times 10^{6}$	-50.0	40.0
B	$1.6 \times 10^{7}$	-25.0	15.0
C	$3.2 \times 10^{7}$	-12.5	?

Case question 1 · Calculate

Using the satellite table, what is the satellite's kinetic energy at point C? Give your answer in gigajoules, to the nearest 0.1 GJ.

Your answer GJ

Case question 2 · Multiple choice

Using the satellite table, how does the satellite's speed at point A compare with its speed at point C?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

A student derives the escape speed of a probe launched from the surface of an airless planet of mass $M$ and radius $R$ , with $U_{g} = 0$ at infinite separation. Tap the first line that goes wrong.

Check yourself · Match

A satellite of mass $m$ moves in a circular orbit of radius $r$ around a planet of mass $M$ , with $U_{g} = 0$ at infinite separation. Tap a quantity, then the expression for it.

Check yourself · Fill the statement

Tap a blank, then an answer.

A probe launched at exactly the escape speed gives the probe-planet system a total mechanical energy that is . As the probe moves away, the system's gravitational potential energy becomes negative. The probe's speed reaches zero only when it is from the planet.

Part 7 Keep it Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 3.3 Potential Energy · Multiple choice

A satellite orbits a planet at a distance $r$ from the planet's center. With the zero at infinite separation, the gravitational potential energy of the satellite-planet system is $- 8.0 \times 10^{9}$ J. The satellite is moved out to a distance $2 r$ from the center. What is the system's gravitational potential energy now?

A new situation: nothing here turns at all. A skater glides straight across a frozen pond past a lamp post at its edge, and the question is her angular momentum about the post.

From 6.3 Angular Momentum and Angular Impulse · Calculate

A 60 kg skater glides in a straight line across level ice at a steady 3.0 m/s, and friction is negligible. At one instant she is 5.0 m from a lamp post, and the angle between the line from the post to her and her velocity is 150°. What is the magnitude of her angular momentum about the post?

Your answer kg·m^2/s

In three lines

The central object's motion is negligible, gravity is the only force, and the conservation laws decide how the satellite moves.
Circular: $v = \sqrt{\frac{G M}{r}}$ , with $E$ , $U_{g}$ , $L$ and $K$ constant. Elliptical: $E$ and $L$ constant while $U_{g}$ and $K$ trade off, and $r_{1} v_{1} = r_{2} v_{2}$ at the closest and farthest points.
$U_{g} = - \frac{G M m}{r}$ , zero at infinite separation; escape makes $E = 0$ , so $v_{esc} = \sqrt{\frac{2 G M}{r}}$ .

Unit 6 check: 10 questions from every lesson in Unit 6

Vocabulary

Escape velocity: The velocity of a satellite at which the mechanical energy of the satellite-central-object system is zero; at a distance $r$ from a central body of mass $M$ its size is $v_{esc} = \sqrt{\frac{2 G M}{r}}$ .
Circular orbit: An orbit at a constant distance from the central object, in which the system's mechanical energy and gravitational potential energy and the satellite's angular momentum and kinetic energy are all constant.
Elliptical orbit: An orbit whose distance from the central object changes, in which the system's mechanical energy and the satellite's angular momentum are constant while the gravitational potential energy and the satellite's kinetic energy each change.
Gravitational potential energy: For a satellite and a massive central object, the energy $U_{g} = - G \frac{m_{1} m_{2}}{r}$ , defined to be zero when the satellite is an infinite distance from the central object.

Course alignment, for teachers

AP Physics 1: Algebra-Based topic 6.6, Unit 6: Energy and Momentum of Rotating Systems.

Skill 1.C: Create qualitative sketches of graphs that represent features of a model or the behavior of a physical system.
Skill 2.A: Derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway.
Skill 2.C: Compare physical quantities between two or more scenarios or at different times and locations in a single scenario.
Skill 3.C: Justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

This lesson maps all 8 essential knowledge statements of the topic to the part that teaches each one: 8 are checked by a scored question. The map is in the lesson file.