Unit 7: Gravitation

Newton’s Universal Law of Gravitation

What gravity is (in the Newtonian model)

In AP Physics C: Mechanics, gravitation is treated primarily using Newton’s classical (non-relativistic) model: every mass attracts every other mass with a force that acts along the line connecting their centers. This single interaction explains why objects have weight near Earth, how satellites and planets move in orbit, and how energy and speed change when an object moves through a gravitational field.

A key idea to keep in mind from the start is that gravity is a long-range, inverse-square interaction. The same law describes the pull between two apples, between Earth and the Moon, and between the Sun and a distant planet. The only difference is the sizes of the masses and the separation.

Historically, Newton proposed this law in 1687. It remains the foundation for most orbit calculations in this course, even though modern physics (Einstein’s general relativity) refines the picture by describing gravity as curvature of spacetime caused by mass-energy.

The force law and how to use it

Newton’s law of universal gravitation says the magnitude of the gravitational force between two point masses (or spherically symmetric masses treated as point masses) is

F_g = G\frac{m_1 m_2}{r^2}

Here:

• F_g is the gravitational force magnitude
• G is the universal gravitational constant
• m_1 and m_2 are the interacting masses
• r is the center-to-center separation

The gravitational constant is approximately

G \approx 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2

The direction is always attractive: each mass pulls the other toward itself along the line joining them.

Two immediate “how it works” details prevent common errors:

1. Use center-to-center distance. If one mass is Earth and the other is a satellite at altitude h, then

r = R_E + h

2. Spherical bodies act like point masses (outside them). If a mass distribution is spherically symmetric (like an idealized planet), then for points outside the sphere you can treat the entire mass as located at its center. This is a consequence of the shell theorem and is why the inverse-square law works so cleanly for planets.

Vector form (direction matters)

If you define \hat{r} as a unit vector pointing from mass 1 to mass 2, then the force on mass 2 due to mass 1 is

\vec{F}_{1\rightarrow 2} = -G\frac{m_1 m_2}{r^2}\hat{r}

The negative sign encodes attraction: the force points opposite \hat{r}, meaning mass 2 is pulled toward mass 1.

Superposition: gravity from multiple masses

Gravity obeys superposition: the net gravitational force on a mass is the vector sum of the forces from every other mass.

\vec{F}_{net} = \sum_i \vec{F}_i

This matters in problems with multiple attracting bodies (Earth plus Moon, two stars, and so on). A practical warning: superposition is simple in principle but can be algebra-heavy, and the most common mistake is adding magnitudes instead of vectors. Sketch directions first.

Worked example: gravity changes with altitude

A satellite of mass m orbits Earth at altitude h. Compare the gravitational force there to its weight at Earth’s surface.

At the surface:

F_{surface} = G\frac{M_E m}{R_E^2}

At altitude h:

F_{alt} = G\frac{M_E m}{\left(R_E + h\right)^2}

Take the ratio:

\frac{F_{alt}}{F_{surface}} = \frac{R_E^2}{\left(R_E + h\right)^2}

This shows gravity does not suddenly vanish in space; it decreases smoothly with distance as an inverse square.

Exam Focus

• Typical question patterns:
  • “Find the gravitational force between…” often with altitude so you must use r = R + h.
  • “Compare forces/weights at two distances” using ratios to avoid heavy computation.
  • Multi-mass superposition setups requiring vector addition.
• Common mistakes:
  • Using surface radius R_E when the problem gives altitude h.
  • Adding force magnitudes instead of vectors in superposition problems.
  • Forgetting the force is mutual: the two bodies exert equal-magnitude forces (Newton’s third law).

Gravitational Field and Gravitational Acceleration

From force to field: why fields are useful

A field is a way to describe an interaction without always talking about two objects at once. Instead of saying “Earth pulls on you with a force,” you say “Earth creates a gravitational field, and any mass placed there experiences a force.” This separates the source (the planet or mass creating the field) from the test mass (the object experiencing the force), which makes many problems cleaner.

Definition of gravitational field strength

The gravitational field at a point is defined as force per unit mass on a small test mass:

\vec{g} = \frac{\vec{F}_g}{m}

For a spherically symmetric mass M, outside the mass distribution:

\vec{g}(r) = -G\frac{M}{r^2}\hat{r}

The magnitude is

g(r) = G\frac{M}{r^2}

Weight vs gravitational force vs the value of g

It helps to keep three related ideas distinct:

• Gravitational force on a mass m due to Earth:

F_g = G\frac{M_E m}{r^2}

• Field strength created by Earth:

g = G\frac{M_E}{r^2}

• Weight in introductory contexts:

W = mg

In Newtonian gravitation, the expression for weight is really shorthand for the gravitational force when the field is approximately uniform over the object’s size.

A subtle but important conceptual point: astronauts in orbit still experience a significant gravitational force. They feel “weightless” because they are in free fall: their spacecraft and bodies accelerate together under gravity, so there is little or no normal force.

Superposition in field form

Fields also superpose:

\vec{g}_{net} = \sum_i \vec{g}_i

Once you have \vec{g}_{net} at a location, the force on any object of mass m is

\vec{F} = m\vec{g}_{net}

Worked example: where is the net field zero between two masses?

Two masses M and m lie on a line separated by distance d. Where (between them) is the net gravitational field zero?

Let the point be at distance x from M and d minus x from m. Between them, the fields point in opposite directions, so cancellation is possible. Set magnitudes equal:

G\frac{M}{x^2} = G\frac{m}{\left(d - x\right)^2}

Cancel G:

\frac{M}{x^2} = \frac{m}{\left(d - x\right)^2}

Take square roots:

\frac{\sqrt{M}}{x} = \frac{\sqrt{m}}{d - x}

Solve:

\sqrt{M}(d - x) = \sqrt{m}x

\sqrt{M}d = x\left(\sqrt{M} + \sqrt{m}\right)

x = \frac{\sqrt{M}}{\sqrt{M} + \sqrt{m}}d

This shows the zero-field point is closer to the smaller mass.

Exam Focus

• Typical question patterns:
  • “Find g at altitude h” or “compare g at two radii.”
  • “Locate the point where net gravitational field is zero” (usually along a line).
  • “Explain weightlessness in orbit” conceptually (free fall vs absence of gravity).
• Common mistakes:
  • Treating g as constant even when distance from the planet changes significantly.
  • Mixing up direction: field points toward the attracting mass.
  • Setting forces equal instead of fields equal when the test mass differs (field method avoids this).

Gravitational Potential Energy and Gravitational Potential

Why potential energy is needed

For many gravitation problems, forces change with position (inverse-square), so constant-acceleration kinematics do not apply. Energy methods become powerful because gravity is a conservative force: the work done by gravity depends only on initial and final positions, not the path.

Gravitational potential energy can be described as the work required to move an object from infinity to a point in the gravitational field (with the standard sign convention discussed below). In SI units, potential energy is measured in joules.

Work done by gravity and the zero of potential energy

You can choose where potential energy is zero, but you must do it consistently. For gravitation, the standard convention is

• U = 0 at r \rightarrow \infty

With that choice, gravitational potential energy for two point masses is

U(r) = -G\frac{Mm}{r}

The negative sign reflects attraction and the convention that bound states have negative potential energy when zero is at infinity.

Connecting force and potential energy

Along the radial coordinate, conservative forces satisfy

F_r = -\frac{dU}{dr}

If

U(r) = -G\frac{Mm}{r}

then

\frac{dU}{dr} = -G Mm\frac{d}{dr}\left(r^{-1}\right)

Since

\frac{d}{dr}\left(r^{-1}\right) = -r^{-2}

you get

\frac{dU}{dr} = G\frac{Mm}{r^2}

and therefore

F_r = -G\frac{Mm}{r^2}

Potential energy changes and energy conservation

The change in gravitational potential energy moving from r_i to r_f is

\Delta U = U_f - U_i = -G\frac{Mm}{r_f} + G\frac{Mm}{r_i}

Total mechanical energy (when only gravity does work) is conserved:

K_i + U_i = K_f + U_f

with kinetic energy

K = \frac{1}{2}mv^2

Near Earth’s surface, the gravitational field is approximately constant over small height changes, so gravitational potential energy is often approximated by

\Delta U \approx mg\Delta y

and you will also see the near-surface form

PE = mgh

This constant-g form is widely used in practical settings (for example, estimating energy changes in roller coasters and other rides), but it is an approximation valid only when the height change is small compared to Earth’s radius.

Gravitational potential (energy per unit mass)

The gravitational potential V is defined as potential energy per unit mass:

V = \frac{U}{m}

For a spherically symmetric mass M:

V(r) = -G\frac{M}{r}

Once you know V, the potential energy of an object of mass m is

U = mV

Relationship between potential and field

In the radial direction,

g_r = -\frac{dV}{dr}

This is the field analog of F_r = -dU/dr.

Worked example: launching upward using energy

You launch a projectile of mass m from Earth’s surface (radius R_E) straight upward with speed v_0. Ignore air resistance. Find its speed at altitude h.

Initial energy:

E_i = \frac{1}{2}mv_0^2 - G\frac{M_E m}{R_E}

Final energy at radius R_E + h:

E_f = \frac{1}{2}mv^2 - G\frac{M_E m}{R_E + h}

Set E_i = E_f and cancel m:

\frac{1}{2}v_0^2 - G\frac{M_E}{R_E} = \frac{1}{2}v^2 - G\frac{M_E}{R_E + h}

Solve for v^2:

\frac{1}{2}v^2 = \frac{1}{2}v_0^2 - G\frac{M_E}{R_E} + G\frac{M_E}{R_E + h}

v^2 = v_0^2 - 2G M_E\left(\frac{1}{R_E} - \frac{1}{R_E + h}\right)

Exam Focus

• Typical question patterns:
  • “Find the change in gravitational potential energy between radii r_i and r_f.”
  • “Use energy conservation to find speed at a different radius.”
  • “Interpret graphs of U(r) or V(r) and connect slope to force/field.”
• Common mistakes:
  • Using mgh for large altitude changes (where g is not constant).
  • Losing the negative sign in U = -G Mm/r and concluding bound objects have positive energy.
  • Mixing up U (depends on test mass) and V (does not).

Circular Orbits, Orbital Speed, and Period

Why orbits are a “gravity plus centripetal acceleration” story

A circular orbit happens when gravity provides exactly the inward (centripetal) acceleration needed to keep an object moving in a circle at constant speed. For an object of mass m in a circular orbit of radius r around a central mass M, the required centripetal acceleration magnitude is

a_c = \frac{v^2}{r}

and the required centripetal force magnitude is

F_c = m\frac{v^2}{r}

Gravity supplies that force:

G\frac{Mm}{r^2} = m\frac{v^2}{r}

Cancel m:

G\frac{M}{r^2} = \frac{v^2}{r}

Solve for orbital speed:

v = \sqrt{\frac{GM}{r}}

This encodes an important conceptual fact: in circular orbit, a higher radius means a lower speed.

Orbital period

The orbital period T is the time for one full revolution. For circular motion,

v = \frac{2\pi r}{T}

Combine with v = \sqrt{GM/r} and solve:

T = 2\pi\sqrt{\frac{r^3}{GM}}

This is the circular-orbit form of Kepler’s third law and shows that the period grows like a power of r equal to three halves.

Circular orbit energy

Potential energy:

U = -G\frac{Mm}{r}

Kinetic energy using v^2 = GM/r:

K = \frac{1}{2}mv^2 = \frac{1}{2}m\frac{GM}{r} = \frac{1}{2}G\frac{Mm}{r}

So

K = -\frac{1}{2}U

Total mechanical energy:

E = K + U = -\frac{1}{2}G\frac{Mm}{r}

A common misconception is to think “higher orbit means higher speed because it’s farther and must travel more.” The circumference is larger, but the speed is smaller because the required centripetal acceleration is smaller and gravity is weaker.

Worked example: geostationary orbit (symbolic setup)

A geostationary orbit is a circular orbit in Earth’s equatorial plane with period equal to Earth’s rotation period, so the satellite stays above the same point on Earth.

Given period T, solve for orbital radius r using

T = 2\pi\sqrt{\frac{r^3}{GM_E}}

Rearrange:

\left(\frac{T}{2\pi}\right)^2 = \frac{r^3}{GM_E}

r^3 = GM_E\left(\frac{T}{2\pi}\right)^2

r = \left[GM_E\left(\frac{T}{2\pi}\right)^2\right]^{1/3}

Altitude is then

h = r - R_E

On an exam, you might not need to compute the number; setting up the relationship correctly is often the main task.

Exam Focus

• Typical question patterns:
  • “Find the speed/period of a circular orbit at radius r” using centripetal force equals gravity.
  • “Compare two circular orbits” using ratios like v \propto r^{-1/2} and T \propto r^{3/2}.
  • “Find the orbital radius for a given period” (geostationary-style problems).
• Common mistakes:
  • Using r = h instead of r = R + h.
  • Confusing orbital speed v with escape speed (different concepts, different formulas).
  • Forgetting that the satellite’s mass cancels in orbital speed and period formulas.

Kepler’s Laws (Conceptual and Calculational Use)

What Kepler’s laws say and why they matter

Kepler’s laws are empirical descriptions of planetary motion that Newton later explained using universal gravitation. You should know both the qualitative meaning and how to use the most testable quantitative relationship.

1. Kepler’s First Law: Planets move in ellipses with the Sun at one focus.
2. Kepler’s Second Law: A line from the Sun to a planet sweeps out equal areas in equal times.
3. Kepler’s Third Law: The square of the orbital period is proportional to the cube of the semi-major axis.

In our solar system, most planetary orbits are nearly circular even though they are technically elliptical.

Ellipses and eccentricity (orbit shape)

The deviation of an ellipse from a perfect circle is measured by its eccentricity. If a is the semi-major axis and c is the distance from the center of the ellipse to either focus, then

e = \frac{c}{a}

Small e means “nearly circular,” while larger e means “more stretched.”

Kepler’s third law in Newtonian form

For objects orbiting the same central mass M, the relationship is

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

Here a is the semi-major axis of the ellipse. For a circle, a equals the orbital radius r. For circular orbits, we already derived

T = 2\pi\sqrt{\frac{r^3}{GM}}

and squaring gives the same form.

What Kepler’s second law is really telling you

Kepler’s second law is a statement about conservation of angular momentum for central forces. Gravity is a central force (it always points toward the center), which implies torque about the center is zero and angular momentum is conserved. Even when you are not asked to compute areas, you should be able to reason qualitatively:

• closer to the Sun (smaller r) means faster speed
• farther from the Sun (larger r) means slower speed

Barycenter (center of mass of a two-body orbit)

Kepler’s first law is often stated as “the Sun is at one focus,” but more precisely the focus is at the center of mass of the Sun-planet system. When one body orbits another, both bodies orbit their common center of mass, called the barycenter. For planets much less massive than the Sun, this correction is small because the barycenter lies very close to the Sun’s center.

Worked example: center of mass of the Sun-Earth system

Let the mass of Earth be

m = 5.98 \times 10^{24} \text{ kg}

the mass of the Sun be

M = 1.99 \times 10^{30} \text{ kg}

and the average Sun-Earth distance be

R = 1.496 \times 10^{11} \text{ m}

Measure position along the line from the Sun toward Earth, with the Sun’s center at x equals zero. The center of mass location is

x_{cm} = \frac{mR}{M+m}

Numerically,

x_{cm} \approx \frac{\left(5.98 \times 10^{24}\right)\left(1.496 \times 10^{11}\right)}{1.99 \times 10^{30} + 5.98 \times 10^{24}}

which is approximately

x_{cm} \approx 4.49 \times 10^{5} \text{ m}

This is tiny compared with the Sun-Earth distance, so for many problems it is reasonable to treat the Sun as fixed.

Worked example: comparing periods using Kepler’s third law

Two planets orbit the same star. Planet 1 has orbital radius r_1 (circular) and period T_1. Planet 2 has orbital radius r_2 (circular). Find T_2.

From Kepler’s third law for the same central mass:

\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}

So

T_2 = T_1\left(\frac{r_2}{r_1}\right)^{3/2}

Exam Focus

• Typical question patterns:
  • “Use Kepler’s third law to compare periods/radii” often as a ratio problem.
  • “Identify what changes along an elliptical orbit” (speed increases near periapsis).
  • “Derive or interpret T^2 \propto a^3” by connecting to circular orbit results.
• Common mistakes:
  • Using Kepler’s third law to compare orbits around different central masses without including M.
  • Treating a as the closest approach rather than the semi-major axis.
  • Thinking Kepler’s second law is about constant speed (it’s not).

Escape Speed and Gravitational Binding Energy

What escape speed means (and what it does not mean)

Escape speed is the minimum speed an object must have at a given radius to travel infinitely far away and arrive at infinity with zero kinetic energy (ignoring air resistance and additional forces). Escape speed is about energy, not “breaking through” the atmosphere, and it does not require engines after launch in the idealized model.

Deriving escape speed from energy

Start at radius r from a mass M with speed v_e. At infinity, take U(\infty) = 0 and the minimum escape condition is K(\infty) = 0.

Initial energy:

E_i = \frac{1}{2}mv_e^2 - G\frac{Mm}{r}

Final energy:

E_f = 0

Set E_i = E_f:

\frac{1}{2}mv_e^2 - G\frac{Mm}{r} = 0

Solve:

v_e = \sqrt{\frac{2GM}{r}}

Compare with circular orbital speed v = \sqrt{GM/r}: escape speed is larger by a factor of \sqrt{2} at the same radius.

Gravitational binding energy (system perspective)

Binding energy is the energy required to separate a bound system into pieces at infinite separation (with zero kinetic energy at infinity). For two masses M and m separated by radius r, the magnitude of the gravitational potential energy

\left|U\right| = G\frac{Mm}{r}

represents how much energy you must add (in an idealized process) to take the system from separation r to infinity.

For orbital motion, the sign of total energy is a powerful classifier:

• If total energy E is negative, the object is bound (elliptical or circular orbit).
• If E is zero, the trajectory is parabolic (escape at exactly escape speed).
• If E is positive, the trajectory is hyperbolic (unbound flyby).

Worked example: escape speed from Earth’s surface (symbolic)

From Earth’s surface, r equals R_E:

v_e = \sqrt{\frac{2GM_E}{R_E}}

If the problem gives g instead of GM_E, use

g = G\frac{M_E}{R_E^2}

so

GM_E = gR_E^2

Substitute:

v_e = \sqrt{\frac{2gR_E^2}{R_E}} = \sqrt{2gR_E}

Exam Focus

• Typical question patterns:
  • “Derive escape speed” or “find escape speed at radius r.”
  • “Compare escape speeds” using ratio reasoning v_e \propto r^{-1/2}.
  • “Determine whether an object is bound” using the sign of total energy.
• Common mistakes:
  • Using r = h instead of r = R + h.
  • Confusing escape speed with orbital speed and using the wrong factor of 2.
  • Setting final potential energy to zero at the surface instead of at infinity (inconsistent reference).

Gravitation for Extended Spherical Mass Distributions (Shell Theorem Ideas)

Why you can treat planets as point masses (most of the time)

Many gravitation problems assume planets and stars are spherical and have spherically symmetric density. The shell theorem gives two key conclusions for a spherically symmetric mass distribution:

1. Outside the sphere, the gravitational force is the same as if all mass were concentrated at the center.
2. Inside a thin spherical shell, the net gravitational force is zero.

These results justify why most problems use

F_g = G\frac{Mm}{r^2}

with r measured from the center, even though the mass is spread out.

What happens inside a uniform solid sphere (conceptual result)

Inside an idealized uniform-density sphere, only the mass at smaller radius contributes to the net gravitational force; outer shells cancel. The enclosed mass grows like the cube of r, so the gravitational force inside grows like r rather than shrinking like the inverse square.

Qualitatively:

• at the center, net gravitational force is zero (symmetry)
• moving outward, the force increases approximately linearly (for uniform density)
• at the surface, it transitions to the inverse-square decrease outside

Worked example: field scaling inside a uniform-density sphere

Let a uniform sphere of radius R have total mass M. At radius r less than R, the enclosed mass is

M_{enc} = M\left(\frac{r^3}{R^3}\right)

Then the gravitational field magnitude is

g(r) = G\frac{M_{enc}}{r^2} = G\frac{M}{R^3}r

So the field is proportional to r inside the sphere.

Exam Focus

• Typical question patterns:
  • “Justify treating Earth as a point mass” (spherical symmetry argument).
  • “Sketch/interpret g(r) or U(r) inside vs outside a planet” (conceptual graphing).
  • “Determine how gravity changes inside Earth” using proportional reasoning.
• Common mistakes:
  • Applying g = GM/r^2 inside a hollow shell (should be zero inside the shell).
  • Assuming gravity keeps increasing as you go toward the center (it goes to zero at the center).
  • Confusing “inside a shell” with “inside a solid sphere” (different results).

Two-Body Motion, Center of Mass Ideas, and What AP Usually Assumes

When the “central mass is fixed” approximation is valid

In many orbit problems, you treat the central body (Earth, Sun) as stationary and the satellite as moving. This approximation becomes very good when M is much greater than m. In reality, both bodies orbit their center of mass, and the gravitational forces are equal in magnitude and opposite in direction:

F = G\frac{Mm}{r^2}

Both bodies accelerate:

a_m = \frac{F}{m} = G\frac{M}{r^2}

a_M = \frac{F}{M} = G\frac{m}{r^2}

So the more massive body accelerates much less. This is why treating Earth as fixed is reasonable for a small satellite, but treating one star as fixed is not reasonable in a binary system with comparable masses.

Why this matters conceptually

Even when you do not compute barycenters often, this viewpoint prevents a subtle misunderstanding: gravity is not “something big objects do to small objects.” It is a mutual interaction between masses.

Worked example: where is the center of mass on a line?

Two masses M and m are separated by distance d. Measured from mass M along the line toward m, the center of mass location is

x_{cm} = \frac{md}{M + m}

If M is much larger than m, then x_{cm} is very small compared to d.

Exam Focus

• Typical question patterns:
  • “Explain why satellite mass does not affect orbital speed/period” (it cancels).
  • “Describe the motion of two gravitating bodies” conceptually (both orbit the center of mass).
  • Center-of-mass calculations as a supporting step in multi-concept problems.
• Common mistakes:
  • Treating gravitational force as one-directional rather than mutual.
  • Assuming the larger mass experiences no acceleration at all.
  • Forgetting that “fixed central mass” is an approximation with conditions.

Mixed Gravitation Problems: Strategy, Modeling Choices, and Multi-Concept Links

Choosing the right tool: force, field, energy, or period relations

Gravitation questions become much easier when you choose a method that matches what is asked.

• Use force when you need acceleration or when other forces (tension, normal force) appear along with gravity.
• Use field when comparing effects on different test masses or combining multiple sources.
• Use energy when speed changes with radius or when finding escape conditions.
• Use orbital relations when you are told something about period, radius, or speed for circular orbits.

A frequent exam skill is not heavy computation; it is selecting a model and then executing it cleanly.

Example 1: satellite launched into circular orbit (energy plus orbit condition)

Suppose a satellite starts from rest very far away and falls toward a planet of mass M. At what speed must it be moving at radius r to be captured into a circular orbit at that radius (assuming you could somehow redirect it without changing speed)?

For a circular orbit at radius r, the speed must be

v = \sqrt{\frac{GM}{r}}

The energy perspective tells you that a falling object from infinity would have speed

v_{fall} = \sqrt{\frac{2GM}{r}}

which is larger by a factor of \sqrt{2}. To transition from a parabolic approach into a circular orbit at the same radius, you would need to remove energy (for example, with a burn or atmospheric drag).

Example 2: comparing energies of two circular orbits

A satellite moves from a circular orbit of radius r_1 to another circular orbit of radius r_2. Compare the total mechanical energies.

For a circular orbit:

E = -\frac{1}{2}G\frac{Mm}{r}

So the energy difference is

\Delta E = -\frac{1}{2}G Mm\left(\frac{1}{r_2} - \frac{1}{r_1}\right)

If r_2 > r_1, then 1/r_2 < 1/r_1, making \Delta E positive. You must add energy to raise the orbit even though the satellite’s speed decreases.

Common modeling pitfalls to watch for

• If a problem says “in orbit,” do not automatically assume “no gravity.” In orbit, gravity is essential.
• If a problem involves large altitude changes, do not default to constant g.
• If a problem involves a circular orbit, you can use the circular relations, but only if the orbit is actually circular.

Exam Focus

• Typical question patterns:
  • Multi-step prompts: find r from period, then find speed, then find energy.
  • Conceptual comparisons: “Which orbit has greater total energy?” or “Which has larger speed?”
  • Free-response derivations combining Newton’s law with centripetal motion or energy.
• Common mistakes:
  • Mixing formulas from circular orbits into non-circular motion without justification.
  • Thinking “adding energy always increases speed” (raising a circular orbit adds energy but decreases speed).
  • Sign errors in U and therefore in total energy interpretations.