Mistake Master
Gravitational force
Topic 2.5 named the rule that ties net force to acceleration. This topic names a particular force: gravity. Two masses attract along the line connecting their centers, with size proportional to each mass and inverse-square in the distance. Near Earth's surface that collapses to the simple $\vec F_g = m \vec g$, but the simple form has limits, and missing them costs points.
§1
What gravity is, and isn't.
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Every two masses attract each other. The size follows Newton's law of universal gravitation:
$$F_g = \dfrac{G\, m_1 m_2}{r^2}$$
where $G$ is a universal constant ($6.67 \times 10^{-11}$ N$\cdot$m$^2$/kg$^2$), $m_1$ and $m_2$ are the two masses, and $r$ is the distance between their centers. The equation is simple. The traps are in what it doesn't say.
It doesn't say one mass pulls and the other passively responds. The pair forces are symmetric. Each mass pulls the other with the same size $F_g$. This is a Newton's-third-law pair: equal magnitudes, opposite directions, on different objects. A 60 kg person on Earth's surface feels Earth's pull at about $600$ N. The Earth feels the person's pull at exactly the same $600$ N. The accelerations differ by a factor of $10^{23}$ because the masses do. The forces are equal.
It doesn't say gravity can repel. There is no negative gravitational mass on the AP exam. Gravity is always attractive.
It doesn't say gravity acts at some point on the surface. The force acts along the line connecting the two centers of mass, on each system's center of mass. For an extended object, you treat the gravitational force as if it were applied at the COM. (This is where Topic 2.1 pays off.)
The size of $r$ matters more than either mass; $r$ is squared in the denominator. The next section unpacks that.
§2
The inverse square.
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The form of the equation matters. The two masses sit in the numerator, multiplying. The distance sits in the denominator, squared. That last fact does most of the heavy lifting, and produces most of the wrong answers.
The most common mistake is the "linear in distance" instinct. Students see "double the distance" and write "halve the force." That's how inverse linear works ($1/r$), not inverse square ($1/r^2$). Inverse square punishes distance harder. Doubling $r$ quarters $F_g$. Tripling $r$ ninths it. Quadrupling $r$ sixteenths it.
The distance $r$ is the center-to-center distance between the two masses, not the surface-to-surface distance. For a person standing on Earth's surface, $r$ is the radius of the Earth ($\approx 6.37 \times 10^6$ m), not zero, not the person's height, not the distance from their head to their feet. The "distance to the source" of gravity is a distance to the source mass's center.
$F_g$ is inverse-square in $r$. Double $r$, quarter $F_g$. Triple $r$, ninth $F_g$. And $r$ is always center-to-center.
§3
The field model: $\vec F_g = m \vec g$.
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Inverse-square is right for two distant masses. Near the surface of one big mass, factor it.
The gravitational force on a small object of mass $m$, near a much larger mass $M$, is
$$\vec F_g = m \vec g, \qquad |\vec g| = \dfrac{G M}{r^2}.$$
The right-hand piece, $\vec g$, has nothing to do with the small mass. It depends only on $M$ (the source) and $r$ (where you are). It's a property of the source mass at every point in space: the gravitational field. Put any test mass $m$ in that field and the force on it is $m$ times the field there.
Units check: $[\,\vec g\,] = $ N/kg $= $ (kg$\cdot$m/s$^2$)/kg $= $ m/s$^2$. The field strength in N/kg is equivalent to an acceleration in m/s$^2$. When gravity is the only force, $\vec a = \vec F_g/m = \vec g$. The free-fall acceleration in m/s$^2$ numerically equals the field strength in N/kg. That's why $g \approx 10$ shows up in both contexts.
Near the surface of the Earth, $\vec g$ has size about $9.8$ N/kg (rounded to $10$ for AP problems) and points toward Earth's center. Stand on the Moon and $\vec g$ is smaller (about $1.6$ N/kg). On Mars, smaller still ($3.7$). On Jupiter, larger ($25$). The same object has the same mass everywhere, but the field it lives in varies by body.
Inertial mass and gravitational mass. The $m$ in $\vec F = m \vec a$ is the object's inertial mass: how hard it is to accelerate. The $m$ in $\vec F_g = m \vec g$ is the object's gravitational mass: how strongly it couples to a gravitational field. These are different in principle, but experiments show they're identical to a part in $10^{15}$. That's why every object falls at the same rate in vacuum, no matter its mass: the $m$ in $F_g = mg$ cancels exactly with the $m$ in $F = ma$. Heavier and lighter objects fall together. (This seeds general relativity. For AP, just take the equality as a given.)
§4
Worked example: weight on another planet.
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An astronaut of mass $m$ has weight $W_E$ on Earth, where $W_E = m g_E$ and $g_E \approx 10$ N/kg. She visits a planet whose mass is $\alpha$ times Earth's mass and whose radius is $\beta$ times Earth's radius. What does she weigh on the new planet?
The field strength at the surface of any planet is $g = G M / R^2$, where $M$ and $R$ are the planet's mass and radius. For the new planet, $M_p = \alpha M_E$ and $R_p = \beta R_E$, so
$$g_p = \dfrac{G\, (\alpha M_E)}{(\beta R_E)^2} = \dfrac{\alpha}{\beta^2}\, \dfrac{G M_E}{R_E^2} = \dfrac{\alpha}{\beta^2}\, g_E.$$
The astronaut's mass doesn't change. Mass is intrinsic. The new weight is
$$W_p = m g_p = m \cdot \dfrac{\alpha}{\beta^2}\, g_E = \dfrac{\alpha}{\beta^2}\, W_E.$$
If $\alpha = \beta = 1$ (back on Earth), $W_p = W_E$. If $\alpha = 1$ and $\beta = 2$ (Earth-mass planet, twice the radius), $W_p = W_E / 4$: pure inverse-square at fixed mass. If $\beta = 1$ and $\alpha = 2$ (same radius, double the mass), $W_p = 2 W_E$: linear in mass at fixed radius. For Mars-like values ($\alpha \approx 1/9$, $\beta \approx 1/2$), $W_p = (1/9)/(1/4)\, W_E = (4/9) W_E \approx 0.44 W_E$, which is in the ballpark of Mars's actual $0.38 W_E$.
Mass doesn't depend on which planet you're on. Weight does. Mass is intrinsic; weight is what the local field does to that mass.
§5
When can gravity be treated as constant?
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The factored form $\vec F_g = m \vec g$ assumes $\vec g$ is the same at every point of the object's trajectory. That can't be exactly true. As the object moves, $r$ changes, so $g$ changes. So when is $g$ approximately constant?
Answer: when $r$ doesn't change much. (Inverse-square means a small fractional change in $r$ doubles into the change in $g$.) In practice, compute $g$ at two altitudes and compare.
Two cases.
Earth's radius is $R_E \approx 6.37 \times 10^6$ m. The ball's $r$ changes from $R_E$ at the ground to $R_E + 30$ m at the top. The fractional change is
$$\dfrac{\Delta r}{r} = \dfrac{30}{6.37 \times 10^6} \approx 5 \times 10^{-6}.$$
The corresponding fractional change in $g$ is about $10^{-5}$, or roughly one part in a hundred thousand. Far below any precision you'd care about in a kinematics problem. Treat $\vec g$ as constant.
The ISS orbits at $400$ km above Earth's surface. Its $r$ is
$$r_{\text{ISS}} = R_E + 400 \text{ km} \approx 6.37 \times 10^6 + 4 \times 10^5 = 6.77 \times 10^6 \text{ m},$$
so $\Delta r / r \approx 0.06$, about $6\%$. The corresponding $g$ at the ISS is
$$g_{\text{ISS}} = g_E \left( \dfrac{R_E}{r_{\text{ISS}}} \right)^2 \approx (10) (0.94)^2 \approx 8.8 \text{ N/kg},$$
about $88\%$ of surface $g$. The constant approximation breaks at orbital altitudes.
Rule of thumb. If your problem stays within a few kilometers of Earth's surface, treat $\vec g$ as constant at $\approx 10$ N/kg pointing toward Earth's center. If the problem reaches orbital altitudes, you need the inverse-square form.
§6
Worked example: apparent weight.
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"Weight" gets used two different ways. The strict definition: weight is the gravitational force on you, $W = mg$. The colloquial use: weight is what the bathroom scale reads. These agree when the scale isn't accelerating. They disagree when it is.
What the scale actually reads is the normal force pushing back on you. Call that the apparent weight. It can be larger than $mg$, smaller than $mg$, or even zero, depending on how the scale (and you) are accelerating.
Three cases for a person of mass $m$ on a scale, with $+y$ chosen upward.
Acceleration is zero. Newton's second law along $y$:
$$F_N - mg = 0 \quad \Longrightarrow \quad F_N = mg.$$
Apparent weight equals actual weight. The familiar case.
Acceleration is $+a$ (upward). Newton's second law:
$$F_N - mg = ma \quad \Longrightarrow \quad F_N = m(g + a).$$
Apparent weight exceeds actual weight. You feel heavier.
Acceleration is $-g$ (downward at the full $g$). Newton's second law:
$$F_N - mg = m(-g) \quad \Longrightarrow \quad F_N = 0.$$
Apparent weight is zero. The scale reads nothing. You feel weightless. Gravity, $W = mg$, hasn't changed. The contact force has.
Apparent weightlessness on the ISS. Astronauts on the International Space Station "float," but not because gravity is absent. At ISS altitude $g$ is still about $90\%$ of surface $g$ (Section 5). They appear weightless because they and the ISS are in continuous free fall around Earth. The ISS doesn't crash because it's moving sideways fast enough to keep missing Earth as it falls. The astronaut and the station fall together. No contact force between them, so any scale reads zero.
The equivalence principle. The free-fall case sets up a deeper observation, sometimes called Einstein's "happiest thought." An observer in a freely falling elevator can't tell, by any local experiment, whether they're actually falling toward Earth or drifting in deep space. Flip it: an observer in an upward-accelerating spaceship in deep space can't tell whether they're accelerating or sitting on a planet with matching surface gravity. Acceleration and gravity are locally indistinguishable. (This motivates general relativity. AP Physics 1 just asks you to know it.)
Apparent weight is the size of the normal force, $F_N$. It equals $mg$ only when the system isn't accelerating along the direction of gravity. Accelerate up, $F_N$ grows. Accelerate down, $F_N$ shrinks. Accelerate down at the full $g$, $F_N = 0$.
§7
Three mistakes that cost real points.
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Three failure modes show up over and over on gravitational-force problems. Each one has a family of distractors built around it on the AP exam.
"Mass and weight are the same thing."
Everyday English uses "weight" for what's really mass. The conflation gets baked in early and is hard to shake.
The AP exam tests it directly. The trap is usually set on a different planet. "An astronaut weighs $800$ N on Earth. What is her mass on Mars?" The wrong-but-tempting answer is "still $800$ N"; the right answer is "$80$ kg, same as on Earth." Mass is intrinsic. Weight is a force that depends on the local $g$.
Fix. Mass is in kilograms and never changes when you travel. Weight is in newtons and changes with $g$. If a problem gives you a weight, divide by $g$ to recover the mass. The mass is the part you'd carry to Mars. The spring-scale reading isn't.
"Double the distance, halve the force."
Inverse linear, not inverse square. It's an instinct that treats distance like a dial: turn it up by $2$, response down by $2$. Gravity doesn't work that way. The denominator is squared, so the response goes down by $4$.
The rule, generalized: scale $r$ by a factor of $k$, and $F_g$ scales by $1/k^2$. Double $r$ ($k = 2$): $F_g$ goes to a quarter. Triple $r$ ($k = 3$): $F_g$ goes to a ninth. Halve $r$ ($k = 1/2$): $F_g$ goes to four times. The squaring is what makes orbits work; inverse-linear gravity wouldn't give closed orbits.
Fix. Change the distance? Square the factor and divide. Change a mass? Just multiply. Build the ratio $F_{\text{new}}/F_{\text{old}}$ from three factors (one for each mass, one squared for the distance), and you'll never miss this trap.
"The scale always reads $mg$."
True only when the scale (and you) aren't accelerating along the direction of gravity. In an elevator, on a banked surface, in free fall, in centripetal motion, the scale reads $F_N$, not $mg$. The two coincide only when $\sum F_y = 0$.
ISS astronauts appear weightless. A student writing $F_N = mg$ reflexively concludes "$g$ at the ISS must be near zero," which is off by a factor of $10$: $g_{\text{ISS}} \approx 9$ N/kg (Section 5). The right call: $F_N = 0$ because the ISS is in free fall, not because $g$ is.
Fix. Don't quote $F_N = mg$ as a rule. Write Newton's second law along the gravity-aligned axis for the specific situation, then solve for $F_N$. At rest: $F_N = mg$. Elevator accelerating up: $F_N = m(g + a)$. Free fall: $F_N = 0$. Same procedure each time.