Question 1

What is Newton's law of universal gravitation and how does it work?

Accepted Answer

Newton’s law of universal gravitation: every two masses attract with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers of mass: |Fg| = G m1 m2 / r^2 (G ≈ 6.67×10^−11 N·m^2/kg^2). The force is always attractive and acts along the line joining the centers. You can think in two equivalent ways: - Force form: use Fg to find interactions between two bodies. - Field form: the gravitational field (strength) produced by mass M at distance r is g = G M / r^2 (units N/kg). If gravity is the only force, an object’s acceleration = g. Near Earth’s surface g ≈ 10 N/kg, so weight = mg. Useful AP points: you should be able to apply the inverse-square form, treat forces as acting at centers of mass, and use g ≈ 10 N/kg when gravity is effectively constant (CED 2.6.A–B). For a concise topic review, see the Fiveable study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps).

Question 2

Why is gravitational force always attractive and never repulsive?

Accepted Answer

Because of how mass appears in Newton’s law and in our observations: Newton’s law of universal gravitation (|Fg| = G m1 m2 / r^2) gives a magnitude that’s proportional to the product m1·m2, which is always positive for ordinary (positive) mass. The force vector points along the line joining the centers and is defined toward each other, so the interaction pulls masses together—that’s 2.6.A.1.i in the CED. In field language (2.6.A.2), the gravitational field g created by a mass points toward that mass, so a test mass feels acceleration toward the source (always attractive). We’ve never measured “negative mass” that would produce repulsion; all experimental tests treat gravitational mass as positive and equal to inertial mass (2.6.D.3). On a deeper level, general relativity explains gravity as mass–energy curving spacetime so free-falling objects follow paths that converge (also an attractive effect), but for AP Physics 1 you should rely on Newton’s law and the fact gravitational force is always attractive (see the Topic 2.6 study guide for review: https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps). For the exam, be ready to state that gravitational force is attractive and acts along the line between centers of mass (CED 2.6.A.1.i–ii).

Question 3

What's the difference between gravitational mass and inertial mass?

Accepted Answer

Inertial mass and gravitational mass are two ways of measuring “how much stuff” an object has, but they show up in different laws. Inertial mass is the property that resists acceleration—it’s the m in F = ma (Newton’s second law). Gravitational mass is the property that determines how strongly an object is pulled by gravity—it’s the m that appears in Newton’s law of universal gravitation (Fg = G m1 m2 / r^2) and in weight (Fg = m g). Physically they act the same in everyday experiments: inertial and gravitational mass have been measured to be equivalent (this is the equivalence principle), which is why all objects fall with the same g when gravity is the only force. Units are the same (kg), but conceptually one measures resistance to acceleration and the other measures coupling to the gravitational field. For AP review, see Topic 2.6.D in the Course and Exam Description and the Fiveable study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps).

Question 4

I'm confused about gravitational fields - can someone explain what they actually are?

Accepted Answer

Think of a gravitational field as a way to describe how mass "tells" space to pull on other masses without touching them. Instead of describing the force only when two objects interact, the field g at a point gives the gravitational force per unit mass you would feel there: g = Fg/m = G M / r^2 (direct from the CED). Its direction is toward the mass creating the field and its magnitude (in N/kg) equals the acceleration an object would have if gravity were the only force (so near Earth g ≈ 10 N/kg). Weight is just mg—the gravitational force on your mass m in that field. Fields make noncontact forces easy to map out across space, let you treat forces on “test masses” independently, and are used a lot on the exam for inverse-square problems and shell-sphere reasoning (see Topic 2.6). For a focused review, check the Topic 2.6 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps). More unit review and practice problems are at the unit page (https://library.fiveable.me/ap-physics-1-revised/unit-2) and the practice bank (https://library.fiveable.me/practice/ap-physics-1-revised).

Question 5

How do you calculate the gravitational force between two objects using that G formula?

Accepted Answer

Use Newton’s law of universal gravitation: |Fg| = G m1 m2 / r^2. Steps: - Identify m1 and m2 (in kg) and r, the distance between their centers of mass (in m). - Use G = 6.67×10^(-11) N·m^2/kg^2. - Compute F = G·m1·m2 / r^2. The force is attractive and acts along the line joining the centers (each mass feels the same magnitude toward the other). Example: two 1000-kg objects 10 m apart → F = (6.67e-11)(1000)(1000)/(10^2) ≈ 6.67e-9 N (very tiny). Remember special cases from the CED: a large spherical body can be treated as if its mass is concentrated at its center (shell theorem), so for an object near Earth’s surface weight ≈ mg with g = GM_earth / R_earth^2 ≈ 9.8–10 N/kg. This formula (and when g is effectively constant) is tested on the AP exam—review Topic 2.6 (Gravitational Force) on Fiveable (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps) and practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Question 6

What does it mean when we say gravity acts on the center of mass?

Accepted Answer

Saying “gravity acts on the center of mass” means you can treat the net gravitational force on an extended object as if a single force were applied at its center of mass. For translational motion this is exact: the object’s center of mass accelerates as if all of its mass were concentrated there and the force pointed along the line joining the two centers of mass (CED 2.6.A.1–1.iii). Practically, that lets you use Fg = G m1 m2 / r^2 or near Earth weight mg and Newton’s second law for the whole object without worrying about internal mass distribution. It doesn’t mean gravity actually only pulls at one tiny point—parts of the object feel slightly different pulls, which can cause torque or tidal effects if the field varies across the object (see shell theorem/Topic 2.6.E). For AP review, see the Topic 2.6 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps) and practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Question 7

Why can we treat Earth's gravity as constant near the surface when the formula has distance squared in it?

Accepted Answer

Good question—the inverse-square law still holds, but "near the surface" the distance change is tiny compared to Earth's radius. Gravity at distance r is g(r) = GM/r^2. If you move up a height h above the surface, g(R + h) ≈ g(R)·(1 − 2h/R) (from a first-order expansion). Since Rearth ≈ 6.37×10^6 m, even h = 10 km gives only ~0.3% change; h = 100 m gives ~0.003% change. That’s why AP Physics treats g as constant (g ≈ 9.8 ≈ 10 N/kg) for motions near Earth’s surface—the change in g is negligible over typical classroom/planetary heights (CED 2.6.B.1 and 2.6.A.2). For problems involving large altitude changes or orbital motion you must use GM/r^2 instead. Want extra practice or a quick refresher on this topic? Check the Topic 2.6 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps) or try related practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Question 8

What's the difference between weight and apparent weight?

Accepted Answer

Weight is the gravitational force Earth (or another body) exerts on you: Fg = mg (CED 2.6.A.3). Apparent weight is what a scale reads—i.e., the magnitude of the normal force exerted on you (CED 2.6.C.1). They’re equal only when gravity is the only vertical force and you’re not accelerating (e.g., standing still or moving at constant velocity near Earth: N = mg). If you accelerate, apparent weight ≠ mg: in an elevator with upward acceleration a, N = m(g + a); with downward acceleration a, N = m(g − a). In free fall N = 0, so you feel weightless even though gravity still acts (CED 2.6.C.3). The equivalence principle notes an observer in a noninertial frame can’t locally distinguish between apparent weight and gravitational force (CED 2.6.C.4). For a quick review, see the Topic 2.6 study guide on Fiveable (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps) and try practice problems at (https://library.fiveable.me/practice/ap-physics-1-revised).

Question 9

I don't understand why astronauts are weightless in space if gravity is still acting on them.

Accepted Answer

Gravity is still pulling on astronauts—that pull (their weight) doesn’t vanish in orbit. “Weightless” means their apparent weight (the normal force you feel) is zero. In orbit both the spacecraft and astronauts are in free fall: gravity provides the centripetal force that keeps them falling around Earth, but nothing pushes up on their bodies, so the normal force is zero and they float. This matches CED 2.6.C: apparent weight = normal force, and a system appears weightless when gravity is the only force acting. Quick numbers: at 400 km up, g ≈ 8.7 m/s² (not zero), but both astronaut and ship accelerate together toward Earth, so no contact forces act between them. Small residual accelerations (tides, thruster firings) cause tiny motions—that’s why we call it microgravity, not “no gravity.” For AP review see Topic 2.6 (gravitational force) on Fiveable (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps) and try practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Question 10

How does Newton's shell theorem work and why don't you feel gravity inside a hollow sphere?

Accepted Answer

Newton’s shell theorem says two handy things: for a thin, uniform spherical shell, (1) any mass outside the shell feels the shell’s gravity as if all the shell’s mass were concentrated at the center (so F = G Mshell m / r^2), and (2) any mass located anywhere inside that thin shell feels zero net gravitational force from the shell. Physically this works because every small patch of mass on the shell pulls you one way, and there's an exactly cancelling patch on the opposite side—when you integrate all contributions the vector sum is zero. That’s why you don’t “feel” gravity inside a hollow sphere: contributions from the whole shell cancel. For a solid uniform sphere, only the mass at radii smaller than your distance from the center contributes; inside the sphere the net force grows linearly with r (Fg ∝ r), so at the center it’s zero. (CED students aren’t expected to derive the theorem mathematically.) For more review, see the Topic 2.6 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps) and unit resources (https://library.fiveable.me/ap-physics-1-revised/unit-2). Want extra practice? Check the AP Physics 1 practice set (https://library.fiveable.me/practice/ap-physics-1-revised).

Question 11

What happens to gravitational force when you're inside a planet instead of on its surface?

Accepted Answer

If you go inside a planet (modeled as a uniform sphere), gravity gets weaker the closer you get to the center—not stronger. Two key CED facts to use: Newton’s shell theorem and the inverse-square law. A thin spherical shell produces zero net gravitational force inside it (2.6.E.2.i). So only the mass at radii ≤ your distance r from the center contributes. For uniform density ρ that partial mass is M_partial = ρ(4/3)πr^3, so the field g(r)=G M_partial / r^2 = G(4/3)πρ r. Thus g (and your weight F = m g) is proportional to r and goes to zero at the center. Outside the planet you see the usual inverse-square behavior (g ∝ 1/r^2). Useful CED refs: shell theorem and the inside-sphere result (2.6.E.2–3). For practice, check the Topic 2.6 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps) and more problems at (https://library.fiveable.me/practice/ap-physics-1-revised).

Question 12

Why do we use g = 10 m/s² instead of 9.8 m/s² in AP Physics problems?

Accepted Answer

Short answer: AP problems usually use g = 10 m/s² as a convenient approximation so you can do algebra and mental math quickly without losing the physics. The CED even states that near Earth’s surface g ≈ 10 N/kg (Topic 2.6.B). Using 10 instead of 9.8 is fine when the question is about conceptual relationships, kinematics, or forces and the expected precision is low. Why that’s okay: - It keeps calculations simple (e.g., mg → 10m) and matches AP scoring expectations for routine problems. - AP often tests understanding of functional dependence (mg, g in free fall, weight vs apparent weight) rather than high-precision numerics. When to use 9.8 (or more precise values): if the problem asks for a precise numerical value, involves gravitational-field calculations using G and distances, or when small differences matter (orbital mechanics, measured g experiments). For more review and examples, see the Topic 2.6 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps) and unit resources (https://library.fiveable.me/ap-physics-1-revised/unit-2).

Question 13

Can someone explain the equivalence principle in simple terms?

Accepted Answer

Think of the equivalence principle like this: inside a small sealed room you can’t tell whether the force you feel is due to gravity or due to acceleration. If the room sits on Earth you feel a downward gravitational force mg and the floor pushes up with a normal force N = mg. If the room is in deep space but accelerating upward at g, you feel the same push from the floor (N = ma = mg). So an observer inside can’t distinguish “real” gravity from a matching acceleration of their frame. In AP terms this links apparent weight (the normal force) and gravitational force (mg)—they’re only equal when the frame is inertial or not accelerating (CED 2.6.C). It also connects to inertial vs gravitational mass: experiments show they’re equivalent, which is why gravitational acceleration acts the same on all masses (CED 2.6.D, 2.6.A). For a short review, see the Topic 2.6 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps). For more practice, Fiveable’s unit resources and practice problems can help (https://library.fiveable.me/ap-physics-1-revised/unit-2, https://library.fiveable.me/practice/ap-physics-1-revised).

Question 14

How do you know when to use the full gravitational force equation versus just F = mg?

Accepted Answer

Use the full law |Fg| = G m1 m2 / r^2 whenever the distance between the two masses or the mass distribution matters—e.g., gravity between planets, satellites, or when you’re not very close to a planet’s surface. That formula comes from Newton’s universal gravitation and the inverse-square dependence (CED 2.6.A.1). Use F = mg (weight) when one mass is much larger (an astronomical body) and you’re near its surface so g ≈ constant (CED 2.6.A.2, 2.6.B). Near Earth’s surface g ≈ 9.8 ≈ 10 N/kg and the field doesn’t change much over typical lab/household distances, so weight = m g is fine (CED 2.6.A.3, 2.6.B.1–2). Quick checks: - If r changes significantly (orbital altitude, interplanetary), use G m1 m2 / r^2. - If object is “near” a planet’s surface and g can be treated constant, use F = m g. - For spherical bodies, apply shell theorem to treat mass as concentrated at center when object is outside the sphere (CED 2.6.E). For a focused review, see the Topic 2.6 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps) and try practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Question 15

What's a gravitational field and how is it different from gravitational force?

Accepted Answer

A gravitational field is a way to describe how a massive object (like Earth) affects space around it—it assigns a vector g at every point that tells you the gravitational force per unit mass there. Gravitational force is what you get when you multiply that field by an object's mass: Fg = m g. More generally, Newton’s law gives the force between two masses: |Fg| = G m1 m2 / r^2, while the field from mass M is |g| = G M / r^2 (so |Fg| = m · |g|). The field is a property of the source mass and location; force depends on the test mass you place in the field. If gravity is the only force, the object’s acceleration equals g (g in N/kg numerically equals m/s^2). For AP review, see Topic 2.6 in the CED and the Fiveable study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/6-gravitational-force/study-guide/Xtm92y3jgBBJXDps).