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United StatesPhysics C: MechanicsSyllabus dot point

How does Newton's law of universal gravitation describe the attraction between masses, and how does it give the gravitational field and weight near a planet?

Topic 2.6 Gravitational Force: apply Newton's law of universal gravitation, define the gravitational field strength, relate it to weight, and treat gravity inside and outside a spherical mass.

A focused answer to AP Physics C: Mechanics Topic 2.6, covering Newton's law of universal gravitation and its inverse-square character, the gravitational field strength and its relation to weight, the field outside and inside a uniform sphere, and apparent weightlessness, with calculus-aware worked examples.

Generated by Claude Opus 4.811 min answer

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  1. What this topic is asking
  2. Newton's law of universal gravitation
  3. The gravitational field and weight
  4. Gravity inside a uniform sphere
  5. Try this

What this topic is asking

The College Board (Topic 2.6) wants you to apply Newton's law of universal gravitation, recognize its inverse-square character, define the gravitational field strength gg, relate it to weight, and handle gravity both outside and inside a spherical mass. The 2024-25 revision sharpened the expectation that you can treat the field within an extended body, which is a distinctive calculus-flavoured part of AP Physics C.

Newton's law of universal gravitation

The defining feature is the inverse-square dependence: tripling the separation cuts the force to one ninth. The force is always attractive and acts along the line of centers. A useful theorem (which Newton proved with calculus) is that a uniform spherical mass attracts external objects exactly as if all its mass were concentrated at its center, so you use the distance to the center, not to the surface.

The gravitational field and weight

It is often cleaner to think of one mass creating a gravitational field that the other responds to. The field strength at distance rr from a mass MM is the force per unit test mass:

g=Fm=GMr2.g = \frac{F}{m} = \frac{GM}{r^2}.

Near a planet's surface (r=Rr = R) this gives the familiar g=GM/R2g = GM/R^2, about 9.89.8 m/s squared on Earth, and the weight of a mass mm is W=mgW = mg. Because gg depends on MM and rr, weight varies from planet to planet and falls with altitude, while the mass itself is unchanged. The field is a vector pointing toward the attracting mass, and its units (m/s squared, equivalently N/kg) match an acceleration: a freely falling object accelerates at exactly gg.

Gravity inside a uniform sphere

A hallmark AP Physics C result concerns the field inside a uniform sphere. By the shell theorem, the mass in shells outside your radius rr contributes nothing; only the mass enclosed within radius rr pulls you. For uniform density the enclosed mass is Menc=Mr3R3M_{enc} = M\dfrac{r^3}{R^3}, so the field inside is

g(r)=GMencr2=GMR3r(rR),g(r) = \frac{GM_{enc}}{r^2} = \frac{GM}{R^3}\,r \quad (r \le R),

which grows linearly from zero at the center to GM/R2GM/R^2 at the surface. Outside, it falls off as 1/r21/r^2. The full graph of gg versus rr therefore rises as a straight line up to the surface and then decays as an inverse square, a shape the exam asks you to sketch.

Try this

Q1. The Moon's mass is about 1/811/81 of Earth's and its radius about 1/3.71/3.7. Estimate the Moon's surface gg as a fraction of Earth's. [3 points]

  • Cue. gM/R2g \propto M/R^2, so gMoongEarth=1/81(1/3.7)2=1/811/13.7=13.7810.17\dfrac{g_{Moon}}{g_{Earth}} = \dfrac{1/81}{(1/3.7)^2} = \dfrac{1/81}{1/13.7} = \dfrac{13.7}{81} \approx 0.17, about one sixth.

Q2. Explain why an astronaut in a low orbit feels weightless even though gravity there is nearly as strong as at the surface. [2 points]

  • Cue. The astronaut and station are in free fall together, accelerating at the local gg, so there is no normal force; gravity still acts strongly.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)5 marksSection II (FRQ). A planet has mass MM and radius RR. (a) Derive an expression for the gravitational field strength gg at its surface. (b) Determine gg at a height h=Rh = R above the surface, as a fraction of the surface value. (c) For a uniform-density planet, derive the field strength at a distance r<Rr < R from the center, and sketch gg versus rr from the center outward.
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A 5-point gravitation FRQ including the field inside a sphere.

(a) Surface field (1 point): a mass mm at the surface feels F=GMmR2F = \dfrac{GMm}{R^2}, and g=F/m=GMR2g = F/m = \dfrac{GM}{R^2}.
(b) At h=Rh = R (2 points): the distance from the center is 2R2R, so g=GM(2R)2=GM4R2=g4g' = \dfrac{GM}{(2R)^2} = \dfrac{GM}{4R^2} = \dfrac{g}{4}, one quarter of the surface value.
(c) Inside (2 points): only the mass within radius rr attracts. For uniform density, that enclosed mass is Menc=Mr3R3M_{enc} = M\dfrac{r^3}{R^3}, so g(r)=GMencr2=GMR3rg(r) = \dfrac{GM_{enc}}{r^2} = \dfrac{GM}{R^3}r, which grows linearly from 00 at the center to GM/R2GM/R^2 at the surface, then falls off as 1/r21/r^2 outside.

Markers reward using only the enclosed mass inside the sphere and getting the linear-then-inverse-square shape.

AP 2021 (style)1 marksSection I (multiple choice). A satellite orbits at a height equal to the Earth's radius above the surface (so r=2Rr = 2R from the center). The gravitational field there compared with the surface value is... (A) the same (B) one half (C) one quarter (D) one eighth. Justify your reasoning.
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A 1-point inverse-square MCQ. The answer is (C).

The field outside a planet follows g=GM/r2g = GM/r^2. Going from r=Rr = R to r=2Rr = 2R multiplies r2r^2 by four, so the field drops to one quarter. The trap (B) treats the dependence as 1/r1/r rather than 1/r21/r^2.

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