Unit 2 · Topic 2.6

2.6 Gravitational Force

Every pair of masses attracts with a force $F_g = \frac{Gm_1m_2}{r^2}$ . The gravitational field, g, is that force per kilogram, about 9.8 N/kg at Earth's surface and weaker farther out. Your apparent weight is the normal force you feel, which changes when you accelerate, and inside a uniform sphere gravity grows in proportion to your distance from the center. Since the 2024 course update, gravitation is part of Unit 2; it used to be its own unit (Unit 7), so older videos and review books list it last.

Key terms

universal gravitation
gravitational field
weight
apparent weight
inertial mass
shell theorem

Newton's law of universal gravitation

Any two objects with mass pull on each other along the line between their centers:

$F_g = \frac{Gm_1m_2}{r^2}$

Here $G = 6.67 \times 10^{-11}$ N·m²/kg² is the universal gravitational constant, and r is the distance between the centers. A uniform sphere (or spherical shell) pulls on anything outside it as if all its mass were at its center.

The force follows an inverse-square law: double the distance and the force drops to 1/4; triple it and it drops to 1/9. The pull on each object is the same size, by Newton's third law, even if one is a planet and the other is a person.

Gravitational field and weight

The gravitational field at a point is the gravitational force per unit mass on an object placed there. Outside a sphere of mass M:

$g = \frac{GM}{r^2}$

Its units, N/kg, are the same as m/s², because g is also the free-fall acceleration. At Earth's surface it's about 9.8 N/kg. Near the surface, a change in height of a few kilometers barely changes r, so g is nearly constant and your weight is simply $F_g = mg$ .

Mass measured by how hard it is to accelerate (inertial mass) and mass measured by how hard gravity pulls (gravitational mass) have always been found to be equal. That's why all objects fall with the same acceleration in a vacuum.

Apparent weight

Your apparent weight is what a scale reads: the normal force pushing up on you. In an elevator with acceleration a (up positive), Newton's second law gives $F_N - mg = ma$ , so $F_N = m(g + a)$ .

Accelerating upward (starting up, or slowing while going down), you feel heavier. Accelerating downward, you feel lighter. In free fall, a = −g, the scale reads zero, and you feel weightless. Astronauts in orbit feel weightless for this reason, not because gravity is missing; at the space station's height, g is still about 90% of its surface value.

The equivalence principle says you can't tell the difference from the inside. Sealed in a windowless rocket accelerating at 9.8 m/s² in deep space, you'd feel exactly the same as standing at rest on Earth, and no experiment inside the rocket could tell which it was.

Gravity inside a uniform sphere

Inside a uniform spherical shell, the pulls from all parts of the shell cancel exactly. So at distance r from the center of a uniform solid sphere (r < R), only the mass inside radius r pulls on you. That mass is $M\frac{r^3}{R^3}$ , so:

$g(r) = \frac{GMr}{R^3} \quad (r \le R)$

The field grows in proportion to r inside, reaches its biggest value at the surface, then falls off as $\frac{1}{r^2}$ outside. You need to use this result but not prove the shell theorem.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Gravity at an altitude of one Earth radius
Earth has mass $5.97 \times 10^{24}$ kg and radius $6.37 \times 10^6$ m. Find g at a height of 6370 km above the surface, and the weight there of a 100 kg satellite.
Show the solutionHide the solution
1. Step 1: The distance from Earth's center is r = 2R, twice the surface distance.
2. Step 2: $g = \frac{GM}{(2R)^2} = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(1.274 \times 10^7)^2} \approx 2.45$ N/kg.
3. Step 3: Shortcut: doubling r cuts g to 1/4 of 9.8, which is 2.45 N/kg. ✓
4. Step 4: Weight: $(100)(2.45) \approx 245$ N.
Answer: g ≈ 2.45 N/kg; weight ≈ 245 N.
Example 2Calculator allowed
A scale in an elevator
A 60 kg student stands on a scale in an elevator. What does the scale read when the elevator (a) accelerates upward at 2.0 m/s² and (b) moves upward while slowing down at 2.0 m/s²? Use g = 9.8 m/s².
Show the solutionHide the solution
1. Step 1: The scale reads the normal force: $F_N = m(g + a)$ with up positive.
2. Step 2: (a) a = +2.0 m/s²: $F_N = 60(11.8) = 708$ N.
3. Step 3: (b) Moving up but slowing means the acceleration points down: a = −2.0 m/s². $F_N = 60(7.8) = 468$ N.
4. Step 4: The trap in (b) is using +2.0 because the elevator moves up. The direction of acceleration, not of motion, sets the reading.
Answer: (a) 708 N; (b) 468 N (the student's true weight is 588 N).
Example 3Calculator allowed
Halfway to the center
Treat Earth as a uniform sphere with surface gravity 9.8 N/kg. What is g halfway from the surface to the center? What would it be at the same distance, R/2, from the center of a planet with Earth's mass compressed into a sphere of radius R/2?
Show the solutionHide the solution
1. Step 1: Inside a uniform sphere, g is proportional to r: at r = R/2, $g = \frac{1}{2}(9.8) = 4.9$ N/kg.
2. Step 2: For the compressed planet, r = R/2 is at its surface, so all the mass is inside: $g = \frac{GM}{(R/2)^2} = 4(9.8) = 39.2$ N/kg.
3. Step 3: Same distance from the center, very different fields, because different amounts of mass are inside that radius.
Answer: 4.9 N/kg inside Earth; 39.2 N/kg at the surface of the compressed planet.

Common mistakes

Measuring r from the surface instead of from the center. r is center to center.
Using $\frac{1}{r^2}$ inside a planet. Inside a uniform sphere, g is proportional to r.
Saying astronauts are weightless because there's no gravity in orbit. They're in free fall.
Using the direction of motion to decide whether you feel heavier. Only the direction of acceleration matters.

On the exam

Ratio questions are common: how does the force change if one mass doubles and the distance triples? Write the new force over the old one and cancel.
Expect a sketch of g against r for a uniform planet: a straight line up from zero at the center, a peak at the surface, then an inverse-square drop.

Connected topics

Videos

Gravitational forces and fields | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Topic 2.6 - Gravitational Force
Lessons With LondotWatch on YouTube (opens in a new tab)
Newton's Universal Law of Gravitation Introduction (The Big G Equation)
Flipping PhysicsWatch on YouTube (opens in a new tab)
AP Physics C Mechanics - Unit 2 - Lesson 18C - Gravity within Planets (Shells)
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
AP Physics C - Gravity
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
Weight, apparent weight, and weightlessness | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 2.6 Gravitational Force. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

Two spheres attract each other with a gravitational force F. If the distance between their centers is doubled, what is the new force?

Question 2 of 4Calculator allowed

The gravitational field at Earth's surface is about 9.8 N/kg. What is it at a height above the surface equal to Earth's radius?

Question 3 of 4Calculator allowed

A planet is a uniform solid sphere with gravitational field g₀ at its surface. Imagine a narrow tunnel to its center. What is the gravitational field halfway from the center to the surface?

Question 4 of 4Calculator allowed

Planet X has twice Earth's mass and twice Earth's radius. How does the gravitational field at its surface compare with Earth's?

0 of 4 answered