Gravitational Force and Weight

AP Physics 1· AP Physics 1 CED — Dynamics· 14 min read

1. Core Definitions★★☆☆☆⏱ 3 min

Gravitational force is the fundamental attractive force between any two objects with mass, one of the four fundamental forces of nature. Weight is a specific case of gravitational force: it is the gravitational force exerted on an object by a much larger nearby body (like Earth or the Moon), so it is not an intrinsic property of the object.

📘 Definition

Mass vs Weight

mass = $m$ , weight = $W$ or $F_{g}$

Mass is an intrinsic measure of an object's inertia, with units of kilograms, and is constant regardless of location. Weight is a gravitational force, with units of newtons, and changes with the local gravitational field.

Example:

A 60 kg object has the same mass on Earth and the Moon, but 1/6 the weight on the Moon.

2. Newton’s Law of Universal Gravitation★★★☆☆⏱ 4 min

Newton’s law of universal gravitation describes the gravitational force between any two point masses (or spherical masses, where all mass can be treated as concentrated at the object’s center of mass). The magnitude of the attractive force is proportional to the product of the two masses and inversely proportional to the square of the distance between their centers of mass.

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

Where $G = 6.67 \times 1 0^{- 11} N \cdotp m^{2} / kg^{2}$ is the universal gravitational constant, $m_{1}$ and $m_{2}$ are the masses of the two interacting objects, and $r$ is the distance between the centers of mass of the two objects. Gravitational force is always attractive, follows an inverse-square law, and obeys Newton’s third law: the force each mass exerts on the other is equal in magnitude and opposite in direction. Most AP Physics 1 questions on this topic use proportional reasoning, so you will rarely need to plug in the value of $G$ .

📐 Worked Example

Object A (mass $m$ ) and Object B (mass $2 m$ ) are separated by distance $d$ , and exert a gravitational force of magnitude $F$ on each other. If the mass of A is doubled to $2 m$ , the mass of B is tripled to $6 m$ , and the separation distance is increased to $2 d$ , what is the new gravitational force between the objects?

1
Write the original force equation to relate $F$ to the given constants:
$F = G \frac{(m)(2m)}{d^2} = G \frac{2m^2}{d^2} \implies \frac{G m^2}{d^2} = \frac{F}{2}$
2
Substitute the new values into the universal gravitation formula:
$F_{\text{new}} = G \frac{(2m)(6m)}{(2d)^2} = G \frac{12m^2}{4d^2} = G \frac{3m^2}{d^2}$
3
Substitute the relationship from the original force to get the new force in terms of $F$ :
$F_{\text{new}} = 3\left(\frac{F}{2}\right) = \frac{3}{2}F$
4
Confirm the inverse-square term was correctly applied: doubling the distance squares the denominator, giving a factor of 1/4, which matches the calculation.

Exam tip:

For proportional reasoning questions on gravitational force, always square the distance proportionality factor before substituting; it is the most frequently missed step on AP MCQs.

3. Gravitational Field and Near-Surface Weight★★☆☆☆⏱ 3 min

Near the surface of a large spherical body like Earth, the distance from a small object to the center of the large body is approximately constant, so we can simplify the universal gravitation formula to get a simple expression for weight.

🔬 Derivation

Goal:

Derive the simplified near-surface weight formula $W = m g$ from universal gravitation

Starting from:

Newton's law of universal gravitation between a large body of mass $M$ and radius $R$ , and a small object of mass $m$ near the surface

1
Substitute $r = R$ into the universal gravitation formula:
2
$F_g = G \frac{M m}{R^2} = m \left( \frac{GM}{R^2} \right)$
3
Define the local gravitational field strength $g = \frac{GM}{R ^{2}}$ , which is constant near the surface.

Result:

For objects near the surface of a large body, weight simplifies to $W = m g$ , where $g$ is the local gravitational field strength.

On Earth’s surface, $g \approx 9.8 m/s^{2} = 9.8 N/kg$ . The AP exam repeatedly tests the core distinction between mass (intrinsic inertia, constant) and weight (force, dependent on local gravity).

📐 Worked Example

A student has a mass of 65 kg on Earth, where $g_{Earth} = 9.8 N/kg$ . On the Moon, the gravitational field strength is $g_{Moon} = 1.6 N/kg$ . What are the student’s mass and weight on the Moon, respectively?

1
Recall that mass is intrinsic and does not change with location: the student’s mass remains 65 kg on the Moon.
2
Use the weight formula $W = m g$ for the Moon’s gravitational field:
$W_{\text{Moon}} = m g_{\text{Moon}} = (65 \text{ kg})(1.6 \text{ N/kg}) = 104 \text{ N}$
3
Check consistency: on Earth, the student’s weight is $(65) (9.8) = 637$ N, which is much larger, matching the Moon’s weaker gravity. Units are correct: mass in kilograms, weight in newtons.

Exam tip:

Always check units on the answer to distinguish mass vs. weight in conceptual questions: if the question asks for a force (weight), it must have units of newtons; mass is always in kilograms.

4. Apparent Weight★★★☆☆⏱ 4 min

Apparent weight is the normal force exerted on an object by a supporting surface (like a bathroom scale in an elevator), which is what the scale actually measures. It is not equal to the object’s actual weight when the object and supporting scale are accelerating vertically.

To solve apparent weight problems, draw a free-body diagram with actual weight $W = m g$ downward and normal force $n$ (apparent weight) upward. Apply Newton's second law with upward as positive:

\sum F = n - mg = ma \implies n = m(g + a)

If acceleration is upward, $a$ is positive, so apparent weight is larger than actual weight. If acceleration is downward, $a$ is negative, so apparent weight is smaller than actual weight. In free fall, $a = - g$ , so $n = 0$ , which is the experience of weightlessness.

📐 Worked Example

A 50 kg person stands on a scale in an elevator that is accelerating downward at 2.0 m/s². What is the person’s apparent weight, as measured by the scale?

1
Draw the free-body diagram: downward force is actual weight $m g$ , upward force is the normal force $n$ (the apparent weight the scale reads). Upward is positive, so acceleration $a = - 2.0 m/s^{2}$ .
2
Apply Newton's second law and rearrange to solve for $n$ :
$n - mg = ma \implies n = m(g + a)$
3
Substitute values:
$n = 50 \text{ kg} (9.8 \text{ m/s}^2 - 2.0 \text{ m/s}^2) = 50 (7.8) = 390 \text{ N}$
4
Check consistency: accelerating downward means apparent weight should be less than actual weight. Actual weight is $50 * 9.8 = 490$ N, so 390 N matches this expectation.

Exam tip:

Always assign the correct sign to acceleration based on its direction, not the direction the elevator is moving; an elevator moving downward can accelerate upward when slowing to stop, which would increase apparent weight.

5. AP-Style Concept Check★★★☆☆⏱ 2 min

✓ Quick check

Test your understanding of gravitational force net vectors:

Two identical asteroids of mass $M$ are separated by distance $r$ , and exert a gravitational force of magnitude $F$ on each other. A third asteroid of mass $M$ is placed exactly halfway between the two original asteroids, along the line connecting them. What is the magnitude of the net gravitational force on the center asteroid?
- A) $0$
- B) $\frac{2 G M ^{2}}{r ^{2}}$
- C) $\frac{4 G M ^{2}}{r ^{2}}$
- D) $\frac{8 G M ^{2}}{r ^{2}}$
Reveal answer
A) $0$ —
Each original asteroid exerts an equal magnitude, opposite direction attractive force on the center asteroid. Each force has magnitude $\frac{4 G M ^{2}}{r ^{2}}$ , so they cancel out completely for a net force of zero.

6. Common Pitfalls

Wrong move:

Using $W = m g$ with $g = 9.8 N/kg$ to calculate gravitational force between two planets separated by many Earth radii.

Why:

Students memorize the simple near-surface weight formula and forget it is only an approximation for constant distance to the large body's center.

Correct move:

Always use Newton’s universal gravitation $F_{g} = G m_{1} m_{2} / r^{2}$ for problems where the distance between objects is large relative to the radius of the larger body.

Wrong move:

When distance between two masses doubles, dividing the original force by 2 instead of 4 in proportional reasoning problems.

Why:

Students remember force depends inversely on distance, but miss the squared term in the inverse-square law.

Correct move:

Always write out the full proportionality $F_{g} \propto 1/ r^{2}$ before substituting the distance factor, and square the factor first.

Wrong move:

Claiming an object’s mass changes when it moves to the Moon, or that weight is constant regardless of location.

Why:

Students mix up the definitions of mass (inertia) and weight (gravitational force).

Correct move:

Always default to: mass is intrinsic and constant, weight is a force that changes with local gravitational field.

Wrong move:

Claiming a larger mass exerts a larger gravitational force on a smaller mass than vice versa.

Why:

Students see the larger mass produces a larger acceleration on the smaller mass, so incorrectly assume the force is also larger.

Correct move:

Always remember gravitational force between two masses is equal in magnitude on both masses, per Newton’s third law.

Wrong move:

Taking acceleration as positive when it is downward, leading to the wrong conclusion that downward acceleration increases apparent weight.

Why:

Students forget to align acceleration sign to their chosen coordinate system.

Correct move:

Always use upward = positive for vertical acceleration problems, and assign sign based on acceleration direction, not motion direction.

7. Quick Reference Cheatsheet

Category	Formula/Definition	Key Notes
Universal Gravitational Force	$F_{g} = G \frac{m _{1} m _{2}}{r ^{2}}$	Applies to any two masses; $r$ = distance between centers of mass; force always attractive
Universal Gravitational Constant	$G = 6.67 \times 1 0^{- 11} N \cdotp m^{2} / kg^{2}$	Constant across the entire universe
Weight (near large body surface)	$W = m g$	Only valid near the surface of a large body; $g$ = local gravitational field
Gravitational Field Strength	$g = \frac{GM}{R ^{2}}$	$M$ = large body mass, $R$ = large body radius; units: $N/kg = m/s^{2}$
Apparent Weight	$n = m (g + a)$	Upward = positive; $a$ = acceleration of supporting surface; $n$ = scale reading
Mass vs Weight	Mass: $m$ (kg), Weight: $W$ (N)	Mass is intrinsic (constant), Weight is a gravitational force (changes with location)
Gravitational Force Proportionality	$F_{g} \propto \frac{m _{1} m _{2}}{r ^{2}}$	Used for proportional reasoning, no need to plug in $G$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Proportional reasoning gravity problem
2023 · FRQ
Apparent weight elevator problem

What's Next

This topic is the foundation for all gravitational interactions in AP Physics 1, and is a core prerequisite for Unit 3: Circular Motion and Gravitation, where you will use gravitational force as the centripetal force for orbiting objects. Without mastering the inverse-square law, the mass-weight distinction, and apparent weight calculations, you will struggle to connect force concepts to orbital motion and equilibrium problems. Gravitational force also appears repeatedly in momentum and energy problems, where weight is a common force acting on systems, so mastering this topic simplifies all subsequent dynamics work on the exam.