Gravitational Force and Weight — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Covers Newton’s law of universal gravitation, gravitational force calculation, the mass vs. weight distinction, gravitational field strength, apparent weight, and force diagram techniques for gravitational forces in AP Physics 1 dynamics.

You should already know: Newton’s three laws of motion, free-body force diagram construction, vector addition of force vectors.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Gravitational Force and Weight?

Gravitational force is the fundamental attractive force between any two objects with mass, one of the four fundamental forces in nature. For AP Physics 1, aligned with the CED for Unit 2: Dynamics, this topic makes up roughly 4-6% of the total exam weight, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, most often paired with other dynamics concepts like net force, acceleration, or circular motion. Weight is a specific type of gravitational force: it is the gravitational force exerted on an object by a much larger body (like Earth or the Moon), so it is not an intrinsic property of the object. Common notation: gravitational force between two masses is written $F_g$, weight is also often written $F_g$ or $W$, with direction always pointing toward the center of mass of the larger attracting body. A core point the AP exam repeatedly tests is the distinction between mass (an intrinsic measure of inertia, units of kilograms, constant for an object regardless of location) and weight (a force, units of newtons, changes with the gravitational field of the nearby large mass).

2. Newton’s Law of Universal Gravitation

Newton’s law of universal gravitation describes the gravitational force between any two point masses (or spherical masses, where we can treat all mass as concentrated at the object’s center). The law states that 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. The formula is: $$F_g=G\frac{m_1{m}_2}{r^2}$$ Where $G=6.67\times 10^{-11}{\text{Ncdotpm}}^2/{\text{kg}}^2$ 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. Key properties tested on the AP exam: gravitational force is always attractive, it follows an inverse-square law (doubling the distance quarters the force), and it obeys Newton’s third law: the force $m_1$ exerts on $m_2$ is equal in magnitude and opposite in direction to the force $m_2$ exerts on $m_1$. Most AP questions on this law use proportional reasoning, so you will rarely need to plug in the value of $G$ for a correct answer.

Worked Example

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

1. Write the original force equation to relate $F$ to the constants: $F=G\frac{(m)(2m)}{d^2}=G\frac{2m^2}{d^2}$, which rearranges to $\frac{G{m}^2}{d^2}=\frac{F}{2}$.
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 step 1: $F_{\text{new}}=3\left(\frac{F}{2}\right)=\frac{3}{2}F$.
4. Confirm the inverse-square term was correctly applied: distance doubled, so the denominator is squared to $4d^2$, 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. Mass vs. Weight and Gravitational Field Strength

Near the surface of a large spherical body like Earth, the distance from the 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. Let $M$ be the mass of the large body, $R$ its radius, and $m$ the mass of a small object near the surface. Substituting into the universal gravitation formula gives: $$F_g=W=G\frac{Mm}{R^2}=m\left(\frac{GM}{R^2}\right)=mg$$ Where $g=\frac{GM}{R^2}$ is the gravitational field strength near the surface of the large body. On Earth’s surface, $g\approx 9.8{\text{ m/s}}^2=9.8\text{ N/kg}$, which is why weight is simply $W=mg$ for objects near Earth’s surface. The core distinction the AP exam tests here is: mass is a measure of inertia (how much an object resists acceleration, per Newton’s second law), it is intrinsic, units of kilograms, and does not change with location. Weight is a force, units of newtons, and depends on the local gravitational field, so it changes if you move to a different planet or a different altitude.

Worked Example

A student has a mass of 65 kg on Earth, where $g_{\text{Earth}}=9.8\text{ N/kg}$. On the Moon, the gravitational field strength is $g_{\text{Moon}}=1.6\text{ N/kg}$. 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 $W=mg$ 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. Compare to Earth values for consistency: on Earth, the student’s weight is $(65)(9.8)=637$ N, which is much larger, as expected for the Moon’s weaker gravity.
4. Confirm units: mass is reported in kilograms, weight in newtons, which matches the definition of each quantity.

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

Apparent weight is the normal force exerted on an object by a supporting surface (like a bathroom scale in an elevator), which is what a scale actually measures. It is not equal to the object’s actual weight when the object and the supporting scale are accelerating vertically. To solve apparent weight problems, draw a free-body diagram for the object: the two vertical forces are actual weight $W=mg$ downward, and the normal force $n$ (apparent weight) upward from the scale. Apply Newton’s second law $\sum F=ma$, taking upward as the positive direction: $$\sum F=n-mg=ma$$ Rearranged, this gives: $$n=m(g+a)$$ Where $a$ is the acceleration of the object/scale. If acceleration is upward, $a$ is positive, so apparent weight is larger than actual weight (you feel heavier). If acceleration is downward, $a$ is negative, so apparent weight is smaller than actual weight (you feel lighter). If the object is 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 $mg$, upward force is the normal force $n$ (the apparent weight the scale reads).
2. Assign signs: upward is positive, so acceleration is $a=-2.0{\text{ m/s}}^2$ (downward). Write Newton’s second law: $n-mg=ma$.
3. Substitute values and solve: $n=m(g+a)=50\text{ kg}(9.8{\text{ m/s}}^2-2.0{\text{ m/s}}^2)=50(7.8)=390\text{ N}$.
4. Check consistency: accelerating downward means apparent weight should be less than actual weight. Actual weight is $50\ast 9.8=490$ 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. Common Pitfalls (and how to avoid them)

• Wrong move: Using $W=mg$ with $g=9.8\text{ N/kg}$ 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.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

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) $2G{M}^2/r^2$ C) $4G{M}^2/r^2$ D) $8G{M}^2/r^2$

Worked Solution: Each original asteroid exerts an attractive force on the center asteroid, so the force from the left asteroid pulls left, and the force from the right asteroid pulls right. The distance from each original asteroid to the center asteroid is $r/2$, so the magnitude of each force is $F_i=G{M}^2/(r/2{)}^2=4G{M}^2/r^2$. Since the two forces are equal in magnitude and opposite in direction, their vector sum is zero. The correct answer is A.

Question 2 (Free Response)

A 70 kg astronaut stands on a bathroom scale in a rocket launching vertically from Earth’s surface. (a) When the rocket moves upward at constant speed, what does the scale read, in newtons? What concept does this measurement represent? (b) When the rocket has an upward acceleration of 12 m/s², what does the scale read? (c) After launch, when the rocket is two Earth radii above Earth’s surface (so the distance from the rocket to Earth’s center is $3R$, where $R$ is Earth’s radius), what is the astronaut’s weight in terms of their weight $W_0$ on Earth’s surface?

Worked Solution: (a) Constant speed means acceleration $a=0$. Newton’s second law gives $n-mg=0$, so $n=mg=(70\text{ kg})(9.8{\text{ m/s}}^2)=686\text{ N}$. The scale measures the astronaut’s apparent weight, which equals actual weight when acceleration is zero. (b) For upward acceleration $a=12{\text{ m/s}}^2$, use $n=m(g+a)=70(9.8+12)=70(21.8)=1526\text{ N}$. (c) Weight follows $W\propto 1/r^2$. On Earth’s surface, $W_0=GMm/R^2$. At $r=3R$, $W=GMm/(3R{)}^2=W_0/9$. The astronaut’s weight is 1/9 of their surface weight.

Question 3 (Application / Real-World Style)

The James Webb Space Telescope (JWST) orbits the Sun at a distance of approximately $1.5\times 10^{11}\text{ m}$, the same as Earth’s orbital distance. The mass of the Sun is $1.99\times 10^{30}\text{ kg}$, and the mass of JWST is 6200 kg. What is the magnitude of the gravitational force exerted by the Sun on JWST at this orbit? Use $G=6.67\times 10^{-11}{\text{ Ncdotpm}}^2/{\text{kg}}^2$.

Worked Solution: Use Newton’s universal gravitation formula: $$F_g=G\frac{m_{\text{Sun}}{m}_{\text{JWST}}}{r^2}$$ Substitute values: $$F_g=(6.67\times 10^{-11})\frac{(1.99\times 10^{30})(6200)}{(1.5\times 10^{11}{)}^2}\approx 36.5\text{ N}$$ This surprisingly small force (equal to the weight of a ~3.7 kg object on Earth) is enough to keep the 6200 kg telescope in a stable circular orbit because the centripetal acceleration required at JWST’s orbital speed is very low.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundation for all future work involving gravitational interactions in AP Physics 1, and it 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.

Next topics to study: Newton's Laws of Motion Free-Body Force Diagrams Circular Motion and Orbits Equilibrium and Net Force