Overview

Unit 3.3 | The Gravitational Force

Picture of Jason Kuma
Jason Kuma

Writer | Coach | Builder | Fremont, CA

Article Content
This lessons will cover gravitation — another linear force responsible for pulling two masses together.

Unit 3 Breakdown

You are on Lesson 1 of 4

  1. Unit 3.1 | Understanding circular motion and centripetal forces
  2. Unit 3.2 | Solving circular motion problems using FBDs
  3. Unit 3.3 | The gravitational force (Current Lesson)
  4. Unit 3.4 | Combining circular motion and gravitation (satellites, orbits, and more)

In this lesson: 

  • We will learn about gravitational force
  • Where the gravitational force applies in common scenarios
  • The relationship between centripetal and gravitational forces and solving problems that involve both

The force between any two object

It turns out there’s an attractive force that exists between any two or more masses.

For example, you and your computer’s screen; The desk and your notebook; You and the earth; The Earth and the Sun.

Any two masses will share an equally attractive force between them.

This is the gravitational force.

Gravitational force equation

[katex] F_g = \frac{Gm_1m_2}{r^2} [/katex]

In this equation:

  • [katex] G [/katex] is the gravitational constant [katex] 6.69 \times 10^{-11} [/katex] NOT to be confused with the gravitational field [katex] 9.81 \, m/s^2[/katex].
  • [katex] m_1 [/katex] and [katex] m_2 [/katex] are the masses of each object in no particular order
  • [katex] r [/katex] is the distance between the centers of each mass

A force you already know

Consider a person and the earth. These two masses are pulled together by the gravitational force.

And we already have a name for this attraction: the force due to gravity ([katex] F_g [/katex]. We also call it weight ([katex] mg [/katex]).

So why do we have two names for the same force?

It’s because they are not used in the same cases. You can only use [katex] mg [/katex] if you know the value of the [katex]g[/katex] the gravitational field.

On the surface of earth, [katex] g [/katex] is [katex] 9.81 \, m/s^2[/katex].

But what if you are not on the surface? What happens to the value of gravity, [katex]g[/katex] if you’re in the international space station (ISS)?

Diminishing Gravitational Fields

As you move further away from earth, the magnitude of the gravitational feild decreases.

A satellite in orbit for example, experiences a much lesser gravitational field compared to when at the surface ([katex] 9.81 \, m/s^2[/katex].

This means that the satellite weighs more on the surface than it does in orbit!

To prove this we can calculate weight of the satellite using [katex]mg[/katex]. But only if we know the value of [katex]g[/katex].

So instead we use the [katex]F_g[/katex] (gravitation) formula. Plug the mass of the satellite, the mass of the earth, and the distance between the two to find the weight of the person.

Bonus Question: What happens to your weight if you move extremely far away from earth?

Answer: As you move far away from earth, the gravitational field acting on you would decrease to 0. This means you would weight nothing as [katex] g = 0 [/katex].

Calculating gravitational fields

To find the gravitational forces we simply use the gravitation formula, but can we also calculate the value for [katex]g[/katex] as well?

It turns out we can. Here’s a quick derivation:

  1. As discussed, weight is equal to the gravitational force. Thus, we can set the two equal to each other. Let [katex] m_p[/katex] be the mass of the person and [katex] m_E[/katex] be the mass of the earth (or any other planet)
    • [katex] m_pg = G\frac{m_pm_E}{r^2} [/katex]
  2. In this step we can cancel out [katex] m_p[/katex] from both sides to get our final equation for [katex]g[/katex].
    • [katex] g = G\frac{m_E}{r^2} [/katex]

You can use this equation to find the magnitude of the gravitational field any distance anywhere from the center of the earth.

Question: What value for g would you get if you plugged in the radius of earth?

Answer: [katex] 9.81 \, m/s^2[/katex]!

To recap, you can find the gravitational field any distance away from any mass (not just the earth) using this equation:

[katex] g = G\frac{m_E}{r^2} [/katex]

PS: Finding the equilibrium distance between two masses

In this video we’ll solve the following question:

Open in UBQ
Question 1
Advanced
Mathematical
A spacecraft somewhere in between the Earth and the Moon experiences zero net force acting on it. This is because the Earth and the Moon pull the spacecraft in equal but opposite directions. Find the distance \(D\) away from Earth such that the spacecraft experiences zero net force. The distance between the Moon and Earth is \(\sim 3.844 \times 10^8 \, \text{m}\). Note: You may need the mass of the Earth and the Moon. You can find this in the formula table.

Locked Video Content

You Try – 4 practice questions

Open in UBQ
Question 2
Advanced
Mathematical

Find the net gravitational force on a \(2.0 \, \text{kg}\) sphere midway between a \(4.0 \, \text{kg}\) sphere and a \(7.0 \, \text{kg}\) sphere that are \(1.2 \, \text{m}\) apart.

Open in UBQ
Question 3
Intermediate
Proportional Analysis

The exoplanet HD 69830b has a mass 10 times that of the Earth and a radius 5 times that of the Earth. The value of g on HD 69830b is most nearly

Open in UBQ
Question 4
Intermediate
Proportional Analysis

The magnitude of the gravitational field on the surface of a new planet is \(20 \, \text{N/kg}\). The planet's mass is half the mass of Earth. The radius of Earth is \(6400 \, \text{km}\). What is the radius of the new planet?

Open in UBQ
Question 5
Intermediate
Mathematical
A \(1.5 \, \text{kg}\) object is located at a distance of \(1.7 \times 10^{6} \, \text{m}\) from the center of a larger object whose mass is \(7.4 \times 10^{22} \, \text{kg}\).
Part (a)3 pts

What is the magnitude of the force acting on the smaller object?

Part (b)3 pts

What is the magnitude of the force acting on the larger object?

Part (c)3 pts

What is the acceleration of the smaller object when it is released?

Preview Lesson 3.4

The last lesson of this Unit will make the connection between gravitational and centripetal forces. After this you’ll be able to solve every type of circular motion problem.

Nerd Notes

Discover the world's best Physics resources

Continue with

By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

Metric Prefixes

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

Phy Pro

One price to unlock most advanced version of Phy across all our tools.

$11.99

per month

Billed Monthly. Cancel Anytime.

Physics is Hard, But It Does NOT Have to Be

We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.

Trusted by 10k+ Students
Try UBQ

1000+ Physics test questions sorted by topic. 100% free with AI powered problem solving help. 

Stuck on a problem? Try Phy AI

The world’s most accurate AI Physics and Math problem solver.

Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed

Thanks for reading Nerd-Notes!

If you got 2 seconds, log in for more free tools.

By continuing, you agree to the updated Terms of Sale, Terms of Use, and Privacy Policy.

We use cookies to improve your experience. By continuing to browse on Nerd Notes, you accept the use of cookies as outlined in our privacy policy.