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Unit 3.3 | The Gravitational Force

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Jason Kuma

Writer | Coach | Builder | Fremont, CA

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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

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

In this equation:

  • G is the gravitational constant 6.69 \times 10^{-11} NOT to be confused with the gravitational field 9.81 \, m/s^2.
  • m_1 and m_2 are the masses of each object in no particular order
  • r 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 ( F_g . We also call it weight ( mg ).

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 mg if you know the value of the g the gravitational field.

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

But what if you are not on the surface? What happens to the value of gravity, g 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 ( 9.81 \, m/s^2.

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 mg. But only if we know the value of g.

So instead we use the F_g (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 g = 0 .

Calculating gravitational fields

To find the gravitational forces we simply use the gravitation formula, but can we also calculate the value for g 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 m_p be the mass of the person and m_E be the mass of the earth (or any other planet)
    • m_pg = G\frac{m_pm_E}{r^2}
  2. In this step we can cancel out m_p from both sides to get our final equation for g.
    • g = G\frac{m_E}{r^2}

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: 9.81 \, m/s^2!

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

g = G\frac{m_E}{r^2}

PS: Finding the equilibrium distance between two masses

In this video we’ll solve the following question:

Difficulty - Advanced
Solve Type - Mathematical
A spacecraft somewhere in between the earth and the moon experiences 0 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 ~3.844 x 108 m. NOTE: You may need the mass of the earth and moon. You can find this in the formula table.
View Full Question and Explanation

Locked Video Content

You Try – 4 practice questions

Question 1
Difficulty - Advanced
Solve Type - Mathematical

Find the net gravitational force on a 2.0 kg sphere midway between a 4.0 kg sphere and a 7.0 kg sphere that are 1.2 m apart.

View Full Question and Explanation
Question 2
Difficulty - Intermediate
Solve Type - 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

  1. 2 m/s2

  2. 4 m/s2

  3. 8 m/s2

  4. 10 m/s2

  5. 15 m/s2

View Full Question and Explanation
Question 3
Difficulty - Intermediate
Solve Type - Proportional Analysis

The magnitude of the gravitational field on the surface of a new planet is 20 N/kg. The planet's mass half the mass of Earth. The radius of Earth is 6400 km. What is the radius of the new planet?

View Full Question and Explanation
Question 4
Difficulty - Intermediate
Solve Type - Mathematical
A 1.5 kg object is located at a distance of 1.7 x106 m from the center of a larger object whose mass is 7.4 x 1022 kg.
View Full Question and Explanation

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.

Picture of Jason Kuma
Jason Kuma

Writer | Coach | Builder | Fremont, CA

Programs

Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy 

Unit 5 – Momentum 

Unit 6 – Torque 

Unit 7 – Oscillations 

Unit 8 – Fluids

Reading Key

LRN
RE
PS
PQ
Black
White
Blue
Orange

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g} 
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
 KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\omega (Angular Velocity)\text{radians per second (rad/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

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

  1. Start with the given measurement: \text{5 km}

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

  3. Perform the multiplication: \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}

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

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

  1. Some answers may be slightly off by 1% depending on rounding, etc.
  2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
  3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
  4. Bookmark questions that you can’t solve so you can come back to them later. 
  5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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