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Unit 3.4 | Combining circular motion and gravitation (satellites, orbits, & more)

Circular Motion and Gravitation
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Jason Kuma

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This final lesson will help you bridge the connection between circular motion and gravitation.

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
  4. Unit 3.4 | Combining circular motion and gravitation – satellites, orbits, and more (Current Lesson)

In this lesson: 

  • We will learn how centripetal and gravitational forces are related
  • Learn how to solve problems that involve both forces

Making the connection

To see the connection between circular motion and gravitation, let’s use a satellite as an example.

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Common Equations to Know

Now that you understand how to use circular motion and gravitational forces together, it’s a good idea to know which equations you need to know.

These are not equations that will be given to you.

Instead, you need to know how to derive them.

Equation 1: General Relationships (you can manipulate these equations as you wish)

F_c = \frac{mv^2}{r} = G \frac{M m}{r^2}

Equation 2: Speed of an orbiting object

v = \sqrt{G \frac{M}{r}}

Equation 3: Period of an orbiting object

T = 2\pi \sqrt{\frac{r^3}{GM}}

PS – More satellite and gravitational questions

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Your Turn – Practice Circular Motion and Gravitation

Here are six questions to help you really understand how to apply the equations and solve problems.

Remember to use the framework: Draw an FBD; Apply Newton’s second law; Solve for the variable in question.

Question 1
Difficulty - Intermediate
Solve Type - Conceptual

Which of the following best explains why astronauts experience weightlessness while orbiting the earth?

  1. The spaceship is traveling at a constant velocity.

  2. The centripetal force on the astronaut is zero.

  3. The force of gravity is very small because the astronaut is far away from the planet.

  4. The spaceship is in free fall.

  5. The gravitational pull from the sun cancels out the gravitational pull from the earth.

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Question 2
Difficulty - Intermediate
Solve Type - Mathematical
A satellite circling Earth completes each orbit in 132 minutes.
View Full Question and Explanation
Question 3
Difficulty - Intermediate
Solve Type - Mathematical

Consider a neutron star with a mass equal to the sun, a radius of 10 km, and a rotation period of 1.0 s. What is the radius of a geosynchronous orbit about the neutron star? The mass of the sun can be found in the formula sheet above.

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Question 4
Difficulty - Intermediate
Solve Type - Mathematical
A communications satellite orbits the Earth at an altitude of 35,000 km above the Earth’s surface. Take the mass of Earth to be 6 \times 10^{24} \text{ kg} the the radius of Earth to be 6.4 \times 10^6 \text{ m}. What is the satellite’s velocity?
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Question 5
Difficulty - Intermediate
Solve Type - Mathematical

In 2014, the European Space Agency placed a satellite in orbit around comet 67P/Churyumov-Gerasimenko and then landed a probe on the surface. The actual orbit was elliptical, but we can approximate it as a 50 km diameter circular orbit with a period of 11 days.

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Question 6
Difficulty - Advanced
Solve Type - Mathematical

A 2.2 \times 10^{21} \, \text{kg} moon orbits a distant planet in a circular orbit of radius 1.5 \times 10^8 \, \text{m}. It experiences a 1.1 \times 10^{19} \, \text{N} gravitational pull from the planet. What is the moon's orbital period in earth days?

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Unit 3 Recap

Take a second to read the Unit 3 Speed Recap. This goes over everything you learned (in 5 minutes). Moreover, be sure to get in tons of practice in using UBQ. These questions are hand selected for you to master the concepts you learned!

Unit 4 Preview

Congrats! You finished Unit 3.

It’s time to move on to Unit 4 Energy. This unit marks the halfway point and is one of the easier to understand topics!

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
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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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