Dynamics.

Dynamics

Dynamics The laws and causes of motion were first properly formulated in the book _________ by ____________ ___________ in the year __________. These laws were later modified in 1905 and 1915 by _____________ ______________.

Dynamics The laws and causes of motion were first properly formulated in the book Principia by Isaac Newton in the year These laws were later modified in 1905 and 1915 by Albert Einstein.

Word Definition for Dynamics: a study of _______objects…
Accelerate or… Don’t accelerate…

Word Definition for Dynamics: a study of WHY objects…
Accelerate or… Don’t accelerate…

Accelerate or… Don’t accelerate… 3 possible forms ?… 2 possible states of motion

Accelerate or… Don’t accelerate… 3 possible forms… Speeding up Slowing down Changing direction 2 possible states of motion…

Accelerate or… Don’t accelerate… 3 possible forms… Speeding up Slowing down Changing direction Newton says: Any of these forms of acceleration are caused by a non-zero ___________ or ________________ force. 2 possible states of motion…

Accelerate or… Don’t accelerate… 3 possible forms… Speeding up Slowing down Changing direction Newton says: Any of these forms of acceleration are caused by a non-zero net or unbalanced force. 2 possible states of motion…

Accelerate or… Don’t accelerate… 3 possible forms… Speeding up Slowing down Changing direction Newton says: Any of these forms of acceleration is caused by a non-zero net or unbalanced force. The symbols for net force are … 2 possible states of motion…

Accelerate or… Don’t accelerate… 2 possible states of motion…

Accelerate or… Don’t accelerate… 2 possible states of motion… ?

Accelerate or… Don’t accelerate… 2 possible states of motion… at rest or at constant velocity

Accelerate or… Don’t accelerate… 2 possible states of motion… at rest or at constant velocity Newton’s first law : If an object is not speeding up, not slowing down or not changing direction, then it has an acceleration of ______ and a net force of ______ acting on it.

Accelerate or… Don’t accelerate… 2 possible states of motion… at rest or at constant velocity Newton’s first law : If an object is not speeding up, not slowing down or not changing direction, then it has an acceleration of zero and a net force of ______ acting on it.

Accelerate or… Don’t accelerate… 2 possible states of motion… at rest or at constant velocity Newton’s first law : If an object is not speeding up, not slowing down or not changing direction, then it has an acceleration of zero and a net force of zero acting on it.

Accelerate or… Don’t accelerate… 2 possible states of motion… at rest or at constant velocity Newton’s first law : If an object is at ________ or __________ _________, then it has an acceleration of zero and a net force of zero acting on it.

Accelerate or… Don’t accelerate… 2 possible states of motion… at rest or at constant velocity Newton’s first law : If an object is at rest or constant velocity, then it has an acceleration of zero and a net force of zero acting on it.

Accelerate or… Don’t accelerate… 2 possible states of motion… at rest or at constant velocity Newton’s first law : If an object is at rest or constant velocity, then it has an acceleration of zero and a net force of zero acting on it. If the net force is zero, there could be ____ forces acting or the forces are _____________.

Accelerate or… Don’t accelerate… 2 possible states of motion… at rest or at constant velocity Newton’s first law : If an object is at rest or constant velocity, then it has an acceleration of zero and a net force of zero acting on it. If the net force is zero, there could be no forces acting or the forces are _____________.

Accelerate or… Don’t accelerate… 2 possible states of motion… at rest or at constant velocity Newton’s first law : If an object is at rest or constant velocity, then it has an acceleration of zero and a net force of zero acting on it. If the net force is zero, there could be no forces acting or the forces are balanced.

Accelerate or… Don’t accelerate… 2 possible states of motion… at rest orat constant velocity Newton’s first law : If an object is at rest or constant velocity, then it has an acceleration of zero and a net force of zero acting on it. If the net force is zero, there could be no forces acting or the forces are balanced. The converse is also true If the net force on an object is zero, then the acceleration is ______ and the object is either at ______ or moving at __________ ____________.

Accelerate or… Don’t accelerate… 2 possible states of motion… at rest orat constant velocity Newton’s first law : If an object is at rest or constant velocity, then it has an acceleration of zero and a net force of zero acting on it. If the net force is zero, there could be no forces acting or the forces are balanced. The converse is also true If the net force on an object is zero, then the acceleration is zero and the object is either at ______ or moving at __________ ____________.

Accelerate or… Don’t accelerate… 2 possible states of motion… at rest orat constant velocity Newton’s first law : If an object is at rest or constant velocity, then it has an acceleration of zero and a net force of zero acting on it. If the net force is zero, there could be no forces acting or the forces are balanced. The converse is also true If the net force on an object is zero, then the acceleration is zero and the object is either at rest or moving at constant velocity

The three elements of Newton’s second law in symbols

The size of the acceleration is directly proportional to the net force

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ What does this mean?

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass.

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass. ǀaǀ α 1/m

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass. ǀaǀ α 1/m What does this mean?

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass. ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass. ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y Acceleration is in the same direction as the ________ ___________.

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass. ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y Acceleration is in the same direction as the net force.

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass. ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y Acceleration is in the same direction as the net force. Note: #1 holds true only if the __________ is constant.

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass. ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y Acceleration is in the same direction as the net force. Note: #1 holds true only if the mass is constant.

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass. ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y Acceleration is in the same direction as the net force. Note: #1 holds true only if the mass is constant. #2 holds true only if the _______ ______ is constant.

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass. ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y Acceleration is in the same direction as the net force. Note: #1 holds true only if the mass is constant. #2 holds true only if the net force is constant.

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass. ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y Acceleration is in the same direction as the net force. Note: #1 holds true only if the mass is constant. #2 holds true only if the net force is constant. In SI mks, all three elements can be combined to form Newton’s second law vector equation a = ??????

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass. ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y Acceleration is in the same direction as the net force. Note: #1 holds true only if the mass is constant. #2 holds true only if the net force is constant. In SI mks, all three elements can be combined to form Newton’s second law vector equation a = Fnet / m

The size of the acceleration is directly proportional to the net force ǀaǀ α ǀFnetǀ If ǀFnetǀ ↑ y , then ǀaǀ ↑ y The magnitude of the acceleration is inversely proportional to the mass. ǀaǀ α 1/m If m ↑ y , then ǀaǀ ↓ y Acceleration is in the same direction as the net force. Note: #1 holds true only if the mass is constant. #2 holds true only if the net force is constant. In SI mks, all three elements can be combined to form Newton’s second law vector equation a = Fnet / m Memorize please!

Other Forms of Newton’s Second Law equation

a = Fnet / m What does Fnet = ?

a = Fnet / m Fnet = m a

a = Fnet / m Fnet = m a Friday = Monday afternoon!

a = Fnet / m Fnet = m a Friday = Monday afternoon! What does m = ?

a = Fnet / m Fnet = m a Friday = Monday afternoon! m = ǀFnet ǀ / ǀaǀ

a = Fnet / m Fnet = m a Friday = Monday afternoon! m = ǀFnet ǀ / ǀaǀ Note: Once we know “a” from the second law dynamics equation, we can use _________ kinematics equations to find _____, _____, ____, and_____.

a = Fnet / m Fnet = m a Friday = Monday afternoon! m = ǀFnet ǀ / ǀaǀ Note: Once we know “a” from the second law dynamics equation, we can use four kinematics equations to find v1 , v2, Δd, and Δt.

a = Fnet / m Fnet = m a Friday = Monday afternoon! m = ǀFnet ǀ / ǀaǀ Note: Once we know “a” from the second law dynamics equation, we can use four kinematics equations to find v1 , v2, Δd, and Δt. Also, if we know three kinematics quantities, v1 , v2, Δd, or Δt, we can find “____” and then find Fnet using the second law dynamics equation.

a = Fnet / m Fnet = m a Friday = Monday afternoon! m = ǀFnet ǀ / ǀaǀ Note: Once we know “a” from the second law dynamics equation, we can use four kinematics equations to find v1 , v2, Δd, and Δt. Also, if we know three kinematics quantities, v1 , v2, Δd, or Δt, we can find “ a ” and then find Fnet using the second law dynamics equation.

a = Fnet / m Fnet = m a Friday = Monday afternoon! m = ǀFnet ǀ / ǀaǀ Note: Once we know “a” from the second law dynamics equation, we can use four kinematics equations to find v1 , v2, Δd, and Δt. Also, if we know three kinematics quantities, v1 , v2, Δd, or Δt, we can find “ a ” and then find Fnet using the second law dynamics equation. v1 , v2, Δd, and Δt ↔ a ↔ Fnet , m 4 kinematics equation dynamics 2nd law equation

Definition of the Newton: Unit of Force

Start with

Start with Fnet = m a

Start with Fnet = m a UNIT ?

Start with Fnet = m a Newton

Start with Fnet = m a Newton =

Start with Fnet = m a Newton = UNIT ?

Start with Fnet = m a Newton = kilogram

Start with Fnet = m a Newton = kilogram UNIT ?

Start with Fnet = m a Newton = kilogram meter/sec2

Start with Fnet = m a Newton = 1 kilogram meter/sec2

Start with Fnet = m a In words, one Newton of unbalanced force is applied when a one ___________ mass is accelerated at __________ Newton = kilogram meter/sec2

Start with Fnet = m a In words, one Newton of unbalanced force is applied when a one kilogram mass is accelerated at 1 meter/sec2 Newton = kilogram meter/sec2

Mass vs Force

Mass vs Force Mass Force ? ?

Mass vs Force Mass Force Amount of matter ?

Mass vs Force Mass Force Amount of matter A push or pull

Mass vs Force Mass Force Amount of matter SI unit = kilogram (kg)
A push or pull

A push or pull SI unit = Newton = kg m/s2

scalar A push or pull SI unit = Newton = kg m/s2

scalar A push or pull SI unit = Newton = kg m/s2 vector

Scalar Use a balance or collision experiments to measure A push or pull SI unit = Newton = kg m/s2 Vector

Scalar Use a balance or collision experiments to measure A push or pull SI unit = Newton = kg m/s2 Vector Use a spring scale to measure

Newton’s Third Law of Motion

For every action force on body A by body B, there is an ___________ but ______________ reaction force on body B by body A.

For every action force on body A by body B, there is an equal but opposite reaction force on body B by body A.

For every action force on body A by body B, there is an equal but opposite reaction force on body B by body A. Mr. BIG-A BIG’s Spring scale = more, less or equal to 122 N ? Little’s Spring scale = 122 N Mr. Little-B rope

For every action force on body A by body B, there is an equal but opposite reaction force on body B by body A. Mr. BIG-A Little’s Spring scale = 122 N BIG’s Spring scale = 122 N Mr. Little-B rope

For every action force on body A by body B, there is an equal but opposite reaction force on body B by body A. Mr. BIG-A Little’s Spring scale = 122 N BIG’s Spring scale = 122 N Mr. Little-B rope Action Force Statement: Mr BIG-A pulls on Mr Little-B with a force of 122 N [W]

For every action force on body A by body B, there is an equal but opposite reaction force on body B by body A. Mr. BIG-A Little’s Spring scale = 122 N BIG’s Spring scale = 122 N Mr. Little-B rope Action Force Statement: Mr BIG-A pulls on Mr Little-B with a force of 122 N [W] Reaction force statement: Mr Little-B pulls on Mr BIG-A with a force of 122 N [E]

For every action force on body A by body B, there is an equal but opposite reaction force on body B by body A. Suppose the men are on ice. Which man will have the greater acceleration and why? Mr. BIG-A Little’s Spring scale = 122 N BIG’s Spring scale = 122 N Mr. Little-B rope Action Force Statement: Mr BIG-A pulls on Mr Little-B with a force of 122 N [W] Reaction force statement: Mr Little-B pulls on Mr BIG-A with a force of 122 N [E]

For every action force on body A by body B, there is an equal but opposite reaction force on body B by body A. If the men are on ice, Mr. Little will have the greater acceleration because he has the smaller mass, a α 1/m, and as m ↓ a ↑ for the same net force. Mr. BIG-A Little’s Spring scale = 122 N BIG’s Spring scale = 122 N Mr. Little-B rope Action Force Statement: Mr BIG-A pulls on Mr Little-B with a force of 122 N [W] Reaction force statement: Mr Little-B pulls on Mr BIG-A with a force of 122 N [E]

Important Notes on Newton’s Third Law

This law only applies when there are two bodies

This law only applies when there are two bodies The action and reaction forces never act on the same body. What would happen if action and reaction forces did happen on the same body?

This law only applies when there are two bodies The action and reaction forces never act on the same body. Forces would always be balanced on the same body, and there would be no acceleration ever!

This law only applies when there are two bodies The action and reaction forces never act on the same body. Forces would always be balanced on the same body, and there would be no acceleration ever! The action force acts on one body and the equal, reaction force acts on the other.

This law only applies when there are two bodies The action and reaction forces never act on the same body. The action force acts on one body and the equal, reaction force acts on the other. Forces would always be balanced on the same body, and there would be no acceleration ever! If the action and reaction forces are the only forces acting, each body will accelerate according to |a| α 1/m so that |a1| / |a2| = m2 / m1

Example: Here is an action force statement: A foot exerts an applied force on a soccer ball at 83 N [forward]. What is the reaction force statement? Two-body diagram: Reaction force statement: ? Action: 83 N [forward]

Example: Here is an action force statement: A foot exerts an applied force on a soccer ball at 83 N [forward]. What is the reaction force statement? Two-body diagram: Reaction force statement: The ball exerts an equal applied force on the foot at 83 N [backward]. Action: 83 N [forward] Reaction: 83 N [backward]

Example: Here is an action force statement: A foot exerts an applied force on a soccer ball at 83 N [forward]. What is the reaction force statement? Two-body diagram: Reaction force statement: The ball exerts an equal applied force on the foot at 83 N [backward]. What is the effect of the reaction force of the ball on the foot? Action: 83 N [forward] Reaction: 83 N [backward]

Example: Here is an action force statement: A foot exerts an applied force on a soccer ball at 83 N [forward]. What is the reaction force statement? Two-body diagram: Reaction force statement: The ball exerts an equal applied force on the foot at 83 N [backward]. The applied net force and acceleration is backward on the foot, but the velocity of the foot is forward, so the foot slows down. Action: 83 N [forward] Reaction: 83 N [backward]

Reaction force statement: ?
You try this one!: Here is an action force statement: The planet earth exerts a gravitational force of 1.4 N [down] on a falling apple. What is the reaction force statement? Reaction force statement: ? Action: 1.4 N [down] apple Planet earth

Reaction force statement: The apple exerts an equal
You try this one!: Here is an action force statement: The planet earth exerts a gravitational force of 1.4 N [down] on a falling apple. What is the reaction force statement? Reaction force statement: The apple exerts an equal but opposite gravitational force on the earth of 1.4 N [up] Action: 1.4 N [down] apple Planet earth Reaction: 1.4 N [up]

You try this one!: Here is an action force statement: The planet earth exerts a gravitational force of 1.4 N [down] on a falling apple. What is the reaction force statement? Reaction force statement: The apple exerts an equal but opposite gravitational force on the earth of 1.4 N [up] If the apple exerts the same force up on the earth, why doesn't the earth accelerate up to meet the apple? Action: 1.4 N [down] apple Planet earth Reaction: 1.4 N [up]

You try this one!: Here is an action force statement: The planet earth exerts a gravitational force of 1.4 N [down] on a falling apple. What is the reaction force statement? Reaction force statement: The apple exerts an equal but opposite gravitational force on the earth of 1.4 N [up] The earth doesn't accelerate up to meet the apple because the mass of the earth is so much more than the apple. Note that |a| α 1/m, and as m ↑ a ↓ for the same net force of 1.4 N. Action: 1.4 N [down] apple Planet earth Reaction: 1.4 N [up]

You try this one!: Here is an action force statement: The planet earth exerts a gravitational force of 1.4 N [down] on a falling apple. What is the reaction force statement? Reaction force statement: The apple exerts an equal but opposite gravitational force on the earth of 1.4 N [up] The earth doesn't accelerate up to meet the apple because the mass of the earth is so much more than the apple. Note that |a| α 1/m, and as m ↑ a ↓ for the same net force of 1.4 N. If the earth has a mass of 6.0 X 1024 kg and the apple has a mass of 0.14 kg, what is the ratio of the sizes of the acceleration of the earth : the acceleration of the apple ? Action: 1.4 N [down] apple Planet earth Reaction: 1.4 N [up]

You try this one!: Here is an action force statement: The planet earth exerts a gravitational force of 1.4 N [down] on a falling apple. What is the reaction force statement? Reaction force statement: The apple exerts an equal but opposite gravitational force on the earth of 1.4 N [up] The earth doesn't accelerate up to meet the apple because the mass of the earth is so much more than the apple. Note that |a| α 1/m, and as m ↑ a ↓ for the same net force of 1.4 N. |aearth| / |aapple| = mapple/mearth = 0.14 kg /6.0 x 1024 kg=2.3 X 10-26 Action: 1.4 N [down] apple Planet earth Reaction: 1.4 N [up]

The Fundamental Forces

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces:

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces: Caused by two masses near each other = ?

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces: Caused by two masses near each other = force of gravity

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces: Caused by two masses near each other = force of gravity Caused by two charges near each other = ?

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces: Caused by two masses near each other = force of gravity Caused by two charges near each other = electrical force

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces: Caused by two masses near each other = force of gravity Caused by two charges near each other = electrical force Caused by moving charges = ?

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces: Caused by two masses near each other = force of gravity Caused by two charges near each other = electrical force Caused by moving charges = magnetic force

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces: Caused by two masses near each other = force of gravity Caused by two charges near each other = electrical force Caused by moving charges = magnetic force Charges could be moving or not moving depending on the frame of reference, so the electrical and magnetic forces are actually different forms of the same fundamental force = ?

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces: Caused by two masses near each other = force of gravity Caused by two charges near each other = electrical force Caused by moving charges = magnetic force Charges could be moving or not moving depending on the frame of reference, so the electrical and magnetic forces are actually different forms of the same fundamental force = electromagnetic force

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces: Caused by two masses near each other = force of gravity Caused by two charges near each other = electrical force Caused by moving charges = magnetic force Charges could be moving or not moving depending on the frame of reference, so the electrical and magnetic forces are actually different forms of the same fundamental force = electromagnetic force Caused by nucleons like the proton and neutron being very close together ( < m ) to prevent the protons in the nucleus from repelling each other = ?

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces: Caused by two masses near each other = force of gravity Caused by two charges near each other = electrical force Caused by moving charges = magnetic force Charges could be moving or not moving depending on the frame of reference, so the electrical and magnetic forces are actually different forms of the same fundamental force = electromagnetic force Caused by nucleons like the proton and neutron being very close together ( < m ) to prevent the protons in the nucleus from repelling each other = strong nuclear force

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces: Caused by two masses near each other = force of gravity Caused by two charges near each other = electrical force Caused by moving charges = magnetic force Charges could be moving or not moving depending on the frame of reference, so the electrical and magnetic forces are actually different forms of the same fundamental force = electromagnetic force Caused by nucleons like the proton and neutron being very close together ( < m ) to prevent the protons in the nucleus from repelling each other = strong nuclear force Turns one particle into another, like a neutron into a proton, by emitting “beta” radiation (electrons, positrons, neutrinos) = ?

All the different kinds of pushes and pulls around us, or forces, can be classified as belonging to four fundamental forces: Caused by two masses near each other = force of gravity Caused by two charges near each other = electrical force Caused by moving charges = magnetic force Charges could be moving or not moving depending on the frame of reference, so the electrical and magnetic forces are actually different forms of the same fundamental force = electromagnetic force Caused by nucleons like the proton and neutron being very close together ( < m ) to prevent the protons in the nucleus from repelling each other = strong nuclear force Turns one particle into another, like a neutron into a proton, by emitting “beta” radiation(electrons, positrons, neutrinos)=weak force

The Gravitational Force
Caused by two masses near each other

Caused by two masses near each other Always attractive (according to Newton)

Caused by two masses near each other Always attractive (according to Newton) Fg α mM/d2 M m d

Caused by two masses near each other Always attractive (according to Newton) Fg α mM/d2 M m d Fg (M on m) Fg (m on M)

Caused by two masses near each other Always attractive (according to Newton) Fg α mM/d2 As d ↑ 3 Fg ? M m d Fg (M on m) Fg (m on M)

Caused by two masses near each other Always attractive (according to Newton) Fg α mM/d2 As d ↑ 3 Fg ↓ 9 M m d Fg (M on m) Fg (m on M)

Caused by two masses near each other Always attractive (according to Newton) Fg α mM/d2 As d ↑ 3 Fg ↓ 9 As M ↑ Fg ? M m d Fg (M on m) Fg (m on M)

Caused by two masses near each other Always attractive (according to Newton) Fg α mM/d2 As d ↑ 3 Fg ↓ 9 As M ↑ Fg ↑ 13 M m d Fg (M on m) Fg (m on M)

Caused by two masses near each other Always attractive (according to Newton) Fg α mM/d2 As d ↑ 3 Fg ↓ 9 As M ↑ Fg ↑ 13 Equation? M m d Fg (M on m) Fg (m on M)

Caused by two masses near each other Always attractive (according to Newton) Fg α mM/d2 As d ↑ 3 Fg ↓ 9 As M ↑ Fg ↑ 13 Fg = GmM/d2 G = 6.67 X Nm2/kg2 M m d Fg (M on m) Fg (m on M)

Caused by two masses near each other Always attractive (according to Newton) Fg α mM/d2 As d ↑ 3 Fg ↓ 9 As M ↑ Fg ↑ 13 Fg = GmM/d2 G = 6.67 X Nm2/kg2 If the two masses are small, the force of gravity is also small. If the two masses are large, the force of gravity is also large. M m d Fg (M on m) Fg (m on M)

Example: Find the magnitude of force of gravitational attraction between a 67 kg girl and a 75 kg boy sitting 2.0 m apart.

Example: Find the magnitude of force of gravitational attraction between a 67 kg girl and a 75 kg boy sitting 2.0 m apart. Fg = GmM/d2 G = 6.67 X Nm2/kg2 = ?

Example: Find the magnitude of force of gravitational attraction between a 67 kg girl and a 75 kg boy sitting 2.0 m apart. Fg = GmM/d2 G = 6.67 X Nm2/kg2 = (6.67 X 10-11)(67)(75) / (2.0)2

Example: Find the magnitude of force of gravitational attraction between a 67 kg girl and a 75 kg boy sitting 2.0 m apart. Fg = GmM/d2 G = 6.67 X Nm2/kg2 = (6.67 X 10-11)(67)(75) / (2.0)2 = 8.4 X 10-8 N very, very small force

Try this: Find the magnitude of force of gravitational attraction between the 2.0 x 1030 kg sun and the 6.0 X 1024 kg earth separated by 1.5 X 1011 m.

Try this: Find the magnitude of force of gravitational attraction between the 2.0 x 1030 kg sun and the 6.0 X 1024 kg earth separated by 1.5 X 1011 m. Fg = GmM/d2 G = 6.67 X Nm2/kg2 = (6.67 X 10-11)(2.0 x 1030)(6.0 X 1024) / (1.5 X 1011 )2 = 3.6 X 1022 N very, very large force

Proportionality and the Force of Gravity

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is increased by a factor of three.

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is increased by a factor of three. Note the masses stay constant so... Fg α 1/d2

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is increased by a factor of three. Note the masses stay constant so... Fg α 1/d2 As d ↑ 3, d2 changes how?

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is increased by a factor of three. Note the masses stay constant so... Fg α 1/d2 As d ↑ 3, d2 ↑ 32

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is increased by a factor of three. Note the masses stay constant so... Fg α 1/d2 As d ↑ 3, d2 ↑ 32 , Fg changes how?

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is increased by a factor of three. Note the masses stay constant so... Fg α 1/d2 As d ↑ 3, d2 ↑ 32 , Fg ↓ 32

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is increased by a factor of three. Note the masses stay constant so... Fg α 1/d2 As d ↑ 3, d2 ↑ 32 , Fg ↓ 32 So the new force of gravity = ?

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is increased by a factor of three. Note the masses stay constant so... Fg α 1/d2 As d ↑ 3, d2 ↑ 32 , Fg ↓ 32 So the new force of gravity = N / 32 = ?

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is increased by a factor of three. Note the masses stay constant so... Fg α 1/d2 As d ↑ 3, d2 ↑ 32 , Fg ↓ 32 So the new force of gravity = N / 32 = N

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is divided by a factor of five. Try this.

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is divided by a factor of five. Try this. Note the masses stay constant so... Fg α 1/d2 As d ↓ 5, d2 ↓ 52 , Fg ↑ 52 So the new force of gravity = N X 52 = N

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is divided by a factor of five, one mass is increased by a factor of three, the other mass is divided by four. Try this harder one.

Example: Two objects in deep space exert a force of gravitational attraction of N on each other. Using Newton's Principle of gravitation i.e. Fg α mM/d2 , what is the new force of gravity if the following changes are made? The distance is divided by a factor of five, one mass is increased by a factor of three, the other mass is divided by four. Try this harder one. N X 52 X 3 / 4 = N

You try this: Find the magnitude of force of gravitational attraction between the same 67.0 kg girl on the surface of the earth and the planet earth itself. The distance between the centre of the earth and the centre of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m. The mass of the earth is 6.00 X 1024 kg.

You try this: Find the magnitude of force of gravitational attraction between the same 67.0 kg girl on the surface of the earth and the planet earth itself. The distance between the centre of the earth and the centre of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m. The mass of the earth is 6.00 X 1024 kg. Fg = GmM/d2 G = 6.67 X Nm2/kg2 = (6.67 X ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2 = 657 N

You try this: Find the magnitude of force of gravitational attraction between the same 67.0 kg girl on the surface of the earth and the planet earth itself. The distance between the centre of the earth and the centre of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m. The mass of the earth is 6.00 X 1024 kg. Fg = GmM/d2 G = 6.67 X Nm2/kg2 = (6.67 X ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2 = 657 N You know a short-cut formula to get the force of earth's gravity on an object on the earth's surface. What is it?

You try this: Find the magnitude of force of gravitational attraction between the same 67.0 kg girl on the surface of the earth and the planet earth itself. The distance between the centre of the earth and the centre of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m. The mass of the earth is 6.00 X 1024 kg. Fg = GmM/d2 G = 6.67 X Nm2/kg2 = (6.67 X ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2 = 657 N *Fg = mg m = mass of surface object (kg) g = gravitational field strength or force of earth's gravity on a one kilogram object (N/kg)

You try this: Find the magnitude of force of gravitational attraction between the same 67.0 kg girl on the surface of the earth and the planet earth itself. The distance between the centre of the earth and the centre of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m. The mass of the earth is 6.00 X 1024 kg. Fg = GmM/d2 G = 6.67 X Nm2/kg2 = (6.67 X ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2 = 657 N *Fg = mg m = mass of surface object (kg) g = gravitational field strength or force of a planet's gravity on a one kilogram object (N/kg) On the earth's surface, g = N/kg [d] It is numerically equal to ag , but different.

You try this: Find the magnitude of force of gravitational attraction between the same 67.0 kg girl on the surface of the earth and the planet earth itself. The distance between the centre of the earth and the centre of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m. The mass of the earth is 6.00 X 1024 kg. Fg = GmM/d2 G = 6.67 X Nm2/kg2 = (6.67 X ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2 = 657 N *Fg = mg = ? m = mass of surface object (kg) g = gravitational field strength or force of a planet's gravity on a one kilogram object (N/kg) On the earth's surface, g = N/kg [d] It is numerically equal to ag , but different.

You try this: Find the magnitude of force of gravitational attraction between the same 67.0 kg girl on the surface of the earth and the planet earth itself. The distance between the centre of the earth and the centre of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m. The mass of the earth is 6.00 X 1024 kg. Fg = GmM/d2 G = 6.67 X Nm2/kg2 = (6.67 X ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2 = 657 N *Fg = mg = (67 kg )(9.81 N/kg) [d] = ? m = mass of surface object (kg) g = gravitational field strength or force of a planet's gravity on a one kilogram object (N/kg) On the earth's surface, g = N/kg [d] It is numerically equal to ag , but different.

You try this: Find the magnitude of force of gravitational attraction between the same 67.0 kg girl on the surface of the earth and the planet earth itself. The distance between the centre of the earth and the centre of the girl is very nearly equal to the radius of the earth or 6.39 X 106 m. The mass of the earth is 6.00 X 1024 kg. Fg = GmM/d2 G = 6.67 X Nm2/kg2 = (6.67 X ) (67.0) ( 6.00 X 1024 ) / ( 6.39 X 106 )2 = 657 N *Fg = mg = (67 kg )(9.81 N/kg) [d] = 657 N [d] m = mass of surface object (kg) g = gravitational field strength or force of a planet's gravity on a one kilogram object (N/kg) On the earth's surface, g = N/kg [d] It is numerically equal to ag , but different.

A Special Force of gravity

A Special Force of gravity
The special force of a planet's gravity on a surface object is called __________.

A Special Force of gravity: Weight
The special force of a planet's gravity on a surface object is called weight.

The special force of a planet's gravity on a surface object is called weight. From previous example, the short-cut formula is... ?

The special force of a planet's gravity on a surface object is called weight. From previous example, the short-cut formula is... Fg = mg

The special force of a planet's gravity on a surface object is called weight. From previous example, the short-cut formula is... Fg = mg g depends on the planet

The special force of a planet's gravity on a surface object is called weight. From previous example, the short-cut formula is... Fg = mg g depends on the planet Earth g ≈ 10.0 N/kg [d] Moon g ≈ 1.6 N/kg [d] Mars g ≈ 3.7 N/kg [d]

Try this. Find the weight of the 67.0 kg girl on the surface of the moon (g ≈ N/kg [d]) and on the surface of Mars (g ≈ 3.70 N/kg [d]).

Try this. Find the weight of the 67.0 kg girl on the surface of the moon (g ≈ N/kg [d]) and on the surface of Mars (g ≈ 3.70 N/kg [d]). Moon Fg = mg = (67.0 kg) ( 1.60 N/kg [d] ) = 107 N [d] Mars Fg = mg = (67.0 kg) ( 3.70 N/kg [d] ) = 248 N [d] Remember we calculated the girl's weight on the earth = 657 N

Try this. Find the weight of the 67.0 kg girl on the surface of the moon (g ≈ N/kg [d]) and on the surface of Mars (g ≈ 3.70 N/kg [d]). Moon Fg = mg = (67.0 kg) ( 1.60 N/kg [d] ) = 107 N [d] Mars Fg = mg = (67.0 kg) ( 3.70 N/kg [d] ) = 248 N [d] Remember we calculated the girl's weight on the earth = 657 N Does weight depend on our location in the universe?

Try this. Find the weight of the 67.0 kg girl on the surface of the moon (g ≈ N/kg [d]) and on the surface of Mars (g ≈ 3.70 N/kg [d]). Moon Fg = mg = (67.0 kg) ( 1.60 N/kg [d] ) = 107 N [d] Mars Fg = mg = (67.0 kg) ( 3.70 N/kg [d] ) = 248 N [d] Remember we calculated the girl's weight on the earth = 657 N Yes, weight depends on our location in the universe. Look at our calculations above.

Two formulas for Weight

In our calculation of weight, it is easy to use the short-cut formula: Fg = mg

In our calculation of weight, it is easy to use the short-cut formula: Fg = mg But we also calculated weight using the more general formula: Fg = GmM/d2 Where d = rp (planet radius)

In our calculation of weight, it is easy to use the short-cut formula: Fg = mg But we also calculated weight using the more general formula: Fg = GmM/d2 Where d = rp (planet radius) So m|g| = GmM/rp2

In our calculation of weight, it is easy to use the short-cut formula: Fg = mg But we also calculated weight using the more general formula: Fg = GmM/d2 Where d = rp (planet radius) So m|g| = GmM/rp2 |g| = ?

In our calculation of weight, it is easy to use the short-cut formula: Fg = mg But we also calculated weight using the more general formula: Fg = GmM/d2 Where d = rp (planet radius) So m|g| = GmM/rp2 |g| = GM/rp2

In our calculation of weight, it is easy to use the short-cut formula: Fg = mg But we also calculated weight using the more general formula: Fg = GmM/d2 Where d = rp (planet radius) So m|g| = GmM/rp2 |g| = GM/rp2 Memorize and be able to derive.

Try this: Venus has a planetary radius of 6.05 X 106 m and a mass of 4.83 X 1024 kg. Find the gravitational field strength on Venus. |g| = GM/rp2

Try this: Venus has a planetary radius of 6.05 X 106 m and a mass of 4.83 X 1024 kg. Find the gravitational field strength on the surface of Venus. |g| = GM/rp2 = (6.67 X 10-11)(4.83 X 1024 ) / (6.05 X 106)2 = N/kg

Try this: Venus has a planetary radius of 6.05 X 106 m and a mass of 4.83 X 1024 kg. Find the gravitational field strength on the surface of Venus. |g| = GM/rp2 = (6.67 X 10-11)(4.83 X 1024 ) / (6.05 X 106)2 = N/kg What would our 67.0 kg girl weigh on the surface of Venus?

Try this: Venus has a planetary radius of 6.05 X 106 m and a mass of 4.83 X 1024 kg. Find the gravitational field strength on the surface of Venus. |g| = GM/rp2 = (6.67 X 10-11)(4.83 X 1024 ) / (6.05 X 106)2 = N/kg What would our 67.0 kg girl weigh on the surface of Venus? Fg = mg = (67.0 kg) ( 8.80 N/kg [d]) = 590 N

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface.

Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m
Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m 2rearth rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg[d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: 2rearth rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 2rearth rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r note r varies 2rearth rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r note r varies Which quantities remain constant? 2rearth rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r2 = k / r2 2rearth rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r2 = k / r2 Note r is centre-to-centre 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r2 = k / r2 What is the proportionality? 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r2 = k / r2 |g| α 1 / r2 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r2 = k / r2 |g| α 1 / r2 As we go from the earth's surface to two earth radii above its surface, how does “ r “ change? 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r2 = k / r2 |g| α 1 / r2 r ↑ 3 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r2 = k / r2 |g| α 1 / r2 r ↑ 3 r2 ↑ 32 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r2 = k / r2 |g| α 1 / r2 r ↑ 3 r2 ↑ 32 |g| ↓ 9 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r2 = k / r2 |g| α 1 / r2 r ↑ 3 r2 ↑ 32 |g| ↓ 9 Therefore |g| = ? 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r2 = k / r2 |g| α 1 / r2 r ↑ 3 r2 ↑ 32 |g| ↓ 9 Therefore |g| = 10.0 N/kg / 9 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r2 = k / r2 |g| α 1 / r2 r ↑ 3 r2 ↑ 32 |g| ↓ 9 Therefore |g| = 10.0 N/kg / 9 = N/kg [d] 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Proportionality Method: |g| = GM/rp2 = GM/r2 = k / r2 |g| α 1 / r2 r ↑ 3 r2 ↑ 32 |g| ↓ 9 Therefore |g| = 10.0 N/kg / 9 = N/kg [d] 2rearth Note that the proportionality method did not require that we know the earth radius! r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Equation Method: 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Equation Method: |g| = GM/r2 note r varies 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Equation Method: |g| = GM/r2 note r varies = (6.67 X 10-11)(6.00 X 1024 ) ( ? )2 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Equation Method: |g| = GM/r2 note r varies = (6.67 X 10-11)(6.00 X 1024 ) ( 3 X X 106 )2 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Equation Method: |g| = GM/r2 note r varies = (6.67 X 10-11)(6.00 X 1024 ) ( 3 X X 106 )2 = N/kg [d] very close to other method 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Need to be given Mearth = 6.00 X 1024 kg rearth = 6.39 X 106 m Equation Method: |g| = GM/r2 note r varies = (6.67 X 10-11)(6.00 X 1024 ) ( 3 X X 106 )2 = N/kg [d] very close to other method Note we did have to know the earth radius with this method. 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Using both methods, to two sigs, we found |g| = 1.1 N/kg [d] 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Using both methods, to two sigs, we found |g| = 1.1 N/kg [d] How is this helpful? 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Using both methods, to two sigs, we found |g| = 1.1 N/kg [d] How is this helpful? It tells us that if we have a one kg mass two earth radii above the earth's surface, it will weigh 1.1 N [d]. 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. Using both methods, to two sigs, we found |g| = 1.1 N/kg [d] How is this helpful? It tells us that if we have a one kg mass two earth radii above the earth's surface, it will weigh 1.1 N [d]. It also tells us that if we have a projectile two earth radii above earth's surface, it will have an acceleration of 1.1 m/s2 [d] 2rearth r rearth earth

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. |g| = 1.1 N/kg [d] It tells us that if we have a one kg mass two earth radii above the earth's surface, it will weigh 1.1 N [d]. It also tells us that if we have a projectile two earth radii above earth's surface, it will have an acceleration of 1.1 m/s2 [d] Note the projectile may fall straight down, move in a parabola, or move in a circle or ellipse. It depends on the horizontal constant velocity of the projectile.

Harder Example: The gravitational field strength on the earth's surface is 10.0 N/kg [d]. Find the gravitational field strength two earth radii above the earth's surface. |g| = 1.1 N/kg [d] It tells us that if we have a one kg mass two earth radii above the earth's surface, it will weigh 1.1 N [d]. It also tells us that if we have a projectile two earth radii above earth's surface, it will have an acceleration of 1.1 m/s2 [d] Note the projectile may fall straight down, move in a parabola, or move in a circle in UCM or even an ellipse. It depends on the horizontal constant velocity of the projectile or orbiting body. Your teacher will discuss Newton's thought experiment with you!

More on the gravitational field strength = g

Force of a planet's gravity on a unit mass

Force of a planet's gravity on a unit mass Or, g = Fg / m

Force of a planet's gravity on a unit mass Or, g = Fg / m Si unit is N/kg

Force of a planet's gravity on a unit mass Or, g = Fg / m Si unit is N/kg Also gives the acceleration of a projectile or object moving with UCM in m/s2

Force of a planet's gravity on a unit mass Or, g = Fg / m Si unit is N/kg Also gives the acceleration of a projectile or object moving with UCM in m/s2 There is also an analogous idea called electric field strength = Fe / q or = electrical force per unit + charge

Fields in General

Fields in General Something that has some value at every point in space

Fields in General Something that has some value at every point in space earth

Fields in General Something that has some value at every point in space Could be a scalar field like the temperature of the earth's atmosphere at any point in the air earth

Fields in General Something that has some value at every point in space Could be a scalar field like the temperature of the earth's atmosphere at any point in the air Could be a vector field, like the gravitational field of the earth or the electrical field around a charged particle earth

Fields in General Something that has some value at every point in space Could be a scalar field like the temperature of the earth's atmosphere at any point in the air Could be a vector field, like the gravitational field of the earth or the electrical field around a charged particle earth Some fields, like gravitational and electrical fields are fundamental fields: they are not made of anything else nor are they properties of other quantities like “temperature fields”

Try this: Centre-to centre, the moon is sixty earth radii away from the earth. Given the mass of the earth is 6.00 X 1024 kg and its radius is 6.39 X 106 m, use proportionality and the equation method to find the earth's gravitational field strength at sixty earth radii from its centre. Assuming the moon moves around the earth in UCM, what is the moon's centripetal acceleration about the earth?

Proportionality: Equation: As before, |g| α 1 / r2 |g| = GM/r2
Try this: Centre-to centre, the moon is sixty earth radii away from the earth. Given the mass of the earth is 6.00 X 1024 kg and its radius is 6.39 X 106 m, use proportionality and the equation method to find the earth's gravitational field strength at sixty earth radii from its centre. Assuming the moon moves around the earth in UCM, what is the moon's centripetal acceleration about the earth? Proportionality: Equation: As before, |g| α 1 / r |g| = GM/r2 As r ↑ = (6.67 X 10-11)(6.00 X 1024 ) r2 ↑ (60 X6.39 X 106)2 |g| ↓ = N/kg |g| ↓ 602 |g| = 10.0/ 602 = N/kg This means a one kg mass at the moon's distance from the earth would feel a N force of earth's gravity on it. It is also the value of the centripetal acceleration of the moon in UCM at m/s2.

Mass ≠ Weight Mass Weight

Mass ≠ Weight Mass Amount of matter Weight

Mass ≠ Weight Mass Amount of matter Weight
Force of a planet's gravity on a surface object

Mass ≠ Weight Mass Amount of matter SI unit = kg Weight
Force of a planet's gravity on a surface object

Mass ≠ Weight Mass Amount of matter SI unit = kg Weight
Force of a planet's gravity on a surface object SI unit = Newton

Mass ≠ Weight Mass Amount of matter SI unit = kg
Does not depend on location in the universe Weight Force of a planet's gravity on a surface object SI unit = Newton

Does not depend on location in the universe Weight Force of a planet's gravity on a surface object SI unit = Newton Does depend on location in the universe

Does not depend on location in the universe scalar Weight Force of a planet's gravity on a surface object SI unit = Newton Does depend on location in the universe

Does not depend on location in the universe scalar Weight Force of a planet's gravity on a surface object SI unit = Newton Does depend on location in the universe vector

Homework Practice: New textbook
New textbook: Read p70-p73 Define inertia. Re-do sample problem #1 on p72 do practice Q1 p72 check ans same page Re-do sample problem #1 p73 Q1, Q2 p74 check ans same page Try p 99 Q3,Q10,Q11 p100 Q1-Q5 Q10,Q11 check ans p715 Review p Review and redo sample problems p , Do Q1-Q3 p293 check ans same page review and re-do sample problems p294 Do Q1,2 p295 Do Q1-3, Q5-Q7 Q13 p296 Check ans p715

Homework Practice: Old textbook
Read p77-p86 Define inertia, static equilibrium Review and redo sample problems #1,#2 p79 Do Q2 p80 Q6 p81 check ans same page Review and re-do sample problem #4 p82-83 Do Q10,11,13 p83 check ans same page Do q16 a to d q17,q18 p84 check ans same page Review and re-do sample problem #5 p85 Do Q4,6,8 p87 check ans p 781 Try p115 Q5,6,8,9,12,15 check p782 Read p139-p143 Review and redo sample problem p143 do Q8-Q13 p143 check ans same page p144 Q4,5,6 check p782 Read p274-p276 Review and redo sample problems p275 Do Q2-Q6 p276 Check ans same page Do Q7 p277 chk p783

Dynamics.

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Dynamics.

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Presentation on theme: "Dynamics."— Presentation transcript:

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