April 19, Week 13 Today: Chapter 13, Gravity Homework Assignment - - PowerPoint PPT Presentation

Newton’s Gravity April 19, 2013 - p. 1/8

April 19, Week 13

Today: Chapter 13, Gravity Homework Assignment #10 - Due Today.

Mastering Physics: 7 problems from chapter 9 Written Question: 10.86

From now on, Thursday office hours will be held in room 109 of Regener Hall Exam #4, Next Friday, April 26 Practice Exam on Website.

Newton’s Gravity April 19, 2013 - p. 2/8

Newton’s Law of Gravitation

Newton’s Law of Gravitation - Every object with mass exerts a gravitational force on every other object with mass.

Newton’s Gravity April 19, 2013 - p. 2/8

Newton’s Law of Gravitation

Newton’s Law of Gravitation - Every object with mass exerts a gravitational force on every other object with mass. The magnitude of this force is given by:

Newton’s Gravity April 19, 2013 - p. 2/8

Newton’s Law of Gravitation

Newton’s Law of Gravitation - Every object with mass exerts a gravitational force on every other object with mass. The magnitude of this force is given by: M1 M2 M1 - Mass of first object M2 - Mass of second object

Newton’s Gravity April 19, 2013 - p. 2/8

Newton’s Law of Gravitation

Newton’s Law of Gravitation - Every object with mass exerts a gravitational force on every other object with mass. The magnitude of this force is given by: M1 M2 M1 - Mass of first object M2 - Mass of second object Fg = M1M2

Newton’s Gravity April 19, 2013 - p. 2/8

Newton’s Law of Gravitation

Newton’s Law of Gravitation - Every object with mass exerts a gravitational force on every other object with mass. The magnitude of this force is given by: M1 M2 r M1 - Mass of first object M2 - Mass of second object Fg = M1M2 r - separation distance, center-to-center for spherical objects

Newton’s Gravity April 19, 2013 - p. 2/8

Newton’s Law of Gravitation

Newton’s Law of Gravitation - Every object with mass exerts a gravitational force on every other object with mass. The magnitude of this force is given by: M1 M2 r M1 - Mass of first object M2 - Mass of second object Fg = M1M2 r2 r - separation distance, center-to-center for spherical objects

Newton’s Gravity April 19, 2013 - p. 2/8

Newton’s Law of Gravitation

Newton’s Law of Gravitation - Every object with mass exerts a gravitational force on every other object with mass. The magnitude of this force is given by: M1 M2 r M1 - Mass of first object M2 - Mass of second object Fg = M1M2 r2 r - separation distance, center-to-center for spherical objects Inverse square law

Newton’s Gravity April 19, 2013 - p. 2/8

Newton’s Law of Gravitation

Newton’s Law of Gravitation - Every object with mass exerts a gravitational force on every other object with mass. The magnitude of this force is given by: M1 M2 r M1 - Mass of first object M2 - Mass of second object Fg = GM1M2 r2 r - separation distance, center-to-center for spherical objects Inverse square law

Newton’s Gravity April 19, 2013 - p. 2/8

Newton’s Law of Gravitation

Newton’s Law of Gravitation - Every object with mass exerts a gravitational force on every other object with mass. The magnitude of this force is given by: M1 M2 r M1 - Mass of first object M2 - Mass of second object Fg = GM1M2 r2 r - separation distance, center-to-center for spherical objects Inverse square law Universal Gravitational Constant: G = 6.67 × 10−11 N · m2/kg2

Newton’s Gravity April 19, 2013 - p. 3/8

Inverse Square Law Exercise

When two masses are a distance r1 apart from each other the gravitational force between them is 5 N. They are then moved to a new separation, r2, where the gravitational force between them is 10 N. What is the relation between r1 and r2?

Newton’s Gravity April 19, 2013 - p. 3/8

Inverse Square Law Exercise

When two masses are a distance r1 apart from each other the gravitational force between them is 5 N. They are then moved to a new separation, r2, where the gravitational force between them is 10 N. What is the relation between r1 and r2? (a) r2 = r1

Newton’s Gravity April 19, 2013 - p. 3/8

Inverse Square Law Exercise

When two masses are a distance r1 apart from each other the gravitational force between them is 5 N. They are then moved to a new separation, r2, where the gravitational force between them is 10 N. What is the relation between r1 and r2? (a) r2 = r1 (b) r2 = 2 r1

Newton’s Gravity April 19, 2013 - p. 3/8

Inverse Square Law Exercise

When two masses are a distance r1 apart from each other the gravitational force between them is 5 N. They are then moved to a new separation, r2, where the gravitational force between them is 10 N. What is the relation between r1 and r2? (a) r2 = r1 (b) r2 = 2 r1 (c) r2 = √ 2

• r1 = (1.4) r1

Newton’s Gravity April 19, 2013 - p. 3/8

Inverse Square Law Exercise

When two masses are a distance r1 apart from each other the gravitational force between them is 5 N. They are then moved to a new separation, r2, where the gravitational force between them is 10 N. What is the relation between r1 and r2? (a) r2 = r1 (b) r2 = 2 r1 (c) r2 = √ 2

• r1 = (1.4) r1

(d) r2 = 1 2

• r1

Newton’s Gravity April 19, 2013 - p. 3/8

Inverse Square Law Exercise

When two masses are a distance r1 apart from each other the gravitational force between them is 5 N. They are then moved to a new separation, r2, where the gravitational force between them is 10 N. What is the relation between r1 and r2? (a) r2 = r1 (b) r2 = 2 r1 (c) r2 = √ 2

• r1 = (1.4) r1

(d) r2 = 1 2

• r1

(e) r2 = 1 √ 2

• r1 = (0.707)r1

Newton’s Gravity April 19, 2013 - p. 3/8

Inverse Square Law Exercise

When two masses are a distance r1 apart from each other the gravitational force between them is 5 N. They are then moved to a new separation, r2, where the gravitational force between them is 10 N. What is the relation between r1 and r2? (a) r2 = r1 (b) r2 = 2 r1 (c) r2 = √ 2

• r1 = (1.4) r1

(d) r2 = 1 2

• r1

(e) r2 = 1 √ 2

• r1 = (0.707)r1

Newton’s Gravity April 19, 2013 - p. 3/8

Inverse Square Law Exercise

When two masses are a distance r1 apart from each other the gravitational force between them is 5 N. They are then moved to a new separation, r2, where the gravitational force between them is 10 N. What is the relation between r1 and r2? (a) r2 = r1 (b) r2 = 2 r1 (c) r2 = √ 2

• r1 = (1.4) r1

(d) r2 = 1 2

• r1

(e) r2 = 1 √ 2

• r1 = (0.707)r1

F2 = GM1M2 r2

2

= GM1M2

• r1/

√ 2 2 = GM1M2 r2

1/2

= GM1M2 r2

1

× 2 = 2F1

Newton’s Gravity April 19, 2013 - p. 4/8

Direction

The gravitational force is an “attractive" force ⇒ each object feels a force towards the other.

Newton’s Gravity April 19, 2013 - p. 4/8

Direction

The gravitational force is an “attractive" force ⇒ each object feels a force towards the other. M1 M2

Newton’s Gravity April 19, 2013 - p. 4/8

Direction

The gravitational force is an “attractive" force ⇒ each object feels a force towards the other. M1 M2

Newton’s Gravity April 19, 2013 - p. 4/8

Direction

The gravitational force is an “attractive" force ⇒ each object feels a force towards the other. M1 M2 − → F

1

− → F

1 - Force on 1 due to 2

Newton’s Gravity April 19, 2013 - p. 4/8

Direction

The gravitational force is an “attractive" force ⇒ each object feels a force towards the other. M1 M2 − → F

1

− → F

1 - Force on 1 due to 2

− → F

2 - Force on 2 due to 1

Newton’s Gravity April 19, 2013 - p. 4/8

Direction

The gravitational force is an “attractive" force ⇒ each object feels a force towards the other. M1 M2 − → F

1

− → F

2

− → F

1 - Force on 1 due to 2

− → F

2 - Force on 2 due to 1

Newton’s Gravity April 19, 2013 - p. 4/8

Direction

The gravitational force is an “attractive" force ⇒ each object feels a force towards the other. M1 M2 − → F

1

− → F

2

− → F

1 - Force on 1 due to 2

− → F

2 - Force on 2 due to 1

Geometry determines direction

Newton’s Gravity April 19, 2013 - p. 4/8

Direction

The gravitational force is an “attractive" force ⇒ each object feels a force towards the other. M1 M2 − → F

1

− → F

2

− → F

1 - Force on 1 due to 2

− → F

2 - Force on 2 due to 1

Geometry determines direction φ

Newton’s Gravity April 19, 2013 - p. 4/8

Direction

The gravitational force is an “attractive" force ⇒ each object feels a force towards the other. M1 M2 − → F

1

− → F

2

− → F

1 - Force on 1 due to 2

− → F

2 - Force on 2 due to 1

Geometry determines direction φ

• f. b. d. for M1

Newton’s Gravity April 19, 2013 - p. 4/8

Direction

The gravitational force is an “attractive" force ⇒ each object feels a force towards the other. M1 M2 − → F

1

− → F

2

− → F

1 - Force on 1 due to 2

− → F

2 - Force on 2 due to 1

Geometry determines direction φ

• f. b. d. for M1

− → F

1

Newton’s Gravity April 19, 2013 - p. 4/8

Direction

The gravitational force is an “attractive" force ⇒ each object feels a force towards the other. M1 M2 − → F

1

− → F

2

− → F

1 - Force on 1 due to 2

− → F

2 - Force on 2 due to 1

Geometry determines direction φ

• f. b. d. for M1

− → F

1

φ

Newton’s Gravity April 19, 2013 - p. 5/8

Direction Exercise

Three masses are arranged in a line with the distance between the second and third double that of the distance between the first and second. If the third mass is twice as large as the other two, what direction is the net gravitational force acting on the middle mass? M M 2M d 2d

Newton’s Gravity April 19, 2013 - p. 5/8

Direction Exercise

Three masses are arranged in a line with the distance between the second and third double that of the distance between the first and second. If the third mass is twice as large as the other two, what direction is the net gravitational force acting on the middle mass? M M 2M d 2d (a) Left

Newton’s Gravity April 19, 2013 - p. 5/8

Direction Exercise

Three masses are arranged in a line with the distance between the second and third double that of the distance between the first and second. If the third mass is twice as large as the other two, what direction is the net gravitational force acting on the middle mass? M M 2M d 2d (a) Left (b) Right

Newton’s Gravity April 19, 2013 - p. 5/8

Direction Exercise

Three masses are arranged in a line with the distance between the second and third double that of the distance between the first and second. If the third mass is twice as large as the other two, what direction is the net gravitational force acting on the middle mass? M M 2M d 2d (a) Left (b) Right (c) Up

Newton’s Gravity April 19, 2013 - p. 5/8

Direction Exercise

Three masses are arranged in a line with the distance between the second and third double that of the distance between the first and second. If the third mass is twice as large as the other two, what direction is the net gravitational force acting on the middle mass? M M 2M d 2d (a) Left (b) Right (c) Up (d) Down

Newton’s Gravity April 19, 2013 - p. 5/8

Direction Exercise

Three masses are arranged in a line with the distance between the second and third double that of the distance between the first and second. If the third mass is twice as large as the other two, what direction is the net gravitational force acting on the middle mass? M M 2M d 2d (a) Left (b) Right (c) Up (d) Down (e) The net force is zero

Newton’s Gravity April 19, 2013 - p. 5/8

Direction Exercise

Three masses are arranged in a line with the distance between the second and third double that of the distance between the first and second. If the third mass is twice as large as the other two, what direction is the net gravitational force acting on the middle mass? M M 2M d 2d (a) Left (b) Right (c) Up (d) Down (e) The net force is zero

Newton’s Gravity April 19, 2013 - p. 5/8

Direction Exercise

Three masses are arranged in a line with the distance between the second and third double that of the distance between the first and second. If the third mass is twice as large as the other two, what direction is the net gravitational force acting on the middle mass? M M 2M d 2d − → F

23

− → F

21

(a) Left F21 = G(M)(M) d2 = GM 2 d2 F23 = G(M)(2m) (2d)2 = 2GM 2 4d2 = 1 2 GM 2 d2

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity.

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity:

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Fg = GMEM r2

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Fg = GMEM r2

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Fg = GMEM r2

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Fg = GMEM r2

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Fg = GMEM r2

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Fg = GMEM r2

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Fg = GMEM r2

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Fg = GMEM r2

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Fg = GMEM r2 RE RE - Earth’s radius

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Fg = GMEM r2 RE RE - Earth’s radius r = RE + h r

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Much less than RE Fg = GMEM r2 RE RE - Earth’s radius r = RE + h r

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Much less than RE Fg = GMEM r2 RE RE - Earth’s radius r = RE + h ≈ RE r

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Much less than RE Fg = GMEM r2 RE RE - Earth’s radius r = RE + h ≈ RE r Fg = GMEM R2

E

Newton’s Gravity April 19, 2013 - p. 6/8

Weight

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Much less than RE Fg = GMEM r2 RE RE - Earth’s radius r = RE + h ≈ RE r Fg = GMEM R2

E

= M

• GME

R2

E

Newton’s Gravity April 19, 2013 - p. 7/8

Acceleration due to gravity

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Much less than RE RE r Fg = M

• GME

R2

E

Newton’s Gravity April 19, 2013 - p. 7/8

Acceleration due to gravity

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Much less than RE RE r Fg = M

• GME

R2

E

• = Mg

Newton’s Gravity April 19, 2013 - p. 7/8

Acceleration due to gravity

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Much less than RE RE r Fg = M

• GME

R2

E

• = Mg
• GME

R2

E

• = 9.782 m/s2

Newton’s Gravity April 19, 2013 - p. 7/8

Acceleration due to gravity

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Much less than RP RP r Fg = M

• GME

R2

E

• = Mg
• GME

R2

E

• = 9.782 m/s2

Newton’s Gravity April 19, 2013 - p. 7/8

Acceleration due to gravity

Weight - Force due to gravity. To relate what we used before (Mg) to Newton’s law of gravity: M h Much less than RP RP r Fg = M

• GME

R2

E

• = Mg
• GME

R2

E

• = 9.782 m/s2

Acceleration due to gravity

• n any planet: g = GMP

R2

P

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over.

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2 Wg = −∆Ug = −U2 + U1

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2 Wg = −∆Ug = −U2 + U1

Force not constant ⇒ integration

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2 Wg = −∆Ug = −U2 + U1

Force not constant ⇒ integration Wg =

r2

• r1

− → F

g · −

→ dr

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2 Wg = −∆Ug = −U2 + U1

Force not constant ⇒ integration Wg =

r2

• r1

− → F

g · −

→ dr

− → F

g

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2 Wg = −∆Ug = −U2 + U1

Force not constant ⇒ integration Wg =

r2

• r1

− → F

g · −

→ dr

− → dr − → F

g

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2 Wg = −∆Ug = −U2 + U1

Force not constant ⇒ integration Wg =

r2

• r1

− → F

g · −

→ dr

− → dr − → F

g

=

r2

• r1

Fg dr cos 180◦

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2 Wg = −∆Ug = −U2 + U1

Force not constant ⇒ integration Wg =

r2

• r1

− → F

g · −

→ dr

− → dr − → F

g

=

r2

• r1

Fg dr cos 180◦ Wg = −

r2

• r1

GM1M2 r2 dr

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2 Wg = −∆Ug = −U2 + U1

Force not constant ⇒ integration Wg =

r2

• r1

− → F

g · −

→ dr

− → dr − → F

g

=

r2

• r1

Fg dr cos 180◦ Wg = −

r2

• r1

GM1M2 r2 dr = −GM1M2

r2

• r1

1 r2 dr

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2 Wg = −∆Ug = −U2 + U1

Force not constant ⇒ integration Wg =

r2

• r1

− → F

g · −

→ dr

− → dr − → F

g

=

r2

• r1

Fg dr cos 180◦ Wg = −

r2

• r1

GM1M2 r2 dr = −GM1M2

r2

• r1

1 r2 dr Wg = −GM1M2

• − 1

r

• r2

r1

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2 Wg = −∆Ug = −U2 + U1

Force not constant ⇒ integration Wg =

r2

• r1

− → F

g · −

→ dr

− → dr − → F

g

=

r2

• r1

Fg dr cos 180◦ Wg = −

r2

• r1

GM1M2 r2 dr = −GM1M2

r2

• r1

1 r2 dr Wg = −GM1M2

• − 1

r

• r2

r1 = GM1M2

r2 − GM1M2 r1

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2 Wg = −∆Ug = −U2 + U1

Force not constant ⇒ integration Wg =

r2

• r1

− → F

g · −

→ dr

− → dr − → F

g

=

r2

• r1

Fg dr cos 180◦ Wg = −

r2

• r1

GM1M2 r2 dr = −GM1M2

r2

• r1

1 r2 dr Wg = −GM1M2

• − 1

r

• r2

r1 = GM1M2

r2 − GM1M2 r1

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. M1 r M2 Wg = −∆Ug = −U2 + U1

Force not constant ⇒ integration Wg =

r2

• r1

− → F

g · −

→ dr

− → dr − → F

g

=

r2

• r1

Fg dr cos 180◦ Wg = −

r2

• r1

GM1M2 r2 dr = −GM1M2

r2

• r1

1 r2 dr Wg = −GM1M2

• − 1

r

• r2

r1 = GM1M2

r2 − GM1M2 r1

Newton’s Gravity April 19, 2013 - p. 8/8

Gravitational Potential Energy

Our previous equation, Ug = Mgy, is valid for distance y ≪ RP . For distances large compared to the radius, we have to start over. Wg = −∆Ug = −U2 + U1 M1 r M2

Force not constant ⇒ integration

Wg = −∆Ug = −U2 + U1

Force not constant ⇒ integration Wg =

r2

• r1

− → F

g · −

→ dr

− → dr − → F

g

=

r2

• r1

Fg dr cos 180◦ Wg = −

r2

• r1

GM1M2 r2 dr = −GM1M2

r2

• r1

1 r2 dr Wg = −GM1M2

• − 1

r

• r2

r1 = GM1M2

r2 − GM1M2 r1

Ug = −GM1M2 r