lecture 25 – gravity today’s...

Lecture 25 – Gravity

Week 14 Assignments: (Due 11/24 {Tuesday} by the end of class) Textbook: HW #11 Chp 6: Q9, Q17, Q23, P4, P11, P28, P69, Chp 8: P47, P54, P99

MasteringPhysics: - Assignment 11 (Due 11/24 {Tuesday} by the end of class)

Week 14 Reading: Chapter 17.1-17.3 & 17.6-17.9 -- Giancoli

Today’s Announcements:

* Midterm extra credit (due the last regular thursday of class):

Acceleration due to Gravity, g

F = GMemo/R2 Me

mo

R

F = mog

= GMemo/R2

- So g is related to the size and mass of the earth !

g = GMe/R2

g

- If you know G and Re then g tells you the mass of the earth (“weighing the earth”)

- So how does this relate to g?

Before (chp 4-8):

(toward Earth’s center [of course])

Clicker Question:

b) 3.6x10-7 N

a) 5.4x10-15 N

c) 0.94 N

1) What is gravitational force between Moon and the Earth ? Assume Mmoon = 7.4x1022 kg, Me = 6.0x1024 kg, R = 3.9x108 m and G=6.67x10-11 N m2/kg2.

d) 2.3x1015 N

e) 2.0x1020 N Me

Mm

R

F = GM1M2/R2

Clicker Question:

b) 3.6x10-7 N

a) 5.4x10-15 N

c) 0.94 N

1) What is gravitational force between Moon and the Earth ? Assume Mmoon = 7.4x1022 kg, Me = 6.0x1024 kg, R = 3.9x108 m and G=6.67x10-11 N m2/kg2.

d) 2.3x1015 N

e) 2.0x1020 N Me

Mm

R

F = GM1M2/R2

Clicker Question:

b) 3.6x105 m/s2

a) 1.4x10-5 m/s2

c) 9.8 m/s2

2) What is the acceleration of the Moon due to the Earth ? Assume Mmoon = 7.4x1022 kg, Me = 6.0x1024 kg and F = 2.0x1020 N

d) 2.7x10-3 m/s2

e) 2.1x102 m/s2 Me

Mm

R

Clicker Question:

b) 3.6x105 m/s2

a) 1.4x10-5 m/s2

c) 9.8 m/s2

2) What is the acceleration of the Moon due to the Earth ? Assume Mmoon = 7.4x1022 kg, Me = 6.0x1024 kg and F = 2.0x1020 N

d) 2.7x10-3 m/s2 ≈ g/3600

e) 2.1x102 m/s2 Me

Mm

R

a = 2.7 x 10-3 m/s2 = 1/3600 of acceleration on Earth’s surface (9.8 m/s2)

Rm = 60 Re " am = g/3600

Newton’s Great Insight

g = GMe/R2

a = 2.7 x 10-3 m/s2 = 1/3600 of acceleration on Earth’s surface (9.8 m/s2)

Rm = 60 Re " am = g/3600

Newton’s Great Insight

g = GMe/R2

Eratosthenes of Alexandria (276 – 194 BC):

•  A librarian from Alexandria

•  Learned that on June 21 in Syene, the noon Sun reaches to the bottom of a well –  Where on earth must

Syene be located?

•  In Alexandria, an obelisk cast a shadow at noon on that day

Size of the Earth

Alexandria

Syene

~ 5,000 stadia

Size of the Earth

Rearth ≈ 6400 km

7o

- Correct to better than 1% in the 200’s BC using only a stick and a royal pacer !

•  Circumearth = (360o/7o)x5,000 stadia ~250,000 stadia

~40,000 km

The Moon’s Orbital Radius

Sun

Earth Moon

De

- By timing lunar eclipses it is possible to estimate the diameter of the moon

- By comparing the angular size of the moon to the linear size, its distance can be estimated

Eratosthenes of Alexandria:

Why Doesn’t the Moon Fall to Earth?

- Horizontal Motion: - The Moon falls “around” the earth

a v

•  Why are astronauts “weightless” when in orbit? Are they out of the Earth's gravity? – No! Gravity is what

keeps them in orbit – Astronauts feel

weightless because they are falling at the same speed as their spacecraft

– There is no force pressing them against the floor

Weightlessness

Orbiting Bodies are Falling Bodies

y x

R R - y

In 1 sec: an object falls 4.9m

x2 + (R - y)2 = R2

x2 + R2 - 2yR + y2 = R2

x2 = 2yR - y2

x ≈ (2yR)0.5

If an object is thrown horizontally so it covers 7.9 km in 1 sec., then at the end of the throw it will have fallen exactly the distance that earth has “curved away” and so it will be the same height above the (center of the) earth!

x ≈ (2(4.9m)(6.4x106m))0.5

x ≈ 7.9x103 m

vorbit ≈ 7.9 km/s

x

Orbiting Bodies are Falling Bodies

y x

R R - y (Mobj v2)/R = (GMe Mobj )/R2

Fcent = Fgrav

vorbit ≈ 7.9 km/s

vorb = (GMe/R)0.5

vorbit = ((6.67x10-11 *6.0x1024)/6.4x106)0.5 = 7.9 km/s

vorb = (GMcenter/R)0.5

vorb = (GM/R)0.5

•  The same force makes things fall down on Earth and keeps the planets in their orbits

•  Orbital motion just an example of projectile

motion

•  Orbiting bodies are just falling bodies! •  height is not what makes it hard to achieve orbit, it is the speed

vsurf earth = 7.9 km/s

Torb = (2πR/vorb) = 85 minutes

a = v2/Rm

v = dist/time = 2πRm/T a = (2πRm/T)2/Rm = 4π2Rm/T2

Fm = Mm[4π2Rm/T2 ] = G MeMm/Rm2

Period of an orbit:

Gravity:

4π2Rm/T2 = G Me/Rm2 or: (4π2/ G Me) Rm

3 = T2

Gravity Explains Kepler’s 3rd Law

T2 ≈ 4π2

G(Mcenter) R3

Valid when: Mcenter >> Morb

Clicker Question:

a) 0 gsurface

3) What is the acceleration due to gravity, geff, experienced by an astronaut orbiting the earth as a distance from its center of 1.5Re? Assume Me = 6.0x1024 kg, Re = 6.4x106 m and G=6.67x10-11 N m2/kg2.

Me

geff = GM/R2

Re

geff

1.5Re

b) 0.44 gsurface

c) 0.67 gsurface

d) 1.5 gsurface

e) 2.25 gsurface

Clicker Question:

a) 0 gsurface

3) What is the acceleration due to gravity, geff, experienced by an astronaut orbiting the earth as a distance from its center of 1.5Re? Assume Me = 6.0x1024 kg, Re = 6.4x106 m and G=6.67x10-11 N m2/kg2.

Me

geff = GM/R2

Re

geff

1.5Re

b) 0.44 gsurface

c) 0.67 gsurface

d) 1.5 gsurface

e) 2.25 gsurface

Clicker Question:

b) 3.2x10-2 km/s

a) 1.4x10-6 km/s

c) 5.4 km/s

4) What is the orbital velocity of Neptune? Assume MN = 1.0x1026 kg, MS = 2.0x1030 kg, R = 4.5x1012 m and G=6.67x10-11 N m2/kg2.

d) 7.4x103 km/s

e) 1.6x106 km/s MS

MN

R

vorb = (GMs/R)0.5

vorb

Clicker Question:

b) 3.2x10-2 km/s

a) 1.4x10-6 km/s

c) 5.4 km/s

4) What is the orbital velocity of Neptune? Assume MN = 1.0x1026 kg, MS = 2.0x1030 kg, R = 4.5x1012 m and G=6.67x10-11 N m2/kg2.

d) 7.4x103 km/s

e) 1.6x106 km/s MS

MN

R

vorb = (GMs/R)0.5

vorb

Gravitational Potential Energy

M1

R ΔU = -Wcon = - ∫ Fcon * dr

M2

Fgrav

ΔU = - ∫ F cos(θ) dr

ΔU = - ∫ GM1M2/r2 cos(180) dr r2

r1

ΔU = -GM1M2(1/r2 - 1/r1)

r2

U = -GM1M2/r2

- Potential energy at infinity = 0

- As move closer to massive object, acceleration increases, kinetic energy increases, so potential energy must decreases (U = [-] )