msu physics 231 fall 2015 1 physics 231 topic 9: gravitation alex brown october 30, 2015

MSU Physics 231 Fall 2015 1

Physics 231Topic 9: Gravitation

Alex BrownOctober 30, 2015

MSU Physics 231 Fall 2015 2

What’s up? (Friday Oct 30)

1) The correction exam is now open and is due at 10 pm Tuesday Nov 3th.

The exam grades will available on Wednesday Nov 4..

2) Homework 07 is due Tuesday Nov 10th and covers Chapters 9 and 10. It is

a little longer that usual so you may want to start early.

MSU Physics 231 Fall 2015 3

MSU Physics 231 Fall 2015 4

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Key Concepts: Gravitation

Newton’s Law of Gravitation

Gravitational Acceleration

Planetary MotionKepler’s Laws

Gravitational Potential Energy

Conservation of ME

Artificial Satellites

Covers chapter 9 in Rex & Wolfson

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The gravitational force

221

r

mmGF

G=6.673·10-11 N m2/kg2

N m2/kg2 = m3/(kg s2)

Newton:

The gravitational force works between every two massesin the universe.

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Gravitation between two objects

AB

The gravitational force exerted by the sphericalobject A on B can be calculated as if all of A’s mass would is concentrated in its center andlikewise for object B.

Conditions: B must be outside of AA and B must be ‘homogeneous’

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The force of the earth on the moon is equal and opposite to the force of the moon on the earth!

Gravitation between two objects

221

r

mmGF

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Clicker Quiz! Earth and Moon

a) one quarter

b) one half

c) the same

d) two times

e) four times

If the distance to the Moon were

doubled, then the force of attraction

between Earth and the Moon would be:

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The gravitational force depends inversely on the

distance squared. So if you increase the distance by

a factor of 2, the force will decrease by a factor of 4.

Clicker Quiz! Earth and Moon

a) one quarter

b) one half

c) the same

d) two times

e) four times

If the distance to the Moon were

doubled, then the force of attraction

between Earth and the Moon would be:

2r

MmGF

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Gravitational acceleration at the surface

of planet with mass M

2R

mMGF F=m

g

g = GM/R2

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Gravitational acceleration at the surface of the earth

g=GMe/Re2

G = 6.67x10-11 Me=5.97x1024 kg

Radius from Earth’s Center (km)

Gravitational Acceleration (m/s2)

Earth’s Surface 6366 9.81

Mount Everest 6366 + 8.85 9.78

Mariana Trench 6366 - 11.03 9.85

Polar Orbit Satellite 6366 + 1600 6.27

Geosynchronous Satellite

6366 + 36000 0.22

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Gravitational potential energySo far, we used: PEgravity=mgh Only valid for h near

earth’s surface.

More general: PEgravity=-GMm/r

R

Earlier we noted that we could define the zero of PEgravity anywhere we wanted. So the surface of the earth is as good as anywhere!

22

1

R

MmGPE

R

MmGPE

R < R2

Thus…

PE < PE2

R2=R+h

MSU Physics 231 Fall 2015 14

Gravitational potential energySo far, we used: PEgravity=mgh Only valid for h near

earth’s surface.

More general: PEgravity=-GMm/r

R

Earlier we noted that we could define the zero of PEgravity anywhere we wanted. So the surface of the earth is as good as anywhere!

mghhR

MGm

hRRmGM

R

MmG

hR

MmG

R

MmG

R

MmGPEPE

2

212

)11

(

)()(

)(

R2=R+h

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Gravitational potential energyPEgravity=mgh only valid for h near earth’s surface.

More general: PEgravity=-GMm/r

PE = 0 at infinite distance from the center of the earth (r = ∞)Application: what should the minimum initial velocity of a rocket be if we want to make sure it will not fall back to earth?

KEi + PEi = ½mv2 - GMm/R

KEf + PEf = 0 v = (2GM/R) = (2gR) = 11.2 km/s

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Second cosmic speed

Second cosmic speed: speed needed to break free from a planet of mass Mp and radius Rp (gp = GMp/Rp

2)

v2 =(2GMp/Rp) = (2gpRp)

For earth: g = 9.81 m/s2 R = 6.37x106 m v2 = 11.2 km/s

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Orbital Velocities

What does the word orbit mean?

An orbit is the gravitationally curved path of an object around a point in space.

To orbit the object, you need to satisfy the kinematic conditions of that type of orbit (more on this shortly…)

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4 km/s 6 km/s 8 km/slaunchspeed

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First cosmic speed

First cosmic speed: speed of a satellite of mass m on a low-lying circular around a planet with orbit of Mp and radius Rp (gp = GMp/Rp

2)

rsatellite ≈ Rp

F = mac mgp = m v2/Rp so v1 =(gpRp)

For earth: g = 9.81 m/s2 R = 6.37x106 m v1=7.91 km/s

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Period for all orbitsConsider an object in circular motion around a larger one

32

2

332

2

22

2

2

4

4

2

KrT

GMK

KrrGM

T

rT

mrmr

mv

r

GMm

maF c

M

m

r

v

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/10297sun)(

1099.1sun)(3221

30

msK

kgM

Two common cases

Planets and other objects orbiting the sun

Moon and satellites orbiting the earth

/1099 earth)(

1097.5earth)(3215

24

msK

kgM

GM

K24

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e) The Moon does not crash into Earth because of its high

speed. If it stopped moving, it would fall directly into

Earth. With its high speed, the Moon would fly off into

space if it weren’t for gravity providing the centripetal

force.

Clicker Quiz! Averting Disaster

The Moon does

not crash into

Earth because:

a) it’s in Earth’s gravitational field

b) the net force on it is zero

c) it is beyond the main pull of Earth’s gravity

d) it’s being pulled by the Sun as well as by Earth

e) its velocity is large enough to stay in orbit

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The Moon does not crash into Earth because of its high

speed. If it stopped moving, it would fall directly into

Earth. With its high speed, the Moon would fly off into

space if it weren’t for gravity providing the centripetal

force.

Clicker Quiz! Averting Disaster

The Moon does

not crash into

Earth because:

a) it’s in Earth’s gravitational field

b) the net force on it is zero

c) it is beyond the main pull of Earth’s gravity

d) it’s being pulled by the Sun as well as by Earth

e) its velocity is large enough to stay in orbit

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Synchronous orbit

Synchronous orbit of a satellite: rotation period of satellite of mass m is the same as rotation period of the planet

For earth: period T = 24 hours = 86 x 103 s

r3 = T2/K = 75 x 1021

r = 42 x 106 m

Re = 6.4 x 106 m

(r/Re) = 6.6

/1099 earth)( 3215 msK

GMK

KrT2

32

4

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Total mechanical energy for Orbits

Consider a planet in circular motion around the sun:

M

m

r

v

r

GMmTE

r

GMm

r

GMm

r

GMmKEPETE

r

GMmmvKE

r

GMv

r

mv

r

GMm

maF c

2

22

22

2

2

2

2

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launch speed = 10 km/s

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Kepler’s laws

Johannes Kepler(1571-1630)

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Kepler’s First law

An object A bound to another object B by a force that goes with 1/r2 moves in an elliptical orbit around B, with B beingin one of the focus point of the ellipse.

A B

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area abA

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MSU Physics 231 Fall 2015 31

2

1

a

b

a

ce

Eccentricity: e

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2

1

a

b

a

ce

Eccentricity: e

circle when e = 0

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Kepler’s First law

2

1

a

b

a

ce

An object A bound to another object B by a force that goes with 1/r2 moves in an elliptical orbit around B, with B beingin one of the focus point of the ellipse.

Eccentricity: e

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Kepler’s second lawA line drawn from the sun to the elliptical orbit of a planetsweeps out equal areas in equal time intervals.

Area(D-C-SUN) = Area(B-A-SUN)

)area(A 2

revolution onefor 2

momentumangular L

constant 2

abT

mAL

m

L

T

A

m

L

t

A

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Kepler’s second law

rmin speed and kinetic energy are largest

rmax

speed and kinetic energy are smallest

PEgravity=-GMEarthm/r

rmaxrmin

constant vrmL

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Kepler’s third lawPEgravity=-GMEarthm/r

3221

32

/10297 msK

aKT

same as for a circular orbit except r is replaced by

the semi-major axis (red line)

a = ½(rmin+rmax)

a

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An Example

A

Bstar

Two planets are orbiting a star. The orbit of A has a radius of 1x108 km.

The distance of closest approach of B to the star is 5x107 km and its maximum distance from the star is 1x109 km.

If A has a rotational period of 1 year, what is the rotational period of B?

Need to use Kepler’s 3rd Law

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An Example

A

Bstar

RA = 1x108 kmRB = ½(Rmin+Rmax) = ½(5x107 + 1x109) = 5.25x108 km

Rmin = distance of closest approach = 5x10710(perihelion)Rmax = maximum distance = 1x109 (aphelion)

R3/T2 = constant RA3/TA

2= RB3/TB

2 so TB2=(RB

3/RA3)TA

2

So TB=(5.253 x (1 yr)2) = 12 years

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Quantities that are constant for a given orbit

ellipse area

2 momentumangular

abAT

AmL

a

b

32 period aKT

a

GMmTE

2energy total

2/12/32/1)(

)(22

a

b

aK

abm

T

mAL

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Quantities that depend on distance r

a

b

2

2

)2/1(energy kinetic

locity angular ve

velocity

energy potential

mvKE

mr

Lmr

Lv

r

MmGPE

r

MSU Physics 231 Fall 2015 41