UNIT 5 (end of mechanics )Universal Gravitation, Energy, Orbits, & SHM

Did not teach SHM 2012-13 to HP

As previously learned, two masses will attract one another with equal and opposite force given by:

221

r

mGmFg

where ‘r’ is distance from center to center

G = 6.67x10-11 Nm2/kg2

this was measured 71yrs after Newton’s death by Cavendish.

Gravity acts as an inverse square law

This describes how the force of gravity varies on as a particle moves away or towards another

• Assumed gravity was constant near surface• Only calculated Fg at a point in space

• Assumed Earth was uniform density

Previous discussion of calculating force of gravity due to Earth.

Recall from honors…What was true of gravitational force inside the Earth? What happened to your weight if you were placed at center of the Earth?

Mathematical analysis of force of gravity inside a planet. Assume planet of mass M and radius R and uniform density. Particle of mass, m, located at r.

r

R

Since planet is of uniform density, ρ, we can say that:

'

'

V

M

V

M

3

'

3

34

34 r

M

R

M

3

3'

R

rMM

)(3

3

22

'

MR

r

r

mG

r

mMGFg 3R

mMrG

Inside planet, it only matters what mass is enclosed within shell. Fg grows linearly as you move towards surface

This represents the mass within the shell

Gravity as a function of distance for a planet of mass M and radius R.

You are heaviest when you are at the surface of a planet.

What if the density of a planet is NON-uniform?Consider a spherical planet whose mass density is given by ρ = br (where b = constant in kg/m4) and has radius, a. Calculate the gravitational force on an object of mass, m, a distance r (where r < a) from the center of the planet.

ρ = M/V where dM = ρ dV dV = surface area x thickness

r

drrrdMM0

24)(

r

brdrrbrM0

42)(4

dV = (4πr2) dr

22

4

221 )(

bmrGr

brmG

r

mmGFg

for r < a

For the outside of the planet the gravitational force would be

a

badrrbrM0

42)(4

2

4

2

4

221

r

baGm

r

bamG

r

mmGFg

for r > a

The Gravitational Field• During the 19th century, the notion of the “field”

entered physics (via Michael Faraday). • Objects with mass create an invisible disturbance in

the space around them that is felt by other massive objects - this is a gravitational field. In Newton’s time, there was much discussion about HOW gravity worked - how does the Sun, for instance, reach across empty space, with no actual contact at all, to exert a force on the Earth?

• This spooky notion was called “action at a distance.”

• We learned last year about gravity and Einstein.

g-fieldThe region around a massive object in which another object having mass would feel a gravitational force of attraction is called a gravitational field. The gravitational field's strength at a distance r from the center of an object of mass, M, can be calculated with the equation.

M

2

2

r

MGg

r

mMGmg

FF gg

Gravitational Potential Energy (Ug)We must derive an expression that relates the potential energy of an object to its position from Earth since ‘g’ is NOT constant as you move away from Earth. All previous problems dealt with objects ‘close’ to the earth where changes in ‘g’ were negligible.

Also, we have referred to the PE of an ‘object’, but in reality it is the PE associated with 2 objects…PE of a system. We have taken h=0 to be surface of earth.

We pretended that the object possessed PE by itself (mgh), rather it possessed PE due to its relationship with the Earth. It is the change in PE that is meaningful, not the value of PE at a point in space. We always took PE to be zero at surface of earth.

If we were to move that particle closer to the earth to a new distance, r2, then gravity would do + work on the particle.

r1

r2

2

1

)(r

r

drrFW

Consider a particle of mass, m, a very far distance, r1 , away from the center of the earth (M).

drrdrr

GMmWrr

22

1

r

drr

mMGW

2

We are moving the particle from very far away to a distance, r, from the center of the planet. Therefore, our integral setup and limits would be

rrGMmU 1

r

GMmrU )(

U is always negative for a finite r. As you move away, ΔU=+ and increases as it becomes less negative. Zero is defined infinitely far away.

Don’t get hung up on the minus. Even with mgh, you can have negative PE if you assign 0 to be above you on earth

This is the PE associated with any 2 particles. When particles are separated by a distance r, an external agent would have to supply an energy at least equal to +GmM/r in order to separate them by an infinite distance.Where U(∞) = 0

If we had integrated in the opposite direction, then formula would have been positive, however FG would have opposed movement where minus sign would have to be factored back in due to the work done yielding same result.

r

F

All values of the potential energy are negative, approaching the value of zero (becoming less negative) as R (distance from planet) approaches infinity. Because of this we say that the mass is trapped in an "energy well" - that is, it will have to be given additional energy from an external source if it is to escape gravity's attraction.

P.E. formula from before, U = mgh

)11

(of

E rrmGMU

Using new formula, we should be able to get the old expression for PE of mass, m, near the surface of earth. Consider moving from 1st floor to 2nd floor (Δh),

)(fo

ofE rr

rrmGM

)(2E

E R

hmGM

hrr of where2Eof Rrr

2E

EE R

GMg

Recall

!tadah

hmgU Algebra removes negative sign from G

Since distance above Earth is small compared to RE

Another way to see if new U(g) formula makes sense…

dx

dUF RECALL

dr

rdUFG

)(

2r

GMm

r

GMm

dr

dFG

- means FG points opposite of increasing r

Previous problems involving potential energy of ‘an object’ involved an object associated with the earth where each was assumed a point mass, uniform spherical objects, or having large distances between the COM of 2 objects. It turns out that only uniform spherical mass distributions act like point masses where COM can be used. This is what took Isaac Newton years to prove. (you can look up derivations of this in my calc-based textbooks (university physics) which goes beyond scope of this class).

What if we want to find the potential energy of a system between non-spherical situation (instead of usual object and the earth)?

Ug between thin rod and sphere (pt mass)

A thin, uniform rod has length, L, and mass, M. A point mass, m, is placed a distance, x, from one end of the rod, along the axis of the rod. Calculate the gravitational potential energy of the rod-sphere system. X is not >> L.

r

dMGmdU g

)(

dM

Consider just a small slice of the rod (no rod present), so small that we can treat it as a point mass, dM. Since we are essentially dealing with two point masses, the potential energy is given by

Now, imagine we move more slices (dM) from far way into position, creating the full rod again. We could sum up the energy dU associated with each small slice dM by integrating. It takes work to do this.

r

GmdMdU

Can’t use center of mass for rod…only works for spheres and pt masses

dM

r

The reason why COM can be used when dealing with spheres BUT NOT a rod:

You could argue that a rod would be like a slice through the diameter of the sphere. Therefore, why does a sphere reduce to a point mass and rod does not?

Envision a sphere squeezed to a pancake from top to bottom, then squeezed from left to right. The shape would be rod-like, but it would be uneven in density. Meaning, it would be more dense near the center and less dense at the ends assuming diameter stays as the length. The more mass near the middle makes up for the fact that the mass at the ends is not same distance from the pt mass….evens it out to allow for using COM. Rod can’t do that.Since force varies as inverse square with same mass throughout for rd, it doesn’t work for a rod by using COM.

r

dMGm where dM = λdr since λ is linear

mass density

Lx

x r

drGm

x

Lx

L

GmM

x

LxGmU

lnln

B) Using this expression, use calculus to find the expression for the force of gravity between the rod and the sphere using F = - dU / dx

You can’t use center of mass of rod and just plug in…not the same answer. Formula on left reduces to formula on right when ‘x’ is very large where the length of rod is insignificant.

dx

dUF

xLx

GMmFg )(

2r

GMmF

Calc derivation in notes

Kepler's 1st Law: The Law of Elliptical Orbits 

All planets trace out elliptical orbits. When the planet is located at point P, it is at the perihelion position. When the planet is located at point A it is at the aphelion position. The distance PA = RP + RA is called the major axis.

Semi-major axis is half of the major axis. Eccentricity is a measure of how "oval" an ellipse is.

A line from the planet to the sun sweeps out equal areas of space in equal intervals of time where dA/dt = constant

At the perihelion, the planet’s speed is a maximum. At the aphelion, the planet’s speed is a minimum.  

Kepler's 2nd Law: Law of Areas 

Angular momentum is conserved since net torque is zero. Gravity lies along the moment arm or position vector, r.  vARA = vPRP

The square of the period of any planet is proportional to the cube of the avg distance from the sun.

Kepler's 3rd Law: Law of Periods 

r

mv

r

GMm

maF cg

2

2

After some trivial substitution…

32

32

2 4

krT

rGM

T

Since the orbits of the planets in our

solar system are EXTREMELY close to being circular in shape (the Earth's eccentricity is 0.0167), we can set the centripetal force equal to the force of universal gravitation and,

Energy Considerations for circular Satellite Motion

UKE

r

MmGmvE 2

2

1

r

mv

r

GMm

maF cg

2

2

Furthermore, if we consider the system to be isolated, then energy is conserved. We can show that the total energy of a satellite system is negative.

2

2

1

2

1mv

r

GMm

Multiplying both sides by ½ r

r

GMmE

2

Note that K is positive and equal to half the magnitude of U.Subbing above into K

term in the E eqn

K will never be negative, however, both U and E can be negative. No deep significance to this, just a consequence of choosing infinity to be zero PE.

Changing the orbit of a satellite

r

GMmE

2

E

Eo R

mGME

4

E

Ef R

mGME

6

Calculate the work required to move an earth satellite of mass m from a circular orbit of radius 2RE to one of radius 3RE.

Using our expression from before for total energy:

We get the energies at both orbits:

E

Eof R

mGMEEEW

12

Solving for the work done:

Makes sense work by external source is positive since FEXT and d are both away from planet

How was the energy distributed in changing the orbit?

The initial and final potential energies were:

R

GMmUo 2

R

GMmU f 3

R

GMmU

6

Therefore,

The kinetic energies were:

R

GMmKo 4

Kr

GMmmv

22

1 2

R

GMmK f 6

R

GMmK

12

Part of the work goes into increasing U and the other part goes into decreasing K.

Meaning, we only looked at TOTAL E on last slide

Wkshts energy and orbits

*If object is given initial speed of escape speed, its total energy is equal to zero since both K and U will be zero. If vo > vesc than object will have residual K at r = ∞.

max

2

2

1

r

mMG

R

mMGmv p

p

po

pescape R

GMv

2

Escape VelocityIf an object of mass m is launched vertically upwards with speed vo we can use energy to determine minimum speed to escape planet’s gravitational field. Our parameters are when the object can just reach infinity with zero speed.

Rp is the distance between object and earth at launch

Where the rmax = infinity

)11

(2max

2

rRGMv

ppo 0

Example: Two satellites, both of mass m, orbit a planet of mass M. One satellite is elliptical and one is circular. Elliptical satellite has energy E = -GMm/6r.

M

a) Find energy, E, of circular satellite

r

GMmmvE 2

2

1

r

mvF

r

GMmF cg

2

2

r

GMm

r

GMm

r

GMmE

2)(

2

1

Solve for v2 and sub into K term to get

R

r

b) Find speed of elliptical satellite at closest approach (perihelion)

r

GMmmvEelliptical 2

2

1

r

GMmmv

r

GMmr 2

2

1

6

2

2

1

6

5rmv

r

GMm r

GMvr 3

5

Given in problem

R

r

c) Derive an equation that would allow you to solve for R but do not solve it.

Can’t use Fg = Fc since orbit isn’t circular

R

GMmmv

r

GMmR 2

2

1

6

Rmvrmv Rr Rr LL

Rr vvR

r

r

GMvr 3

5

Subbing vr answer from part b

R

r

r

GMvR 3

5

R

GMm

R

r

r

GMm

r

GMm

2

2

3

5

2

1

6

RR

r

r

1

6

5

6

12

Orbit wkshts

Example: A satellite of mass ms is in an elliptical orbit around a planet of mass mp which is located at one focus of the ellipse. The satellite has a velocity va at the distance ra when it is furthest from the planet.

Note: rp and vp not given

Derive expression for the magnitude of the velocity vp of the satellite when it is closest to the planet? Express your answer in terms of ms, mp, G, va, and ra as needed. This will incorpate your algebra skills!

p

aap

aapp

v

rvr

rmvrmv

LL

aa

pp r

GMmmv

r

GMmmv

UKUK

EE

22

2

1

2

1

paap r

GM

r

GMvv 22

2

1

2

1

paap r

GM

r

GMvv

2222

p

aaaap

vrvGM

r

GMvv

2222

aa

p

aa

aapap rv

GMv

rv

GMvvvvv

22))((

aa

apapap rv

vvGMvvvv

)(2))((

aaap rv

GMvv

2

aaa

p vrv

GMv

2

Black hole + into wormhole dvd on hawking dvd

• Phet lab

Periodic Motion & S.H.M.

Repeating oscillation or vibration is called periodic

Since oscillating mass repeats itself after a period, where t = T, it must be at the same position and moving in the same direction as it was at t = 0.

ωt = θ ωT = 2π rads

Since cosine or sine repeats itself every 2πrad,we can equate that with

ω = 2πf

where ω is the angularfrequency

We can relate periodic motion with an object undergoing uniform circular motion. Consider a mass sitting on top of a rotating platform where lights cast a shadow of mass on wall.

Shadow of rotating mass will move back and forth on wall exhibiting periodic motion.

As the mass rotates with a constant angular speed, ω, through an angle θ in a time t, we can describe the linear movement across the screen as

x = Acosθ

Since θ = ωt we can relate the position of the object with respect to time and angular movement as

x(t) = Acos(ωt)

where Ф is called the phase angle which represents the initial angular position at t = 0

(ωt + Ф ) = the phase of motion

A = amplitude from equilibrium

)cos()( tAtx(full version)

NOTE: ALL calculations with above equation must be done in RADIAN MODE.

xxdx

dcos)(sin xx

dx

dsin)(cos

Trig Calculus

Cxxdx cossin Cxxdx sincos

Caxa

dxax cos1

)sin( Caxa

dxax sin1

)cos(

axaaxdx

dcos)(sin axaax

dx

dsin)(cos

)sin()( tAtvdt

dx

)cos()( 2 tAtadt

dv

Expressions for velocity and acceleration:

)cos()( tAtx

NOTE: ALL calculations with above equation must be done in RADIAN MODE.

SIMPLE HARMONIC MOTIONA special case of periodic

When periodic motion possesses a restoring force that is directly proportional to the negative of displacement, this is said to be SHM.

x=0

2

2

dt

xdmkx

maF

Fs v

This is called a 2nd order differential equation.

Will the equation of position, x(t), satisfy the above condition? If so, then those equations of motion are simple harmonic.

Recall minus means acc opposes disp

Consider a mass oscillating on a frictionless surface

)cos()( tAtx)cos()( 2 tAta

2

2

dt

xdmkx

))cos(( tAk )cos(2 tAm

Inserting our expressions for both position and acceleration into the differential equation above…

m

k2Yields…

which is a solution to the differential equation.

Note that frequency and period do not depend on the amplitude. The frequency at which particle oscillates is called the natural frequency or resonant frequency.

m

k

mk /2 xa 2

2

2

dt

xdmkx

Since

And

then for harmonic oscillators,

•At maximum displacement,Fs = max, v = 0 & a = max

•At equilibrium Fs = 0,

v = max & a = 0

•Repeat of part ‘a’

Energy considerations for SHM

A mass displaced a distance A (amplitude) will oscillate about that distance assuming no friction.

E = ½ kA2

Total energy could also be written as

E = ½ mv2max

E = K + U where ½ kA2 = ½mv2 + ½kx2 v at some arbitrary position, x

Example: A 0.500-kg mass is vibrating in a system in which the restoring constant is 100 N/m; the amplitude of vibration is 0.20 m. Find:a. the energy of the systemb. the maximum kinetic energy and maximum velocityc. the PE and KE when x = 0.100 md. the maximum acceleratione. the equation of motion if x = A at t =0

a) E = ½ kA2 = 2.0J

b) Kmax = E = 2.0J

b) vmax = 2.83m/s

c) U = ½ kx2 = 0.5J K = E - U = 1.5J

d) amax =| -ω2 A |= 40m/s2

e) x(t) = 0.20 cos (14.1t)

If a mass is at rest, suspended from a spring, the mass is in equilibrium. The only forces acting on the mass are a gravitational force down, mg, and the spring force up, kxo, where xo is the stretch of the spring at equilibrium.

Does a vertical spring obey SHM due to gravitational force?

If the mass is now displaced from that equilibrium position by some displacement x, the net force acting on the mass would be the extra force due to this further displacement x. Newton's second law would yield

This fits the form for SHM. Gravitational force doesn’t affect the SHM of the mass

makxFnet mkxa /

Mathematically, kd = mg when mass is in equilibrium where d = stretch. If mass is displaced downward a distance x beyond d, the restoring force within spring would be

Fs = k (d + x) 2nd Law would follow as

Fs-mg = ma and substitution yields

k (d + x) – mg = ma

where kd + kx –mg = ma

Since kd = mg from above

mg + kx – mg = ma and kx =ma where it is obvious that –(k/m)x = a where force opposes displacement

A body of mass m is suspended from a spring with spring constant k in configuration (i) and the spring is stretched 0.1m . If two identical bodies of mass m/ 2 are suspended from a spring with the same spring constant k in configuration (ii), how much will the spring stretch? Explain your answer.

Spring on left stretches x=mg/k

Spring on right stretches x/2 =mg/2k. Why?

Recall scale between 2 masses hanging from pulleys. Mass on right holds it in place. Mass on left stretches spring

5 worksheets

A block is moving undergoing SHM having amplitude A0. At instant block passes thru equilibrium, putty of

mass M is dropped vertically onto block and it sticks. Find new amplitude and period

k

Eo = ½ mvo2 = ½ kA0

2, so vo = 0Am

kGet expression for speed at x=0

To find new amplitude

It’s a collision where x-component of momentum is conserved where new speed is… mvo + 0 = (m + M)v ovMm

mv

K is not conserved due to inelastic collision

22

2

)(

)(

2

1ovMm

mmM

E = ½ (M+m) v2 = )2

1(

)(2omv

Mm

m

oEMm

mE

)(

2

2

1kAE )

2

1)((

2

1 22okA

Mm

mkA

Energy in system is now

)( Mm

mAA o

)2

1)((

2

1 22okA

Mm

mkA

k

mM

k

mT

22

m

k

ω = 2πf = 2π/T

1/ω = T/2π where

Amplitude is less

T is greater. More inertia.

Is a simple pendulum a S.H.O.?

It is if the restoring force is proportional to x or θ

ΣF = ma Fgx = mgsinθ

In the above equation, Fgx is proportional to sinθ, not θ, therefore not SHM. However, when θ is small, sinθ is very nearly equal to θ (in radians) where arc length s=Lθ is nearly the same as the chord length (x=Lsin θ)….therefore

x

F = -mgsinθ ≈ -mgθ where θ = x / L

F = - mgx / L

F ≈ -x we can say it is SHM for small angles

L

mgk

We let (mg/L) represent ‘k’ in Hooke’s law equation where

L

g

Lm

mg

m

kthen

T

2since

g

LTp 2then

Where – sign indicates force tends to decrease θ

if

Physical pendulumAny rigid body that is suspended from a fixed point that does not pass through the COM and oscillates about that point.

Pivot Inet dIFr

2

2

sindt

dImgd

dsinθ

Where – sign indicates torque tends to decrease θ

dsinθ

sindr

2

2

dt

d

I

mgd

Again, if we assume small angles

xdt

xd 22

2

Since

then 2)( I

mgd

Imgdand

Finally mgdIT

22

2

2

sindt

dImgd

d = Distance from pivot

Example: A uniform rod of mass M and length L is pivoted about one end and oscillates in a vertical plane. Find the period of oscillation if the amplitude is small.

Mg

mgdIT 2

2

3

1MLI

g

LL

Mg

MLT

3

22

)2

(

31

2

2

d

Torsional PendulumA mass suspended to a fixed support by a thin wire can be made to twist about its axis. This is known as a torsional pendulum. The mass attached to the wire rotates in the horizontal plane where θ is the angle of rotation. It is the twisting of the wire that creates a restoring torque due to the resistance of the wire to the deformation.

For small angles of θ the magnitude of the restoring torque is proportional to the angle θ

Where κ is the torsion constant of support wire

2

2

dt

dII

2

2

dt

dI

22

2

dt

d

I

Again we have a format that matches SHM

IT 2

What happens if you could jump into a tunnel in the earth? Assume the Earth to be a solid sphere of uniform density, ρ, mass, M, and radius R.  A hole is drilled through from pole to pole. A person of mass m is dropped in the hole with zero initial velocity. The force acting on you as derived earlier…

Journey through the Earth!

rR

MmGF

3

Does this force act as simple harmonic?

marR

MmGF 3 2

2

3)(

dt

xdr

R

MG

2

2

3)(

dt

xdr

R

MG

xdt

xdx

R

MG 2

2

2

3

23

R

MG

It turns out that you would behave like a harmonic oscillator. Determine the period of oscillation from pole to pole in minutes.

623

1052.1 xR

MG min3.422/

T

Turns out that it takes the same time to fall through tunnel whether its pole to pole or LA to NY…doesn’t matter…always 42.3 minutes!

Now determine the period of low orbit satellite skimming just above surface of earth, ignoring friction.

32

2 4r

GMT

Using Kepler’s 3rd Law we get…

In fact we end up showing circular motion mimics the motion of mass inside earth

Set up this demo. 2 springs tied together. Another string tied to crossbar is tied to bottom spring. String tied to top spring is tied to mass.

http://video.mit.edu/watch/spring-paradox-6471/