Chapter 11 – Gravity Lecture 2

• Gravitational potential energy• Escape velocity• Gravitational Field of a point mass• Gravitational Field for mass distributions

– Discrete

– Rod

– Spherical shell

– Sphere

• Gravitational potential energy of a system of particles• Black holes

April 8, 2010

From work to gravitational potential energy.

In the last example, it does not matter on what path the person is elevated to 2 Earth radii above. Only the final height (or distance) matters for the total amount of work performed.

The feature of “conservative force”

Potential Energym

M

r

Force (1,2) = −GmMr1,22

r̂1,2 G =6.67 ×10−11 Nm2 / kg2

Work done to bring mass m from initial to final position.

PE =−W =−rFgd

rr

i

f

∫ =− −GmMr2

⎛⎝⎜

⎞⎠⎟

i

f

∫ dr ⇓

=GmM r−2

i

f

∫ dr =−GmM1rf

−1ri

⎛

⎝⎜

⎞

⎠⎟

Zero point is arbitrary. Choose zero at infinity.

rr

12

1

2

U(r) =−GMmr

Gravitational potential energy

Total energy Etot = K+U = ½ m v2 – G (M m)/r

Apollo 14, at lift-off

Escaping GravityE>0: object is not boundE<0: object is bound to gravity. FieldE=0: kinetic energy just enough to escape gravity (K=U)

04/18/23

Escaping Gravity

• Kinetic energy of the object must be greater than its gravitational potential energy– This defines the minimum velocity to escape

• KE+PE = constant– Consider case when speed is just sufficient to escape to infinity

with vanishing final velocity– At infinity, KE+PE=0, therefore, on Earth,

1

2mvesc

2 −GMEmRE

=0 ⇒

vesc =2GME

RE

=11.2 km/s =25000 mph

Quiz

• You are on the moon and you know how to calculate the escape velocity:

• You find that it is 2.37km/s

A projectile from the moon surface will escape even if it is shot horizontally, not vertically with a speed of at least 2.37km/s

A)Correct

B)Not correct

vesc =2GM

R

Gravity near Earth’s surface...

• Near the Earth’s surface:– R12 = RE

• Won’t change much if we stay near the Earth's surface.– since RE >> h, RE + h ~ RE.

RE

mm

MM

hh FFgFFg

Fg =GMEmRE

2

Gravity...

• Near the Earth’s surface...

• So |Fg| = mg = ma

– a = g

All objects accelerate with acceleration g, regardless of their mass!

Or: the equivalence principle: m=mg = mi

Choosing U(RE) = 0, then

U(h) = m g h , for h << RE

=g

g =GME

RE2 =9.81m/ s2

Variation of g with Height

This is twice the Earth radius: RE = 6000km

We know F should drop with r2

Indeed, “g” has dropped to 9.81/4 m/s2

Question

Suppose you are standing on a bathroom scale in your dorm room and it says that your weight is W. What will the same scale say your weight is on the surface of the mysterious Planet X ?

You are told that RX ~ 20 REarth and MX ~ 300 MEarth.

(a)(a) 00.75 .75 W (b)(b) 1.5 W

(c)(c) 2.25 W

E

X

Fg =GmMr2

Fg,X =300202 Fg,Earth

Gravitational Field

• Gravitational force:

it is a function of space-time (r, t).• Definition of the gravitational field that will act on any

masspoint:

Must be a function of space-time (r, t) concept of “field”.• If the field is caused by a mass distribution we need to

sum over all masspoints as the source.

Fur

12 =Gm1m2

r122 r̂12

gr=

Fg

uru

m gr= g

ri∑Fg

uru

m

Gravitational field

• The gravitational field vectors point in the direction of the acceleration for a particle would experience if placed in that field

• The magnitude is that of the freefall acceleration at that location

• The gravitational field describes the “effect” that any source object M has on the empty space around itself in terms of the force that would be present if a second object m were somewhere in that space

independent of m, only on M !

gr= g

ri∑

Gravitational Field

g1 =g2 =GMr2

Two source mass points M, fieldpoint in plane of symmetry

Magnitude of field due to each mass:

Need to add x and y component of g1 and g2

gx =g1x + g2x =2Gmr2 cos θ( ) =2G

mr2

xp

r==2G

mxp

r3

gy =0

X-component:

Y-component is zero for symmetry reasons

Gravitational Field

gr= dg

r∫

dg =−Gdmr2

dm =λdx=ML

dx

Field due to rod of length L on a point along its axis.

Field by one mass element dm:

r =xp −xs

g = dg= −Gdmr2∫ =−∫ G

ML

dxs

xp −xs( )2

−L2

L2

∫ =...=−GM

xp2 − L

2( )2

Integrate over all mass elements dm:

Gravitational FieldField due to spherical symmetric mass distribution, a shell of mass M and radius R:

Field of a spherical shell

gr=−G

Mr2 r$ r > R

gr=0 r < R

Geometry: spherical shellis 0 anywhere inside (see p.384)

Gravitational FieldField due to homogeneous massive sphere

Field inside the sphere

g =−GMR3 r r < R

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Binding Energy

• The absolute value of the potential energy can be thought of as the binding energy

• At infinite separation, binding energy U=0, thus unbound.• If an external agent applies a force larger than the binding

energy, the excess energy will be in the form of kinetic energy of the particles when they are at infinite separation

04/18/23 20

• The total gravitational potential energy of the system is the sum over all pairs of particles:

simple scalar sum• Gravitational potential energy

obeys the superposition principle

• Each pair of particles contributes a term of Uij

• The absolute value of Utotal represents the work needed to separate the particles by an infinite distance

Systems with Three or More Particles

Potential energy of a system of masses

• What is the total potential energy of this mass system?

m

m

mL

L L

U =−3GmmL

Four identical masses, each of mass M, are placed at the corners of a square of side L. The total

potential energy of the masses is equal to –xGM2/L, where x equals

222.E

2

14.D

24.C

224.B

4.A

Four identical masses, each of mass M, are placed at the corners of a square of side L. The total

potential energy of the masses is equal to –xGM2/L, where x equals

222.E

2

14.D

24.C

224.B

4.A

04/18/23 24

Black Holes• A black hole is the remains of a

star that has collapsed under its own gravitational force

• The escape speed for a black hole is very large due to the concentration of a large mass into a sphere of very small radius– If the escape speed

exceeds the speed of light, radiation cannot escape and it appears black

• The critical radius at which the escape speed equals c is called the Schwarzschild radius, RS

• The imaginary surface of a sphere with this radius is called the event horizon– This is the limit of how close

you can approach the black hole and still escape

04/18/23 Physics 201, UW-Madison 25

Black Holes and Accretion Disks

• Although light from a black hole cannot escape, light from events taking place near the black hole should be visible

• If a binary star system has a black hole and a normal star, the material from the normal star can be pulled into the black hole

• This material forms an accretion disk around the black hole

• Friction among the particles in the disk transforms mechanical energy into internal energy

• The orbital height of the material above the event horizon decreases and the temperature rises

• The high-temperature material emits radiation, extending well into the x-ray region

• These x-rays are characteristics of black holes

Physics 201, UW-Madison 26

Black Holes at Centers of Galaxies

• There is evidence that supermassive black holes exist at the centers of galaxies (M=100million solar masses)

• Theory predicts jets of materials should be evident along the rotational axis of the black hole

An Hubble Space Telescope image of the galaxy M87. The jet of material in the right frame is thought to be evidence of a supermassive black hole at the galaxy’s center.