Sect. 13.4: The Gravitational Field

Gravitational Field• A Gravitational Field , exists at all points in

space.• If a particle of mass m is placed at a point where the

gravitational field is , it experiences a force:

• The field exerts a force on the particle.

• The gravitational field is defined as • Gravitational field = Gravitational Force experienced by a “test”

particle placed at that point divided by the mass of the test particle.

– The presence of the test particle is not necessary for the field to exist

• The source particle creates the field

g

g = mF g g

g

g

mF

g

• The gravitational field vectors point in the direction of the acceleration a particle would experience if it were placed in that field. Figure

• The magnitude is that of the freefall acceleration, g, at that location.

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

g

g

2ˆg GM

m r F

g r

Sect. 13.5: Gravitational Potential Energy

• The gravitational force is conservative• Recall Ch. 8 discussion of conservative forces: Only for

conservative forces can a potential energy U be defined & Total Mechanical Energy is Conserved for Conservative Forces ONLY. Just as in Ch. 8, define the change in Gravitational Potential Energy associated with a displacement of a mass m is as the negative of the work done by the gravitational force on m during the displacement. That is:

F(r) is the Gravitational Force. For a mass m in the Earth’s gravitational field,

f

i

r

f i

r

U U U F r dr

• Doing this integral:

This gives

As always, the reference point where the potential energy is zero is arbitrary. Usually choose it at ri , so (1/ri) 0 and Ui 0.

This gives

• For example, as a particle moves from

point A to B, as in the figure its gravitational potential energy

changes by

U =

UÄ U

Ä GM

Em

1rÄ

1ri

( ) EGM mU r

r

Notes on Gravitational Potential Energy in the Earth’s Gravity

• We’ve chosen the zero for the gravitational potential energy at ri where the gravitational force is also zero.

This means that Ui = 0 when ri or that

This is valid only for r≥RE & NOT for r < RE

– That is, it is valid outside the Earth’s surface but NOT inside it!

– U is negative because of the choice of Ui

( ) EGM mU r

r

Gravitational Potential Energy in Earth’s Gravity

• The figure is a graph of the gravitational potential energy U versus r for an object above the Earth’s surface.

• Note that the potential energy goes to zero as r approaches infinity

• For any two particles, masses m1 & m2, the gravitational potential energy function is

• The gravitational potential energy between any two particles varies as 1/r. (Recall that the force varies as 1/r2)

• The potential energy is negative because the force is attractive & we’ve chosen the potential energy to be zero at infinite separation.

• Some external energy must be supplied to do the positive work needed to increase the separation between 2 objects– The external work done produces an increase in gravitational potential

energy as the particles are separated & U becomes less negative

1 2Gm mU

r

Gravitational Potential Energy:General Discussion

• The absolute value of the potential energy for mass m can be thought of as the binding energy of m. (The energy of binding of m to the object which is attracting it gravitationally).

• Consider two masses m1 & m2 attracting each other gravitationally. If an external force is applied to m1 & m2 giving them an energy larger than the binding energy, the excess energy will be in the form of kinetic energy of m1 & m2 when they are at infinite separation.

Binding Energy

Systems with Three or More Particles• For systems with more than two masses, the total gravitational

potential energy of the system is the sum of the gravitational potential energy over all pairs of particles.

• Because of this, gravitational potential energy is said to obey the superposition principle. Each pair of particles in the system contributes a term to Utotal.

• Example; assume 3 particles as in the figure.

The result is shown in the equation

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

total 12 13 23

1 3 2 31 2

12 13 23

U U U U

m m m mm mG

r r r

Sect. 13.6: Energy Considerations in Satellite Motion

Energy and Satellite Motion• Consider an object of mass m moving with a speed v in the

vicinity of a large mass M– Assume that M >>m as is the case for a small object orbiting

a large one. • The total mechanical energy is the sum of the system’s kinetic

and potential energies.

• Total mechanical energy: E = K +U

• In a system in which m is bound in an orbit around M, can show that E must be less than 0

21

2

MmE mv G

r

• Consider an object of mass m moving in a circular orbit about a large mass M, as in the figure.

• The gravitational force supplies the centripetal force:

Fg = G(Mm/r2) = ma = m(v2/r)

Multiply both sides by r & divide by 2:

G[(Mm)/(2r)] = (½)mv2 Putting this into the total mechanical energy & doing some algebra gives:

=

2

GMmE

r

Energy in a Circular Orbit

21

2

MmE mv G

r

• The total mechanical energy is negative in for a circular orbit.

• The kinetic energy is positive and is equal to half the absolute value of the potential energy

• The absolute value of E is equal to the binding energy of the system

• It can be shown that, for an elliptical orbit, the radius of the circular orbit is replaced by the semimajor axis, a, of the ellipse. This gives:

• The total mechanical energy is negative• The total energy is conserved if the system is

isolated

2

GMmE

a

Energy in an Elliptical Orbit

• Consider an object of mass m projected upward from the Earth’s surface with an initial speed, vi as in the figure.

• Use energy to find the minimum value of the initial speed vi needed to allow the object to move infinitely far away from Earth. E is conserved (Ei = Ef), so, to get to a maximum distance away (rmax) & then stop (v = 0):

(½)mvi2 - G[(MEm)/(RE)] = -G[(MEm)/(rmax)]

We want vi for rmax the right side is 0 so

(½)mvi2 = G[(MEm)/(RE)]. Solve for vi = vescape

This is independent of the direction of vi & of the

object mass m!

Escape Speed from Earth

2 Eesc

E

GMv

R

• The result for Earth can be extended to any planet

• The table gives escape speeds from various objects.

• Note: Complete escape from an object is not really possible– Gravitational force extends to infinity so some force

will always be felt no matter how far away you get

• This explains why some planets have atmospheres and others do not– Lighter molecules have higher average speeds and

are more likely to reach escape speeds

esc

2GMv

R

Escape Speed, General