LaPlace's and Poisson's Equations A useful approach to the calculation of electric potentials is to relate that potential to the charge density which gives rise to it. The electric field is related to the charge density by the divergence relationship and the electric field is related to the electric potential by a gradient relationship Therefore the potential is related to the charge density by Poisson's equation In a charge-free region of space, this becomes LaPlace's equation This mathematical operation, the divergence of the gradient of a function, is called the LaPlacian. Expressing the LaPlacian in different coordinate systems to take advantage of the symmetry of a charge distribution helps in the solution for the electric potential V. For example, if the charge distribution has spherical symmetry, you use the LaPlacian in spherical polar coordinates. Since the potential is a scalar function, this approach has advantages over trying to calculate the electric field directly. Once the potential has been calculated, the electric field can be computed by taking the gradient of the potential. Example: sphere of uniform charge

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Potential of a Uniform Sphere of Charge The use of Poisson's and Laplace's equations will be explored for a uniform sphere of charge. In spherical polar coordinates, Poisson's equation takes the form: but since there is full spherical symmetry here, the derivatives with respect to θ and φ must be zero, leaving the form Examining first the region outside the sphere, Laplace's law applies. Since the zero of potential is arbitrary, it is reasonable to choose the zero of potential at infinity, the standard practice with localized charges. This gives the value b=0. Since the sphere of charge will look like a point charge at large distances, we may conclude that so the solution to LaPlace's law outside the sphere is Now examining the potential inside the sphere, the potential must have a term of order r2 to give a constant on the left side of the equation, so the solution is of the form Substituting into Poisson's equation gives Now to meet the boundary conditions at the surface of the sphere, r=R The full solution for the potential inside the sphere from Poisson's equation is

This activity will illustrate how to calculate the escape velocity of planets, satellites and the Sun.
Time requirement: 50 minutes as an activity for older or mathematically advanced students.

Materials

• Calculator (with square root function)

Procedures

You have learned that the escape velocity (v_esc) of a body depends on the mass (M) and the radius (r) of the given body. The formula which relates these quantities is:

v_esc = (2 * G * M / r)^1/2

where G is called the Gravitational constant. The notation

(2 * G * M / r)^1/2

means (2 * G * M / r) to the one-half power, which is equal to the square root of (2 * G * M / r).
You will calculate the escape velocity for a number of bodies using the MKS system where the units for distance are meters, the units for mass are kilograms, and the units for time are seconds. In this system, the gravitational constant has the value:

G = 6.67 * 10^-11 Newton-meter²/kilogram².

As an example, the mass M of the Earth is 5.98 * 10²⁴ kilograms. The radius r of the Earth is 6378 kilometers, which is equal to 6.378 * 10⁶ meters. The escape velocity at the surface of the Earth can therefore be calculated by:

vesc	=	(2 * G * M / r)1/2
	=	( 2 * (6.67 * 10-11) * (5.98 * 1024) / (6.378 * 106) ) 1/2
	=	1.12 * 104 meters/second
	=	11.2 kilometers/second

So, as with surface gravity, a simple Physics equation can be used to calculate the escape velocity for a body (in this case the Earth) if you know the mass of the body and its radius! The assumption in using this formula is that the body is spherical, but this is a pretty good assumption. If the radius of a body at its equator and pole are very different, then the escape velocity is different at those places and should be calculated separately.
The escape velocity for the Earth is therefore 11.2 kilometers per second. This is the velocity that an object (or gas molecule!) needs at the surface of the Earth to be able to overcome the gravitational attraction of the Earth and escape to space.
A table of masses and radii is given below for many bodies in the Solar System. Make sure to convert the radii from kilometers to meters when making the calculation, and make sure that you can calculate the escape velocity of the Earth correctly. Then, calculate the escape velocity at each of the other bodies.

Body	Mass (kg)	Radius (km)
Earth	5.98 * 1024	6378
Mercury	3.30 * 1023	2439
Venus	4.87 * 1024	6051
Mars	6.42 * 1023	3393
Jupiter	1.90 * 1027	71492
Saturn	5.69 * 1026	60268
Uranus	8.68 * 1025	25559
Neptune	1.02 * 1026	24764
Pluto	1.29 * 1022	1150
Moon	7.35 * 1022	1738
Ganymede	1.48 * 1023	2631
Titan	1.35 * 1023	2575
Sun	1.99 * 1030	696000

Note that the Gas Giant planets (Jupiter, Saturn, Uranus and Neptune) do not have solid surfaces. The radii of these planets are specified at the point where the pressure in their atmospheres is approximately equal to that at the surface of the Earth. As one last exercise, convert the escape velocity of the Earth to kilometers per hour (or miles per hour) to get a good feeling for how much initial velocity an object must really have in order to escape the gravitational force of our planet!

Escape Velocity If the kinetic energy of an object launched from the Earth were equal in magnitude to the potential energy, then in the absence of friction resistance it could escape from the Earth. Escape velocity from the Earth If M =  MEarth and r =  rEarth then vescape =  m/s vescape =km/hr vescape =mi/hr.This data corresponds to a surface gravitational acceleration of g =  m/s2 g =  gEarth. Orbit velocity and escape velocity

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Orbit Velocity and Escape Velocity If the kinetic energy of an object m1 launched from the a planet of mass M2 were equal in magnitude to the potential energy, then in the absence of friction resistance it could escape from the planet. The escape velocity is given by To find the orbit velocity for a circular orbit, you can set the gravitational forceequal to the required centripetal force. Note that the orbit velocity and the escape velocity from that radius are related by

Parallel Axis Theorem

The moment of inertia of any object about an axis through its center of mass is the minimum moment of inertia for an axis in that direction in space. The moment of inertia about any axis parallel to that axis through the center of mass is given by

The expression added to the center of mass moment of inertia will be recognized as the moment of inertia of apoint mass - the moment of inertia about a parallel axis is the center of mass moment plus the moment of inertia of the entire object treated as a point mass at the center of mass.

Statement:
The moment of inertia about Z-axis can be represented as:


Where
I_cmis the moment of inertia of an object about its centre of mass
m is the mass of an object
r is the perpendicular distance between the two axes.

Proof: 
Assume that the perpendicular distance between the axes lies along the x-axis and the centre of mass lies at the origin. The moment of inertia relative to z-axis that passes through the centre of mass, is represented as 


Moment of inertia relative to the new axis with its perpendicular distance r along the x-axis, is represented as:


We get,


The first term is I_cm,the second term is mr²and the final term is zero as the origin lies at the centre of mass. Finally, 

For a planar object, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia of two perpendicular axes through the same point in the plane of the object. The utility of this theorem goes beyond that of calculating moments of strictly planar objects. It is a valuable tool in the building up of the moments of inertia of three dimensional objects such as cylinders by breaking them up into planar disks and summing the moments of inertia of the composite disks. Show the development of the relationship.

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Perpendicular Axis Theorem The perpendicular axis theorem for planar objects can be demonstrated by looking at the contribution to the three axis moments of inertia from an arbitrary mass element. From the point mass moment, the contributions to each of the axis moments of inertia are

Apparent Weight: Person on Scale in Elevator

A person with mass, m, who is located at or near the surface of the Earth will always have some

weight W=mg. When a person stands on a scale, the reading (the number of pounds or newtons)

on the scale is actually the Normal Force that the scale exerts back towards the person to support

the person's weight. (Note that the person and the scale are stationary relative to each other, in

other words they are always in contact with each other, so they always have equal and opposite

action and reaction forces acting between them.)

Things get complicated, though, when the scale and the person experience acceleration. This

will change the contact force (the Normal Force) between the person and the scale.

Let's look at several cases. We will assume that Up is the positive direction and Down is the

negative direction.

Case 1: No acceleration of elevator

If the acceleration of the elevator is zero, then there are two possible

scenarios; the elevator can be at rest (stationary, zero velocity) or moving

with a constant speed (no acceleration if velocity does not change).

In this case, the action and reaction force pair between the person and the

scale is just the weight. The person pushes down on the scale with a force of

-W=-mg (negative direction) and the scale pushes back up against the man

with a Normal Force of FN = +W = +mg. Because the reading on the scale is

the magnitude of the normal force, the scale will read the true weight when

the elevator is NOT accelerating.

Case 2: going up & speeding up (acceleration a is positive (up))

In this case, the elevator and the person are starting from rest at a lower

floor. The elevator accelerates upward. The inertia of the person would

prefer to stay stationary, so the elevator floor and scale must push up on

the person to accelerate him upward along with the elevator. (The person

doesn't sink into the floor when the elevator accelerates up. The elevator

and the scale and the person all move together.)

The scale therefore has to push upward with extra force on the person to

accelerate the person's mass upward. This results in a greater contact

force between the scale and the person. Therefore the Normal Force is

larger, so the reading on the scale is a number that is GREATER than the

true weight.

Let's consider Newton's 2nd

 Law (ΣF=ma) acting on the person. The overall acceleration of the

person is upward (with the elevator). So ma is positive (upward). The only external forces acting on the person are the force of gravity acting down (-W=-mg) and the supporting Normal

Force FN that the scale applies upward on the person. So ΣF=ma= -mg+FN . We want to know

FN because that is the number that we read off the scale. FN =mg+ma, which is GREATER than

the true weight.

Case 3: going up & slowing down (acceleration a is negative (down))

In this case, the elevator and the person are initially moving upward at a

constant speed and slowing down to rest at a higher floor. The

acceleration of the elevator is downward (opposite to the upward motion,

which causes a reduction of the velocity). The inertia of the person

would prefer to keep moving upward at a constant speed, so the elevator

floor and scale effectively drop out a little bit from underneath the person

as the elevator slows down.

The person doesn't float upward, because again the elevator and the

person move together, but the contact force between the person and the

scale is reduced. The scale therefore has to push upward with less force

on the person to support the person's weight. Therefore the Normal Force is smaller, so the

reading on the scale is a number that is LESS than the true weight.

Let's consider Newton's 2nd

 Law (ΣF=ma) acting on the person. The overall acceleration of the

person is downward (with the elevator). So ma is negative (downward). The only external

forces acting on the person are the force of gravity acting down (-W=-mg) and the supporting

Normal Force FN that the scale applies upward on the person. So ΣF= -ma= -mg+FN . We want

to know FN because that is the number that we read off the scale. FN =mg - ma, which is LESS

than the true weight.

Case 4: going down & slowing down (acceleration a is positive (up))

In this case, the elevator and the person are initially moving downward at

a constant speed and then slow to rest at a lower floor. The elevator

accelerates upward (opposite direction to negative/downward velocity to

reduce velocity magnitude). The inertia of the person would prefer to

keep moving downward at the constant speed, so the elevator floor and

scale must push up on the person to accelerate him upward, slowing him

down. (The person doesn't sink into the floor here either. Elevator and

scale and person move together.)

The scale therefore has to push upward with extra force on the person to

accelerate the person's mass upward. This results in a greater contact

force between the scale and the person. Therefore the Normal Force is

larger, so the reading on the scale is a number that is GREATER than the

true weight.Let's consider Newton's 2nd

 Law (ΣF=ma) acting on the person. The overall acceleration of the

person is upward (with the elevator). So ma is positive (upward). The only external forces

acting on the person are the force of gravity acting down (-W=-mg) and the supporting Normal

Force FN that the scale applies upward on the person. So ΣF=ma= -mg+FN . (Note that this is

the same equation as we got in case 2.) We want to know FN because that is the number that we

read off the scale. FN =mg+ma, which is GREATER than the true weight.

Case 5: going down & speeding up (acceleration a is negative (down))

In this case, the elevator and the person are initially at rest at a higher

floor. The elevator then speeds up in the downward direction towards a

lower floor. The elevator acceleration of the elevator is

negative/downward (increasing the velocity magnitude in the downward

direction). The inertia of the person would prefer to stay at rest, so the

elevator floor and scale effectively drop out a little bit from underneath

the person as the elevator accelerates down.

The person doesn't float upward here also, because again the elevator and

the person move together, but the contact force between the person and

the scale is reduced. The scale therefore has to push upward with less

force on the person to support the person's weight. Therefore the Normal Force is smaller, so the

reading on the scale is a number that is LESS than the true weight.

Let's consider Newton's 2nd

 Law (ΣF=ma) acting on the person. The overall acceleration of the

person is downward (with the elevator). So ma is negative (downward). The only external

forces acting on the person are the force of gravity acting down (-W=-mg) and the supporting

Normal Force FN that the scale applies upward on the person. So ΣF= -ma= -mg+FN . (Note

that this is the same equation that we got for Case 3.) We want to know FN because that is the

number that we read off the scale. FN =mg - ma, which is LESS than the true weight.

Case 6: freefall (a = -g)

If the elevator cable were to break, the whole elevator-scale-person

system would all begin to accelerate downward due to the force of

gravity. All objects in freefall accelerate downward with the same

magnitude (acceleration due to gravity, g). The scale and the person are

freefalling together, so there is NO contact force (Normal Force)

between the scale and the person. (When they are both falling together,

there is no way that the scale can support any of the person's weight.)

Note that this is a special case of downward acceleration, which we

discussed in Case 3 and Case 5. Just as in Cases 3 and 5, the apparent

weight (which is zero when a=-g) is less than the true weight.A pictorial summary of apparent weight:

Electric Potential Energy Potential energy can be defined as the capacity for doing work which arises from position or configuration. In the electrical case, a charge will exert a force on any other charge and potential energy arises from any collection of charges. For example, if a positive charge Q is fixed at some point in space, any other positive charge which is brought close to it will experience a repulsive force and will therefore have potential energy. The potential energy of a test charge q in the vicinity of this source charge will be: Show where k is Coulomb's constant. In electricity, it is usually more convenient to use the electric potential energy per unit charge, just called electric potential or voltage. Application:Coulomb barrier for nuclear fusion Energy in electron volts

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Zero Potential The nature of potential is that the zero point is arbitrary; it can be set like the origin of a coordinate system. That is not to say that it is insignificant; once the zero of potential is set, then every value of potential is measured with respect to that zero. Another way of saying it is that it is the change in potential which has physical significance. The zero of electric potential (voltage) is set for convenience, but there is usually some physical or geometric logic to the choice of the zero point. For a single point charge or localized collection of charges, it is logical to set the zero point at infinity. But for an infinite line charge, that is not a logical choice, since the local values of potential would go to infinity. For practical electrical circuits, the earth or ground potential is usually taken to be zero and everything is referenced to the earth. Zero of potential at infinity Zero of mechanical potential energy

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Potential Reference at Infinity The general expression for the electric potential as a result of a point charge Q can be obtained by referencing to a zero of potential at infinity. The expression for thepotential difference is: Taking the limit as rb→∞ gives simply for any arbitrary value of r. The choice of potential equal to zero at infinity is an arbitrary one, but is logical in this case because the electric field and force approach zero there. The electricpotential energy for a charge q at r is then where k is Coulomb's constant.

Gravitational Potential Energy Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about 9.8 m/s2. Since the zero of gravitational potential energy can be chosen at any point (like the choice of the zero of a coordinate system), the potential energy at a height h above that point is equal to the work which would be required to lift the object to that height with no net change in kinetic energy. Since the force required to lift it is equal to its weight, it follows that the gravitational potential energy is equal to its weight times the height to which it is lifted. PE = kg x 9.8 m/s2 xm =  joules. PE = lbs xft = ft lb.

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Gravitational Potential Energy The general expression for gravitational potential energy arises from the law of gravity and is equal to the work done against gravity to bring a mass to a given point in space. Because of the inverse square nature of the gravity force, the force approaches zero for large distances, and it makes sense to choose the zero of gravitational potential energy at an infinite distance away. The gravitational potential energy near a planet is then negative, since gravity does positive work as the mass approaches. This negative potential is indicative of a "bound state"; once a mass is near a large body, it is trapped until something can provide enough energy to allow it to escape. The general form of the gravitational potential energy of mass m is: where G is the gravitation constant, M is the mass of the attracting body, and r is the distance between their centers. This is the form for the gravitational potential energy which is most useful for calculating the escape velocity from the earth's gravity.

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Gravitational Potential Energy From the work done against the gravity force in bringing a mass in from infinity where the potential energy is assigned the value zero, the expression for gravitational potential energy is This expression is useful for the calculation of escape velocity, energy to remove from orbit, etc. However, for objects near the earth the acceleration of gravity g can be considered to be approximately constant and the expression for potential energy relative to the Earth's surface becomes where h is the height above the surface and g is the surface value of the acceleration of gravity.

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Gravitational Potential	Index Energy concepts
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Gravitational Potential Integral