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

Index Electric field concepts

HyperPhysics***** Electricity and Magnetism R Nave

Go Back

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

Index Energy concepts Gravity concepts

HyperPhysics***** Mechanics R Nave

Go Back

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.

Index Moment of inertia concepts

HyperPhysics***** Mechanics R Nave

Go Back

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