UNIT-IGRAVITATIONAL FIELDS AND POTENTIALS

NATURE OF MOTION UNDER CENTRAL FORCES In classical mechanics, a central force on an object is a force whose magnitude only

depends on the distance r of the object from the origin and is directed along the line joining them: 

where   is the force, F is a vector valued force function, F is a scalar valued force function, r is the position vector, ||r|| is its length, and   = r/||r|| is the corresponding unit vector.

Equivalently, a force field is central if and only if it is spherically symmetric.

A central force is a conservative field, that is, it can always be expressed as the negative gradient of a potential:

In a conservative field, the total mechanical energy (kinetic and potential) is conserved:

(where ṙ denotes the derivative of r with respect to time, that is the velocity), and in a central force field, so is the angular momentum:

because the torque exerted by the force is zero. As a consequence, the body moves on the plane perpendicular to the angular momentum vector and containing the origin, and obeys Kepler's second law. (If the angular momentum is zero, the body moves along the line joining it with the origin.)

As a consequence of being conservative, a central force field is irrotational, that is, its curl is zero, except at the origin:

NEWTON'S LAW OF UNIVERSAL GRAVITATION 

It states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them:

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,where:

F is the force between the masses, G is the gravitational constant, m1 is the first mass, m2 is the second mass, and r is the distance between the centers

of the masses.

Assuming SI units, F is measured in newtons (N), m1 and m2 in kilograms (kg), r in meters (m), and the constant G is approximately equal to6.674×10−11 N m2 kg−2. The value of the constant G was first accurately determined from the results of the Cavendish experiment conducted by the British scientist Henry Cavendish in 1798, although Cavendish did not himself calculate a numerical value for G. 

Newton's law of gravitation resembles Coulomb's law of electrical forces, which is used to calculate the magnitude of electrical force between two charged bodies. Both are inverse-square laws, in which force is inversely proportional to the square of the distance between the bodies. Coulomb's Law has the product of two charges in place of the product of the masses, and the electrostatic constant in place of the gravitational constant.

Newton's law has since been superseded by Einstein's theory of general relativity, but it continues to be used as an excellent approximation of the effects of gravity. Relativity is required only when there is a need for extreme precision, or when dealing with gravitation for extremely massive and dense objects.

Newton's law of universal gravitation can be written as a vector equation to account for the direction of the gravitational force as well as its magnitude. In this formula, quantities in bold represent vectors.

where

F12 is the force applied on object 2 due to object 1,G is the gravitational constant,m1 and m2 are respectively the masses of objects 1 and 2,|r12| = |r2 − r1| is the distance between objects 1 and 2, and

 is the unit vector from object 1 to 2.It can be seen that the vector form of the equation is the same as the scalar form given

earlier, except that F is now a vector quantity, and the right hand side is multiplied by the appropriate unit vector. Also, it can be seen that F12 = −F21.

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The gravitational field is a vector field that describes the gravitational force which would be applied on an object in any given point in space, per unit mass. It is actually equal to the gravitational acceleration at that point.

It is a generalization of the vector form, which becomes particularly useful if more than 2 objects are involved (such as a rocket between the Earth and the Moon). For 2 objects (e.g. object 2 is a rocket, object 1 the Earth), we simply write r instead of r12 and m instead of m2 and define the gravitational field g(r) as:

so that can write:

This formulation is dependent on the objects causing the field. The field has units of acceleration; in SI, this is m/s2.

Gravitational fields are also conservative; that is, the work done by gravity from one position to another is path-independent. This has the consequence that there exists a gravitational potential field V(r) such that,

If m1 is a point mass or the mass of a sphere with homogeneous mass distribution, the force field g(r) outside the sphere is isotropic, i.e., depends only on the distance r from the center of the sphere. In that case

KEPLER'S LAWS 

In astronomy, Kepler's laws give a description of the motion of planets around the Sun.

Kepler's laws are:

1. The orbit of every planet is an ellipse with the Sun at one of the two foci.2. A line joining a planet and the Sun sweeps out equal areas during equal intervals of time. 3. The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Kepler tried to fit Brahe's data to the Copernican model but consistently arrived at errors of at least eight seconds of arc, small but not insignificant. He was finally forced to abandon the concept of uniform circular orbital paths but it was to take him several years of painstaking, methodical calculations before he arrived at an alternate model that fitted Brahe's 20 years of data on Mars. The results were published in 1609 in his work Astronomica nova (New Astronomy). In it he explained what are now known as his first two laws of planetary motion.

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Kepler's 1st Law: The Law of Ellipses

Every planet moves in an elliptical orbit, with the Sun at one focus.

All planets orbit the Sun in elliptical orbits with the Sun as one common focus. The eccentricity of the ellipse has been greatly exaggerated in the above diagram. For most planets their orbits are almost circular.

Kepler's 2nd Law: The Law of Equal Areas

As a planet moves in its orbit a line drawn from the Sun to the Planet sweeps out equal areas in equal time intervals.

The line between a planet and the Sun (the radius vector) sweeps out equal areas in equal periods of time. In the diagram, the time interval t2-t1 = t4-t3 so the areas swept through in equal times are equal, that is A1 = A2.

This effect is very noticeable in comets such as Comet Halley that have highly elliptical orbits. When in the inner Solar System, close to the Sun at perihelion, they move much faster than when far from the Sun at aphelion.

The graphic precisely displays Kepler's second law. The area of every triangle formed is exactly the same area. Also notice as the planet or satellite moves closer to the sun, the speed of its orbit

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increases, but when it is located farther away from the sun, the speed in which the planet or satellite revolves around the sun is slower.

Kepler actually formulated the law of equal areas first and it then led him to the law of ellipses. His third law was not published until 1618 in Harmonice mundi (The Harmony of the World). This resulted from his attempts to find a relationship between the distance of a planet from the Sun and its orbital period.

Kepler's Third Law: The Law of Periods or the Harmonic Law

It stated that if P is the period and M is the length of semi major axis of a planet's orbit, then the ratio P²/M³ is the same for all the planets.

The square of a planet's period, T, is directly proportional to the cube of its average distance from the Sun, r.Mathematically this can be expressed as:

T2 ∝ r3 or T2/r3 = k where k is a constant and the same for all planets or orbital bodies (such as comets) in a given system.

If T is measured in Earth years and r in astronomical units (AU) then for the Earth, T = 1 and r = 1 so:

T2/r3 = k ∴ 1/1 = kie. k =1

The implication of Kepler's Third Law is that planets more distant from the Sun take longer to orbit the Sun. Let us see how this can be used to determine the mean distance of Mars from the Sun if its orbital period is 1.88 Earth years.

If T2/r3 = k Then rewriting for rr3 = T2/k r = ((1.88)2/1)1/3 so r = 1.524 AU So Mars is 1.524 astronomical units from the Sun.

Kepler's laws of planetary motion were empirical, they could predict what would occur but could not account for why planets behaved in such a manner. His Rudolphine tables of planetary motion published in 1627 were more accurate than nay previous ones. He came close to uncovering the concept of gravitation and corresponded with Galileo and was aware of his telescopic discoveries.

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Newton's Law of Gravitation

Issac Newton witnessed the apple falling onto Earth, there is no way he could not have also noticed the moon revolving around Earth. Why do these things occur? The Earth must have a force that pulls the moon towards its center; if it did not then moon would either be travelling at uniform straight line motion or not moving at all, following Newton's first law. Newton's obsevations were that all objects near the Earth's surface fall towards the Earth's center because of a gravitational force. That Gravitational force was Universal

In Newton's Universal Law of Gravitation, he states

The gravitational force between two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them.

GRAVITATIONAL POTENTIAL AND FIELD

The gravitational field at any point P in space is defined as the gravitational force felt by a tiny unit mass placed at P.  So, to visualize the gravitational field, in this room or on a bigger scale such as the whole Solar System, imagine drawing a vector representing the gravitational force on a one kilogram mass at many different points in space, and seeing how the pattern of these vectors varies from one place to another. It say “a tiny unit mass” because we don’t want the gravitational field from the test mass itself to disturb the system.  This is clearly not a problem in discussing planetary and solar gravity.  To build an intuition of what various gravitational fields look like, we’ll examine a sequence of progressively more interesting systems, beginning with a simple point mass and

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working up to a hollow spherical shell, this last being what we need to understand the Earth’s own gravitational field, both outside and inside the Earth.

GRAVITATIONAL POTENTIAL AND FIELD DUE TO

(i) Field Outside a Spherical ShellTo find the gravitational field at the point P, just add the contributions from all the rings

in the stack. In other words, divide the spherical shell into narrow “zones”: imagine chopping an orange into circular slices by parallel cuts, perpendicular to the axis—but of course our shell is just the skin of the orange.  One such slice gives a ring of skin, corresponding to the surface area between two latitudes, the two parallel lines in the diagram above. The diagram that this “ring of skin” will have radius  , therefore circumference and breadth  , where we’re taking   to be very small.  This means that the area of the ring of skin is 

 

So, if the shell has mass   per unit area, this ring has mass  , and the gravitational force at P from this ring will be 

. Now, to find the total gravitational force at P from the entire shell we have to add the contributions from each of these “rings” which, taken together, make up the shell.  In other words, we have to integrate the above expression in    

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 So the gravitational field is: 

. In fact, this is quite a tricky integral:  , x and s are all varying!  It turns out to be is easiest done by switching variables from   to s. Label the distance from P  to the center of the sphere by r.  Then, from the diagram,  , and a, r are constants, so  , 

and  

Now  , and from the diagram  , so  ,

and, writing  , 

                    The gravitational field outside a uniform spherical shell is GM/r 2 towards the center. And, there’s a bonus: for the ring, we only found the field along the axis, but for the spherical shell, once we’ve found it in one direction, the whole problem is solved—for the spherical shell, the field must be the same in all directions.

FIELD OUTSIDE A SOLID SPHEREThe gravitational field outside a shell of matter is the same as if all the mass were at a

point at the center, it’s easy to find the field outside a solid sphere: that’s just a nesting set of shells, like spherical Russian dolls.  Adding them up, The gravitational field outside a uniform sphere is GM/r 2 towards the center. This result be adding uniform spherical shells, it is still true if the shells have different densities, provided the density of each shell is the same in all directions.   The inner shells could be much denser than the outer ones—as in fact is the case for the Earth.

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FIELD INSIDE A SPHERICAL SHELL

  

Imagine the shell to be very thin, with a mass density   kg per square meter of surface. Begin by drawing a two-way cone radiating out from the point P, so that it includes two small areas of the shell on opposite sides: these two areas will exert gravitational attraction on a mass at P in opposite directions.  It turns out that they exactly cancel.  

This is because the ratio of the areas A1 and A2 at distances r1 and r2 are given

by  : since the cones have the same angle, if one cone has twice the height of the other, its base will have twice the diameter, and therefore four times the area.  Since the masses of the bits of the shell are proportional to the areas, the ratio of the masses of the cone bases is

also  .  But the gravitational attraction at P from these masses goes as  , and that r2 term cancels the one in the areas, so the two opposite areas have equal and opposite gravitational forces at P.  

In fact, the gravitational pull from every small part of the shell is balanced by a part on the opposite side—you just have to construct a lot of cones going through P to see this.  (There is one slightly tricky point—the line fromP to the sphere’s surface will in general cut the surface at an angle.  However, it will cut the opposite bit of sphere at the same angle, because any line passing through a sphere hits the two surfaces at the same angle, so the effects balance, and the base areas of the two opposite small cones are still in the ratio of the squares of the distances r1, r2.)   

Field Inside a Sphere: How Does g Vary on Going Down a Mine?This is a practical application of the results for shells. On going down a mine, if we

imagine the Earth to be made up of shells, we will be inside a shell of thickness equal to the depth of the mine, so will feel no net gravity from that part of the Earth.  However, we will be closer to the remaining shells, so the force from them will be intensified. 

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Suppose we descend from the Earth’s radius rE  to a point distance r from the center of the Earth. What fraction of the Earth’s mass is still attracting us towards the center? Let’s make life simple for now and assume the Earth’s density is uniform, call it   kg per cubic meter. 

 Then the fraction of the Earth’s mass that is still attracting us (because it’s closer to the

center than we are—inside the red sphere in the diagram) is  .  

The gravitational attraction from this mass at the bottom of the mine, distance r from the center of the Earth, is proportional to mass/r2.    We have just seen that the mass is itself proportional to r3, so the actual gravitational force felt must be proportional to  .  

That is to say, the gravitational force on going down inside the Earth is linearly proportional to distance from the center.  Since we already know that the gravitational force on a

mass m at the Earth’s surface   is mg, it follows immediately that in the mine the gravitational force must be

. So there’s no force at all at the center of the Earth—as we would expect, the masses are

attracting equally in all directions.

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BOY’S METHOD OF DTERMINING ‘G’

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