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A rigid body is an aggregation of small elements, which can be treated as point mass.

Gravitational field of rigid bodies

We shall develop few relations here for the gravitational field strength of bodies of particular geometric shape without any reference to Earth’s gravitation.

Newton’s law of gravitation is stated strictly in terms of point mass. The expression of gravitational field due to a particle, as derived from this law, serves as starting point for developing expressions of field strength due to rigid bodies. The derivation for field strength for geometric shapes in this module, therefore, is based on developing technique to treat a real body mass as aggregation of small elements and combine individual effects. There is a bit of visualization required as we need to combine vectors, having directional property.

Along these derivations for gravitational field strength, we shall also establish Newton’s shell theory, which has been the important basic consideration for treating spherical mass as point mass.

The celestial bodies - whose gravitational field is appreciable and whose motions are subject of great interest - are usually spherical. Our prime interest, therefore, is to derive expression for field strength of solid sphere. Conceptually, a solid sphere can be considered being composed of infinite numbers of closely packed spherical shells. In turn, a spherical shell can be conceptualized to be aggregation of thin circular rings of different diameters.

The process of finding the net effect of these elements fits perfectly well with integration process. Our major task, therefore, is to suitably set up an integral expression for elemental mass and then integrate the elemental integral between appropriate limits. It is clear from the discussion here that we need to begin the process in the sequence starting from ring -->spherical shell -->solid sphere.

Gravitational field due to a uniform circular ring

We need to find gravitational field at a point “P” lying on the central axis of the ring of mass “M” and radius “a”. The arrangement is shown in the figure. We consider a small mass “dm” on the circular ring. The gravitational field due to this elemental mass is along PA. Its magnitude is given by :

Gravitational field due to a ring

The gravitational field is measured on axial point "P".

E = G m P A 2 = G m a 2 + r 2

We resolve this gravitational field in the direction parallel and perpendicular to the axis in the plane of OAP.

Gravitational field due to a ring

The net gravitational field is axial.

E | | = E cos θ

E = E sin θ

We note two important things. First, we can see from the figure that measures of “y” and “θ” are same for all elemental mass. Further, we are considering equal elemental masses. Therefore, the magnitude of gravitational field due any of the elements of mass “dm” is same, because they are equidistant from point “P”.

Second, perpendicular components of elemental field intensity for pair of elemental masses on diametrically opposite sides of the ring are oppositely directed. On integration, these perpendicular components will add up to zero for the whole of ring. It is clear that we can assume zero field strength perpendicular to axial line, if mass distribution on the ring is uniform. For uniform ring, the net gravitational intensity will be obtained by integrating axial components of elemental field strength only. Hence,

Questions & Answers

What are the system of units
Jonah Reply
A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
Samson Reply
58asagravitasnal firce
Amar
water boil at 100 and why
isaac Reply
what is upper limit of speed
Riya Reply
what temperature is 0 k
Riya
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
Mustapha
How MKS system is the subset of SI system?
Clash Reply
which colour has the shortest wavelength in the white light spectrum
Mustapha Reply
how do we add
Jennifer Reply
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
Abhyanshu Reply
x=5.8-3.22 x=2.58
sajjad
what is the definition of resolution of forces
Atinuke Reply
what is energy?
James Reply
Ability of doing work is called energy energy neither be create nor destryoed but change in one form to an other form
Abdul
motion
Mustapha
highlights of atomic physics
Benjamin
can anyone tell who founded equations of motion !?
Ztechy Reply
n=a+b/T² find the linear express
Donsmart Reply
أوك
عباس
Quiklyyy
Sultan Reply
Moment of inertia of a bar in terms of perpendicular axis theorem
Sultan Reply
How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Yara Reply

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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