<< Chapter < Page Chapter >> Page >
Gravitational potential to rigid body at a point is scalar sum of potentials due to elemental mass.

We have derived expression for gravitational potential due to point mass of mass, “M”, as :

V = - G M r

We can find expression of potential energy for real bodies by considering the same as aggregation of small elements, which can be treated as point mass. We can, then, combine the potential algebraically to find the potential due to the body.

The derivation is lot like the derivation of gravitational field strength. There is, however, one important difference. Derivation of potential expression combines elemental potential – a scalar quantity. As such, we can add contributions from elemental parts algebraically without any consideration of direction. Indeed, it is a lot easier proposition.

Again, we are interested in finding gravitational potential due to a solid sphere, which is generally the shape of celestial bodies. As discussed earlier in the course, a solid sphere is composed of spherical shells and spherical shell, in turn, is composed of circular rings of different radii. Thus, we proceed by determining expression of potential from ring -->spherical shell -->solid sphere.

Gravitational potential due to a uniform circular ring

We need to find gravitational potential 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 potential due to this elemental mass is :

Gravitational potential due to a uniform circular ring

Gravitational potential at an axial point

V = - G m P A = - G m a 2 + r 2 1 2

We can find the sum of the contribution by other elements by integrating above expression. We note that all elements on the ring are equidistant from the point, “P”. Hence, all elements of same mass will contribute equally to the potential. Taking out the constants from the integral,

V = - G a 2 + r 2 1 2 0 M m

V = - G M a 2 + r 2 1 2 = - G M y

This is the expression of gravitational potential due to a circular ring at a point on its axis. It is clear from the scalar summation of potential due to elemental mass that the ring needs not be uniform. As no directional attribute is attached, it is not relevant whether ring is uniform or not? However, we have kept the nomenclature intact in order to correspond to the case of gravitational field, which needs to be uniform for expression as derived. The plot of gravitational potential for circular ring is shown here as we move away from the center.

Gravitational potential due to a uniform circular ring

The plot of gravitational potential on axial position

Check : We can check the relationship of potential, using differential equation that relates gravitational potential and field strength.

E = V r = r { G M a 2 + r 2 1 2 }

E = G M x 1 2 X a 2 + r 2 1 2 1 X 2 r

E = - G M r a 2 + r 2 3 2

The result is in excellent agreement with the expression derived for gravitational field strength due to a uniform circular ring.

Gravitational potential due to thin spherical shell

The spherical shell of radius “a” and mass “M” can be considered to be composed of infinite numbers of thin rings. We consider one such thin ring of infinitesimally small thickness “dx” as shown in the figure. We derive the required expression following the sequence of steps as outlined here :

Questions & Answers

what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
many many of nanotubes
what is the k.e before it land
what is the function of carbon nanotubes?
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
can anyone tell who founded equations of motion !?
Ztechy Reply
n=a+b/T² find the linear express
Donsmart Reply
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
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate
Rama Reply
A balloon is released from the ground which rises vertically up with acceleration 1.4m/sec^2.a ball is released from the balloon 20 second after the balloon has left the ground. The maximum height reached by the ball from the ground is
Lucky Reply
work done by frictional force formula
Sudeer Reply
Misthu Reply
Why are we takingspherical surface area in case of solid sphere
Saswat Reply
In all situatuons, what can I generalize?
Cart Reply
the body travels the distance of d=( 14+- 0.2)m in t=( 4.0 +- 0.3) s calculate it's velocity with error limit find Percentage error
Clinton Reply
Explain it ?Fy=?sN?mg=0?N=mg?s
Admire Reply

Get the best Physics for k-12 course in your pocket!

Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Physics for k-12' conversation and receive update notifications?