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The Creative Laws 115.43 KB - From the beginning of time man has been on a permanent journey to discover the true meaning of life.

Within the pages of this book you will find the Authors own journey of discovery into a deeper understanding of the Universal Laws that govern every living thing on this planet.

Related Books The Electronic Funds Transfer laws The Eight Laws of Health In Animal Naturopathy The Divine Laws Anatomy of Frustration: Discover How to Transform Frustration into Creative Opportunities & Unlock the Power to Succeed: A Straight-Talking Guide to Frustrationproof Your Life Health and Wealth Magnetism - Using the Law of Attraction to Create Health and Wealth The 8 Laws of Fat-loss The 48 Laws of Power Reversal The Secret Laws of Social Wisdom Discover How to Improve Your Health & Life Abundance Love The Main Occult Laws My Creative Writing Swag My Creative Life It's All About the Money! The Law of Universal Gravitation states that every object in the universe attracts every other object in the universe with a force that has a magnitude which is directly proportional to the product of their masses and inversely proportional to the distance between their centers squared. Our first example involves calculating the gravitational force between a point mass M and an extended rod of mass m, length L, and mass per unit length,. To begin, divide the rod into a finite (countable) number of segments of mass each located at a distance x from M.

Our next investigation will be to calculate the gravitational force exerted by two equal, equidistant massive particles, M on a point mass, m, located at point P.

Our result will allow us to calculate the gravitation force at any position P along the x-axis. Suppose we now allow the two masses in the previous example to be rotated about the x-axis to create a thin ring in the yz-plane having a radius of a and a mass of 2M. This substitution along with our original supposition that "each mass, M" was stretched out to form half of our ring will allow us to set up the limits of the integral and evaluate it. As we complete our work also notice that G, m, a, x, and are all constants and can be removed from the integral. Notice that this is the EXACT same expression that we calculated for the net gravitational force in the previous example involving two equidistant particles. At this junction in the lesson our goal to determine what the gravitational force would be on a point mass, m, if the point mass were to be placed (1) outside of a thin spherical shell, (2) inside a thin spherical shell, and then (3) inside a solid spherical mass.

Complications arise from the number of variables that need to be related and then integrated.

Although the actual calculus for this derivation is beyond the scope of our lesson, remember that in general for a thin ring the gravitational attraction between it and an external point mass, m, is directed along the x-axis towards the center of the ring. If our point mass, m, were a distance x from a solid sphere, we would expect the same outcome since we can envision our solid sphere as being composed of tightly nested shells.

Through point P, draw two intersecting chords that terminate on opposite sides of our thin shell, crossing at the center of m. Our chords, with their equal apex angles, have created two similar right circular cones having base areas A1 and A2.

This leads us to our final substitution showing that the gravitational force between the interior mass and our point mass is proportional to r's percentage of the original radius R. Written in an easy to follow format the Author shares his experiences and insights into how the Law of Attraction really works and effects each and every one of us. Downloading ebooks comes from the site of the source, but not all, e-books are free, there are electronic books that you need to buy, and then only download or read if you liked the book and want to buy it and then read or download in description of the book there is a link "source" - click on it and then buy or download the ebook. If we pair up segments based on the requirement that they are on opposite sides of any selected diameter, then we can calculate the force the ring exerts on the point mass, m. No matter where the mass segment is located on the ring, only the force components pointing along the x-axis, towards the center of the ring at the origin, will contribute towards the force of gravitational attraction between the ring and the point mass. This symmetry will allow us to calculate the gravitational force on the mass by the ring. Note that we must also pass to the limit to convert our summation into the desired integral.

Stretching the masses, M, out into a ring did NOT affect the magnitude of the gravitational force on the point mass. To do this, we can consider our uniform, spherical shell to be made up of thin rings symmetric to the x-axis. As you can see, the rings will have different radii and circumferences, have different values for angles ? and ?, as well as different values of r and x.

What makes this result all the more beautiful is that it shows that the attractive force between the Earth and a softball arching through the air does not need to become a complicated calculation depending on local terrain. The mass segments represented by these areas are gravitationally attracted to our point mass, m. If we repeated this process for every surface segment, we would always discover that the forces are always balanced. All of the sphere's mass above its location can be considered to consist of nested shells (shown in gray). Download The Creative Laws today and discover for yourself the natural laws and life principles that lead to an abundance of Health, Wealth, Happiness and Success!

If you have difficulty in downloading, please contact us and we will help you in this matter. For those of you interested in this advanced solution, a derivation can be found in section 11-5 of your text (Tipler, Physics for Scientists and Engineers, 5th ed, pages 358-360). That is, our point mass would feel NO gravitational force at point P, no matter where P is inside the shell. From our earlier discussions, we know that the gravitational force on a point mass located anywhere inside a shell equals zero. The rings are going to increase in radius, from r = 0 to r = a, and then decrease back to 0 as we take our slices.

Within the pages of this book you will find the Authors own journey of discovery into a deeper understanding of the Universal Laws that govern every living thing on this planet.

Related Books The Electronic Funds Transfer laws The Eight Laws of Health In Animal Naturopathy The Divine Laws Anatomy of Frustration: Discover How to Transform Frustration into Creative Opportunities & Unlock the Power to Succeed: A Straight-Talking Guide to Frustrationproof Your Life Health and Wealth Magnetism - Using the Law of Attraction to Create Health and Wealth The 8 Laws of Fat-loss The 48 Laws of Power Reversal The Secret Laws of Social Wisdom Discover How to Improve Your Health & Life Abundance Love The Main Occult Laws My Creative Writing Swag My Creative Life It's All About the Money! The Law of Universal Gravitation states that every object in the universe attracts every other object in the universe with a force that has a magnitude which is directly proportional to the product of their masses and inversely proportional to the distance between their centers squared. Our first example involves calculating the gravitational force between a point mass M and an extended rod of mass m, length L, and mass per unit length,. To begin, divide the rod into a finite (countable) number of segments of mass each located at a distance x from M.

Our next investigation will be to calculate the gravitational force exerted by two equal, equidistant massive particles, M on a point mass, m, located at point P.

Our result will allow us to calculate the gravitation force at any position P along the x-axis. Suppose we now allow the two masses in the previous example to be rotated about the x-axis to create a thin ring in the yz-plane having a radius of a and a mass of 2M. This substitution along with our original supposition that "each mass, M" was stretched out to form half of our ring will allow us to set up the limits of the integral and evaluate it. As we complete our work also notice that G, m, a, x, and are all constants and can be removed from the integral. Notice that this is the EXACT same expression that we calculated for the net gravitational force in the previous example involving two equidistant particles. At this junction in the lesson our goal to determine what the gravitational force would be on a point mass, m, if the point mass were to be placed (1) outside of a thin spherical shell, (2) inside a thin spherical shell, and then (3) inside a solid spherical mass.

Complications arise from the number of variables that need to be related and then integrated.

Although the actual calculus for this derivation is beyond the scope of our lesson, remember that in general for a thin ring the gravitational attraction between it and an external point mass, m, is directed along the x-axis towards the center of the ring. If our point mass, m, were a distance x from a solid sphere, we would expect the same outcome since we can envision our solid sphere as being composed of tightly nested shells.

Through point P, draw two intersecting chords that terminate on opposite sides of our thin shell, crossing at the center of m. Our chords, with their equal apex angles, have created two similar right circular cones having base areas A1 and A2.

This leads us to our final substitution showing that the gravitational force between the interior mass and our point mass is proportional to r's percentage of the original radius R. Written in an easy to follow format the Author shares his experiences and insights into how the Law of Attraction really works and effects each and every one of us. Downloading ebooks comes from the site of the source, but not all, e-books are free, there are electronic books that you need to buy, and then only download or read if you liked the book and want to buy it and then read or download in description of the book there is a link "source" - click on it and then buy or download the ebook. If we pair up segments based on the requirement that they are on opposite sides of any selected diameter, then we can calculate the force the ring exerts on the point mass, m. No matter where the mass segment is located on the ring, only the force components pointing along the x-axis, towards the center of the ring at the origin, will contribute towards the force of gravitational attraction between the ring and the point mass. This symmetry will allow us to calculate the gravitational force on the mass by the ring. Note that we must also pass to the limit to convert our summation into the desired integral.

Stretching the masses, M, out into a ring did NOT affect the magnitude of the gravitational force on the point mass. To do this, we can consider our uniform, spherical shell to be made up of thin rings symmetric to the x-axis. As you can see, the rings will have different radii and circumferences, have different values for angles ? and ?, as well as different values of r and x.

What makes this result all the more beautiful is that it shows that the attractive force between the Earth and a softball arching through the air does not need to become a complicated calculation depending on local terrain. The mass segments represented by these areas are gravitationally attracted to our point mass, m. If we repeated this process for every surface segment, we would always discover that the forces are always balanced. All of the sphere's mass above its location can be considered to consist of nested shells (shown in gray). Download The Creative Laws today and discover for yourself the natural laws and life principles that lead to an abundance of Health, Wealth, Happiness and Success!

If you have difficulty in downloading, please contact us and we will help you in this matter. For those of you interested in this advanced solution, a derivation can be found in section 11-5 of your text (Tipler, Physics for Scientists and Engineers, 5th ed, pages 358-360). That is, our point mass would feel NO gravitational force at point P, no matter where P is inside the shell. From our earlier discussions, we know that the gravitational force on a point mass located anywhere inside a shell equals zero. The rings are going to increase in radius, from r = 0 to r = a, and then decrease back to 0 as we take our slices.

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