chapter11

Part I Kepler's Laws  Gravity plays an integral role in the mechanics of our universe. Planetary orbits are possible because of the invisible force of gravity. Planetary orbits are not perfectly circular. Instead, they orbit in an elliptical path with two focuses. Keplers first law states: All planets move in elliptical orbits with the sun at one focus.

The shape of an ellipse is defined by its eccentricity. If the eccentricity is close to 1, its foci are very close to the edge of the ellipse. If it is closer to 0, it becomes more circular. An ellipse with an eccentricity of 0 is simply a circle. For a given point on the edge of an ellipse, there are two radii... one from each focus. The sum of R1 and R2 is always constant. The distance from the outermost edge to the center is called the semimajor axis, or a. From the innermost edge to the center is the semiminor axis, or b. Observe the following animation:

Kepler's second law states that a line joining a planet to the sun will sweep out equal areas in equal times . Since planets move faster while closer to the sun, the area swept out in a given time interval is the same. Kepler's third law has to do with periods. It states that the square of the period of any planet is proportional to the cube of the semimajor axis. If period is T and radius is the mean distance between a planet and the sun.... T^2=Cr^3 In addition to Kepler's Laws, there is Newton's law of gravity. It represents the force of attration between each pair of point particles that is proportional to the product of the masses. The g on a planet can be calculated using the following formula.



=Gravitational Potential Energy = = =

= Potential energy is usually set at 0 at the earth's surface because height, h, plus the earth's radius approximately equals the radius of the earth. On most objects undergoing gravitational force, the radius, r, is usually approximately R of Earth. R(E) + h usually equals about R(E). As we move further from the surface of the earth, the gravitational force exerted by the earth is not uniform. It decreases as 1/r^2. The general definition of potential energy is: = = dU=-F. ds = = F being the conservative force on a particle and ds being displacement. = = = = Manipulating the equation, we get: dU=(G(M e )m)/r^2)dr = M e =Mass of earth = Integrating both sides, we get: = = = U=-(G(M<span style="color: #950909; font-family: 'Lucida Console',Monaco,monospace; font-size: 60%;">e )m)/r+U 0  <span style="color: #950909; font-family: 'Lucida Console',Monaco,monospace; font-size: 40%; line-height: 0px; overflow: hidden;">﻿ <span style="color: #101075; font-family: 'Lucida Console',Monaco,monospace; font-size: 130%;">Where U<span style="color: #101075; font-family: 'Lucida Console',Monaco,monospace; font-size: 60%;">0 is a constant from integration. <span style="color: #101075; font-family: 'Lucida Console',Monaco,monospace; font-size: 130%;">If you use the aforementioned equation to solve a problem pertaining to the earth and the sun, it is easier to set potential energy equal to zero when their separation is infinite, making U<span style="color: #101075; font-family: 'Lucida Console',Monaco,monospace; font-size: 60%;">0 =0. You would do this as opposed to setting it equal to 0 at the earth's surface. <span style="color: #101075; font-family: 'Lucida Console',Monaco,monospace; font-size: 130%;">This gives us U(r)=-GMm/r

<span style="background-color: #91fa14; color: #040795; font-family: 'Lucida Console',Monaco,monospace; font-size: 130%;">Gravitational Potential Energy: <span style="background-color: #91fa14; color: #040795; font-family: 'Lucida Console',Monaco,monospace; font-size: 130%;">U(r)= -GMm/r <span style="background-color: #91fa14; color: #040795; font-family: 'Lucida Console',Monaco,monospace; font-size: 130%;">m represents a particle's mass <span style="background-color: #91fa14; color: #040795; font-family: 'Lucida Console',Monaco,monospace; font-size: 130%;">M represents a spherical object <span style="background-color: #91fa14; color: #040795; font-family: 'Lucida Console',Monaco,monospace; font-size: 130%;">r represents the distance from the objects center

= <span style="background-color: #ffffff; color: #8a0a0a; display: block; font-family: 'Lucida Console',Monaco,monospace; font-size: 150%; text-align: center;">Escape Speed <span style="background-color: #ffffff; color: #b21010; display: block; font-family: 'Lucida Console',Monaco,monospace; font-size: 120%; text-align: center;">With the advanced technology we have today, we are now able to send probes into space and escape the earth's gravitational pull. Some orbit the sun, while others leave the solar system, exploring the vast empty space that lies outside. In order to escape the earth's gravity, an object must obtain a minimum initial speed, also called escape speed. =



Let's test this equation by plugging in the g of earth (9.8) and the radius of earth (6.37*10^6)



Wow, that's 25,000 miles per hour!

Gravitational Field

<span style="color: #34ab07; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 130%;">Until this point, we've been dealing mostly with equations of gravitational force between two objects, treated as point masses. Now let's take a look at the gravitational field.

<span style="color: #34ab07; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 130%;">At point P, we put a point mass of m and calculate the gravitational force on it due to all other particles. The gravitational force, F, divided by the mass m equals the gravitational field at point P.

<span style="color: #34ab07; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 130%;">g=F/m

<span style="color: #34ab07; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 130%;">The gravitational field of the earth points towards the earth and has a magnitude g(r)



<span style="color: #fd0d0d; font-family: Georgia,serif; font-size: 140%;">Gravitational Field Due to a Spherical Shell

<span style="color: #fd0d0d; font-family: Georgia,serif; font-size: 140%;">So far, we've looked at gravitational field due to a planet, now let's observe the gravitational field due to a spherical shell. <span style="color: #fd0d0d; font-family: Georgia,serif; font-size: 140%;">For a spherical shell of mass M, radius R, and a distance away of r:

<span style="color: #0013ff; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 150%;">And for any distance r that is inside of the shell, in other words r<R, g=0

<span style="color: #34ab07; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 160%;">How to find it through integration:

<span style="color: #0013ff; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 150%;">

<span style="color: #0013ff; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 150%;">We will do this in two steps... first find the gravitational field of the ring and then use it to help find that of a spherical shell.

<span style="color: #0013ff; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 150%;">We choose dm as a small point particle on the ring. x is the distance from its center.

<span style="color: #0013ff; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 150%;">The x component of the field due to dm is:

The total field would be:

<span style="color: #0013ff; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 150%; line-height: 0px; overflow: hidden;"> <span style="color: #0013ff; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 150%; line-height: 0px; overflow: hidden;">Now we'll use this to help get the gravitational field of a spherical shell with a

<span style="color: #0013ff; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 150%; line-height: 0px; overflow: hidden;">mass of M and a radius of R.





And there you have it. Thanks for visiting this wikipage.

Sources:  -The Book (Physics for scientists and engineers)   -en.wikipedia.org/wiki/Escape_velocity   -hyperphysics.phy-astr.gsu.edu/hbase/kepler.html   -library.thinkquest.org/11902/physics/newton.html <span style="background-color: #91fa14; color: #040795; font-family: 'Lucida Console',Monaco,monospace; font-size: 130%; line-height: 0px; overflow: hidden;">
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