GRAVITATIONAL FORCE FIELD

Gravitational force and field - Giancoli, sections 5 - 7
1. State Newton's law of universal gravitation.
· By thinking about the force of gravity that causes an apple to fall to the ground and the force necessary to keep the moon in orbit around the earth, Newton was able to come up with his hypothesis. Namely, that a force exists between any two masses.
· The force between any two point masses is given by the formula
· M1 and M2 are the masses of the attracting bodies, r is the separation between the two bodies, and G is a universal constant which has the value G = 6.67 x 10-11 Nm2kg-2.
· The direction of the force is along the line joining the centers of the two masses. This is a force of attraction and the vector form of the equation introduces a minus (-) sign to show that the force is directed towards the attracting object.
· Objects may be considered as point masses if the objects are small in comparison to the distance between them.
· Uniform objects (constant density and spherical symmetry) like the earth may be taken as point masses, with all of the mass of the object located at its center.
2. Define gravitational field strength.
· A mass, M, creates a gravitational field in the space around it. The concept of a "force field" was invented to explain how forces could act over a distance, with no direct contact between the objects involved.
· The idea is that a small mass, m, at some point in space will interact with the field at that point in space created by the larger mass.
· Gravitational field strength is then defined as the ratio of the gravitational force exerted on the mass m to the amount of mass that m possesses.
· In equation form, , where g is the gravitational field strength, FG­ is the gravitational force exerted on m, and m is the mass of the object in question.
· The gravitational field strength describes not only the amount of force exerted on a unit mass, but also the direction of that force.
3. Derive an expression for the gravitational field as a function of distance from a point mass. This includes the field outside a spherical mass.
· Consider a large point mass, M, and a smaller test mass, m. We know that the large point mass will attract the small test mass toward the center of the large point mass with a force given by F = .
·
We can think of this interaction as the interaction between the two masses, m and M, or as the interaction between m and the gravitational field created by M. Using our definition of gravitational field strength, it is very easy to show that: .
· Since uniform objects may be taken as point masses, this result applies also to uniform objects as long as r is greater than robject. That is, we must be physically outside the object to use this result.
· This result shows us that the farther we get from an object the smaller the gravitational field strength becomes. We can show this in two ways.

4. Derive an expression for gravitational field at the surface of a planet.
· If we move to the surface of a planet, taking the planet to be uniform, then we can think of the planet's mass as being located at the center of the planet. We, on the surface of the planet, are located a distance of one planet radius from the center of the planet, so: , where Mp is the mass of the planet and rp is the radius of the planet.
· We should note at this point, that the gravitational force at the surface of the planet is what we have been calling the weight of the object. That is, F­G = mg. The implication of this is that the gravitational field strength g is nothing more than the acceleration due to gravity experienced by an object!
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Near the surface of the earth, we have difficulty recognizing the spherical shape of the earth (due to its large size). Moreover, as we move away from the surface of the earth we are not moving appreciably farther from the center of the earth. Taken together this means that as long as we stay near the surface of the earth, we can ignore the small changes in r that are experienced. Then, we can say that the gravitational field near the surface of the earth is uniform. This means that it has the same value everywhere AND it points in the same direction everywhere. We take the value of the uniform gravitational field strength of the earth to be 9.8 Nkg-1 and the direction to be downward (toward the surface of the earth).
· Finally, it is interesting to note that for uniform objects, like the earth, the gravitational field strength inside the surface of the planet behaves very differently from the gravitational field strength outside the planet. It can be shown using calculus and Gauss' law that inside the earth, the gravitational field strength is directly related to the distance from the center of the earth. A plot of g against r looks like this:

5. Solve problems involving gravitational forces and fields. Vector addition is required to find the gravitational field strength due to more than one mass.
An example: Stars A and B have the same radius but the mass of A is double that of B. Find the gravitational field at point P, a distance of 3 star radii from the center of each star if the stars are separated by a distance of 4 star radii.
Solution: First, note that since gravitational field is a vector quantity, we need to find both the magnitude and the direction of the field at point P. A picture will help!
To begin, let's look at the magnitude of g for each planet. gA =G(2M)/ (3r)2 and gB = GM/ (3r)2. So, it looks like gA = 2 gB. To simplify things, let's call gB =g. Then, gA = 2g. So here's what we've got:
Both vectors make the same angle with the horizontal line. To find the vector components of our two gravitational fields, we will need to know the sin and cosine of this angle. We can figure these out from the first picture. The side opposite the angle has a length of 2r. The hypotenuse has a length of 3r. So, the sine of the angle is 2r/3r = 2/3. The adjacent side has a length of r (from the pythagorean theorem). So this means the cosine of the angle is r / 3r = / 3. Now we are ready:
The horizontal component of gA = 2g cos q = 2 g / 3, to the left.
The horizontal component of gB = g cos q = g / 3, to the left.
So, the net horizontal component = (2 g / 3) + ( g / 3) = 3 g / 3 = g , to the left.
The vertical component of gA = 2g sin q = 2g(2/3) = 4g/3, upwards.
The vertical component of gB = g sin q = g(2/3) = 2g/3, downwards.
So, the net vertical component = 4g/3 - 2g/3 = 2g/3, upwards.

Then, to find the net magnitude of the gravitational field: gnet =
And, to find the angle: tan q = vert. comp / horiz comp = = 0.298. So, q = tan-1(0.298) = 16.6o above the horizontal.

Gravitational energy and potential - This material is not in your textbook.
6. Define gravitational potential energy and gravitational potential. Students should understand that the work done in moving a mass between two points in a gravitational field is independent of the path taken and that gravitational potential energy is taken to be zero at infinity.
· It turns out that our previous definition of gravitational potential energy (EP = mgh) is only good for a uniform gravitational field. When we move far enough away from the earth for g to be non-uniform we need to look at things in a different way.
· However, the concepts of energy conservation and potential energy as energy stored by doing work against gravity are still good. So, we need to look at work done against gravity.

· The form of the law of universal gravitation suggests that the force exists everywhere in space. However, as distances become very large, the force becomes very small. So, if we could move things infinitely far apart, no force would exist between them and no work would need to be done against gravity to move them around. This sounds like a place where there is no gravitational potential energy. So, we define gravitational potential energy to be zero when objects are infinitely far apart.
· But, we still have a problem -- the force of universal gravitation is not a constant force. So, to calculate the work done to separate two objects from each other, we must be looking at an area under a force vs. displacement curve. This area is not a simple area (like a rectangle or a triangle) and requires the rules of calculus. We will just look at the result: Work = - .
· Now, let's think about what we know from before and what we know now. Before, when we lifted an object away from the surface of the earth, it's potential energy increased and we could make the potential energy be zero wherever we wanted. Now, it would be nice if the change in potential energy as objects get farther apart is still positive (i.e. potential energy increases with distance) and the potential energy is zero when objects are infinitely far apart. How can we make this work?
· The answer is to make the potential energy negative! . Try it -- if we move two objects from a distance of r apart to a distance of 2rapart what happens? The initial potential energy is - GMm/r and the final potential energy is -GMm/2r. The change in potential enegy = final - initial = -GMm/2r - (-GMm/r) = +GMm/2r. The change is still positive, so it works!
· Finally, let's think about the work done one more time. The force we are working against always points towards the center of the stationary mass (the one we aren't moving). Any movement that is perpendicular to this direction requires no work be done (definition of work is movement parallel to the direction of force). So, we can make all kinds of crazy paths and it won't matter as long as the net distance from the center of the stationary object is the same. Another way of saying this is "the work done is independent of the path". It only depends on the starting and ending distances from the center of the stationary object.
7. State the expression for gravitational potential due to a point mass.
· Just like we defined the concept of gravitational field to describe the force between two masses separated by a distance r, we can define the concept of gravitational potential to describe the stored energy in the gravitational field when two objects are separated by a distance r.
· Gravitational Potential (V) is defined as the gravitational potential energy for a certain separation divided the mass of the moving object. In equation form: .
· So, just as force can be thought of as the interaction between a small mass and a gravitational field at a location in space, gravitational potential energy can be thought of as the interaction between a small mass and the gravitational potential at a location in space.
· Gravitational potential can be calculated using:
8. Explain the concept of escape speed.
· To talk about escape speed, we first need to look at the total energy of a small mass that is near a larger, fixed mass. E = EP + EK = . Now, we know that as the small object moves away from the larger object, its kinetic energy will decreases and its potential energy will increase. When the objects are infinitely far apart, the potential energy will be zero, the force between the objects will be zero, and the kinetic energy will maintain the value that it had when the force became zero.
· So, escape speed is then defined as the initial speed necessary to launch an object so that it will reach infinity and stay there. This means that its final speed will be exactly zero, or some value larger than zero. We usually take the escape speed to be the value necessary to exactly arrive at infinity with no kinetic energy remaining.
9. Derive an expression for the escape speed of an object from the surface of a planet.
· Using our definition of escape speed from above, it's not too hard to work this out. We will start out with our small object located at the surface of the planet. Let R be the radius of the planet. Then, the two masses can be thought of as initially separated by a distance R (because all of the mass of the planet is thought to be located at its center).
· So, the initial energy of our small object is .
· And, by definition the final energy of the small object when it reaches infinity will be Ef = 0.
· Conservation of energy allows us to set these two equations equal to one another.


·
· So,
· And, . This is the initial speed necessary to escape the gravitational field of the planet.
10. Solve problems involving gravitational potential energy and gravitational potential. These should include problems on escape speed.
Do these problems in Chapter 5: 23, 25, 27, 29, 31, 33, 62, 65, 66
Plus the following:
Gravitational Field
1. What is the magnitude of the gravitational field 2.00 m from a 5.00 kg mass?

2. At what distance above the surface of the earth is the magnitude of its gravitational field 4.90 m/s2 if the gravitational field at the surface has magnitude 9.80 m/s2?

3. What is the magnitude of the gravitational field of the moon at a height of 200 m above the moon’s surface? (Rmoon = 1.74 x 106 m and Mmoon = 7.35 x 1022 kg)

Gravitational Potential Energy
4. Calculate the escape velocity for an object for an object (a) from the surface of the moon (b) from the surface of saturn. (Msaturn = 5.69 x 1026 kg and Rsaturn = 6.03 x 107 m)

5. An artillery shell with mass m is shot vertically upward from the surface of the earth. If the shell’s initial speed is 8.00 x 103 m/s, to what height above the surface of the earth will it rise? (neglect air resistance) (Mearth = 5.98 x 1024 kg and Rearth = 6.38 x 106 m)