Conceptual Physics Chapter 13 1

Universal Gravitation

Conceptual Physics Chapter 13 2

The Falling Moon

¤ Newton recognized that in order for the moon to travel along a circular path, there must be a force acting on the moon.

¤ Newton suggested that the cause for heavenly motion (the orbit of the moon around the earth) was the same as the cause for earthly motion (the apple falling from the tree).

Conceptual Physics Chapter 13 3

The Falling Moon

The reason the moon falls around the earth instead of intothe earth is because it has a tangential velocity that allows it to stay in a circular path.

Conceptual Physics Chapter 13 4

The Falling EarthThe earth (and all of the planets) is attracted to the sun and falls around the sun.If the earth did not have the tangential speed that it does, it would fall from orbit and crash into the sun!

Conceptual Physics Chapter 13 5

Universal Gravitation

Newton concluded that all bodies have an attractive force between them that results from each bodies’ mass. This force is found to be directly proportional to the product of the masses and inversely proportional to the square of the distance between the bodies.

d2FG =

m1·m2G·

Conceptual Physics Chapter 13 6

Law of Universal Gravitation

d2FG =

m1·m2G·

FG is the gravitational force

G is the universal gravitational constant (G = 6.67 x 10-11 N·m2/kg2)m1 and m2 are the two masses

d is the distance between the two masses

Conceptual Physics Chapter 13 7

We can find the gravitational force between two students, each with a mass of 70 kg, sitting 1 meter apart.

Law of Universal Gravitation

d2FG =

m1·m2G·

6.67 x 10-11 N·m2/kg2(70 kg)(70 kg)

(1 m)2=

= 3.3 x 10-7 N

The friction between the floor and the chairs in which these students sit is many times larger than this gravitational force.

Conceptual Physics Chapter 13 8

We can instead find the gravitational force between the earth and a 70-kg student standing on the surface of the earth.

Law of Universal Gravitation

d2FG =

m1·m2G·

6.67 x 10-11 N·m2/kg2(70 kg)(6.0 x 1024 kg)

(6.4 x 106 m)2=

= 684 N

This force is considerable due to the enormous mass of the earth.

this is the radius of the earth

Conceptual Physics Chapter 13 9

We can find the gravitational force acting on any mass positioned on the surface of the earth by multiplying that mass by G and the mass of the earth and then dividing by the square of the radius of the earth.

Law of Universal Gravitation

d2FG =

m1·m2G·

(6.4 x 106 m)2=

= 9.8 N/kg · m

(6.0 x 1024 kg)(70 kg)6.67 x 10-11 N·m2/kg2 ·m

Conceptual Physics Chapter 13 10

Inverse Square Law

Notice the effect of squaring the distance in the denominator. The greater the distance from the earth’s center, the less an object will weigh.

The weight decreases with the square of the distance. The graph of F vs d forms a hyperbola, approaching zero, but never actually reaching zero.

We can show the variation in gravitational force as a function of distance on an F vs d graph.

Conceptual Physics Chapter 13 11

Question

How is the earth able to exert a force on objects that are not in contact with the earth (e.g., the earth exerts a force on the moon)?

Conceptual Physics Chapter 13 12

Gravitational Fields

¤ A gravitational field surrounds the earth and extends infinitely far in all directions. The direction of this gravitational field is towards the earth.

mg =

FG d2

m1·m2G·

m=

d2

mG·=

¤ The strength of the gravitational field at any point is the gravitational force per unit mass and can be described by:

Conceptual Physics Chapter 13 13

Gravitational Fields

The gravitational field strength is dependant only on the mass of the body that generates the field and the distance to the center of that body.

d2

mG·=

g

The gravitational field strength near the earth varies with distance in the same way that the gravitational force does!

Conceptual Physics Chapter 13 14

Gravitational Field inside the Earth

If we drilled a hole through the center of the earth, we could explore the gravitational field strength from the surface of the earth inward to the center.

At the surface of the earth, the entire mass of the earth is positioned beneath you and all of this mass is pulling in the same direction. This leads to the usual gravitational field strength of 9.8 m/s2.

g = 9.8 m/s2

Conceptual Physics Chapter 13 15

Gravitational Field inside the Earth

At the surface of the earth, the entire mass of the earth is positioned beneath you and all of this mass is pulling in the same direction. This leads to the usual gravitational field strength of 9.8 m/s2.

g = 9.8 m/s2

If we jump in the hole, we will fall faster and faster towards the center of the earth. Although we would be speeding up, our acceleration would be decreasing.

At a position halfway to the center of the earth, a considerable portion of the earth’s mass is now pulling upward on us.

g = 4.9 m/s2The remaining mass still pulls downward on us.

This leads to a field strength that is only half of that experienced at the surface of the earth.

Conceptual Physics Chapter 13 16

Gravitational Field inside the Earth

g = 9.8 m/s2

g = 4.9 m/s2This leads to a field strength that is only half of that experienced at the surface of the earth.

As we continue to fall towards the center, our acceleration continues to decrease linearly as our speed increases.

At the very center of the earth, half of the mass of the earth is now above us and pulls upward.

There is also half of the mass of the earth below us pulling downward.

In fact, the surrounding mass of the earth is pulling outward equally in all directions. The net force would be zero – we would experience weightlessness!

g = 0 m/s2

Conceptual Physics Chapter 13 17

In fact, the surrounding mass of the earth is pulling outward equally in all directions. The net force would be zero – we would experience weightlessness!

Gravitational Field inside the Earth

g = 9.8 m/s2

g = 4.9 m/s2

g = 0 m/s2

Although the net force here is zero, the body has its greatest speed and its inertia carries it right past the center of the earth.

As soon as we pass the center, the net force will be upward and will continue to grow as more and more of the earth’s mass is now positioned above us.

Less and less of the earth’s mass is below us as we fall farther towards the surface.

Our acceleration is increasing as our speed is decreasing.

g = 4.9 m/s2

Conceptual Physics Chapter 13 18

Our acceleration is increasing as our speed is decreasing.

Gravitational Field inside the Earth

g = 9.8 m/s2

g = 4.9 m/s2

g = 0 m/s2

g = 4.9 m/s2

We continue to fall towards the surface with an increasing acceleration and a decreasing speed.

At the moment we reach the opposite surface of the earth, our speed falls to zero and we once again experience our usual gravitational field.

g = 9.8 m/s2

Now all of the earth’s mass is above us and we feel a force equal to our weight that pulls us right back into the hole.

Conceptual Physics Chapter 13 19

Now all of the earth’s mass is above us and we feel a force equal to our weight that pulls us right back into the hole.

Gravitational Field inside the Earth

g = 9.8 m/s2

g = 4.9 m/s2

g = 0 m/s2

g = 4.9 m/s2

g = 9.8 m/s2

We would continue to fall back and forth through the hole in simple harmonic motion.

Conceptual Physics Chapter 13 20

Ocean Tides

¤ Envision the entire earth as being covered by a thin layer of water of constant depth.

¤ The moon exerts a gravitational force on the solid earth and the water that covers its surface.

¤ The greatest gravitational force will be exerted on the side of the earth nearest the moon, since F ~

d2

1

Conceptual Physics Chapter 13 21

Ocean Tides

¤ This causes the water on the near side of the earth to bulge toward the moon.

¤ The gravitational force acting on the water on the far side is very weak, so the inertia of the water causes a bulge there as well.

¤ The rotation of the earth beneath these two tidal bulges causes two high tides and two low tides each day.

Conceptual Physics Chapter 13 22

Ocean Tides

¤ While the earth spins, the moon progresses in its orbit and appears at the same position in the sky every 24 hours and 50 minutes.

¤ This is why tides do not occur at the same time every day.

Conceptual Physics Chapter 13 23

Effect of the sun vs the moon

¤ Although the sun pulls on the earth with a force that is 180 times stronger than the moon’s pull on the earth, the moon’s pull still causes the greater tidal effect!

¤ Because of the sun’s great distance from the earth, there is not much difference in the way that it pulls on the oceans of the near side of the earth versus those on the far side of the earth.

¤ This small difference will only slightly elongate the earth and produces tidal bulges that are less than half the size of those produced by the moon.

Conceptual Physics Chapter 13 24

Spring Tides

¤ When the sun, moon and earth all line up, the tidal effects from the sun and moon strengthen each other and we have higher than average high tides and lower than average low tides.

¤ These are called spring tides.

¤ If the alignment is perfect, we have an eclipse.

¤ A lunar eclipse is produced when the earth is between the sun and the moon.

¤ A solar eclipse is produced when the moon is between the earth and the sun.

Conceptual Physics Chapter 13 25

Neap Tides

¤ When the sun, moon and earth are positioned at ninety degrees to one another, the tidal effects from the sun and moon work against one another producing lower than average high tides and higher than average low tides.

Conceptual Physics Chapter 13 26

Lunar Orbit

¤ The moon orbits the earth every 28 days.

¤ During this time, we experience two spring tide conditions and two neap tide conditions.

Conceptual Physics Chapter 13 27

Conceptual Physics Chapter 13 28

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Conceptual Physics Chapter 13 30