Universal gravitation •  CAPA due today. •  Today will finish up with

the hinge problem I started on Wednesday.

•  Will start on Gravity.

Hinge Problem from Wednesday

Hinge Problem cont.

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Fx = 0 = FNx −T cosθ∑Fy = 0 = T sinθ −W1 −W2 − FNy = 0∑

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τ = 0 = T sinθ •L −W2 •L −W1∑ •L2

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T = (W2 +W1

2) /sinθ

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FNx = (W2 +W1

2)cotθ

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FNy = −W1 /2

Guessed wrong on direction

Things to keep in mind Force of gravity acts at the center of mass

Tension force must be in the same direction as the rope

There are often multiple choices for a reasonable axis about which to calculate torques. All of them are OK and you should still be able to solve the problem. A good choice just makes the problem easier.

Clicker question 1 Set frequency to BA

A mass M is placed on a very light board supported at the ends, as shown. The free-body diagram shows directions of the forces, but not their correct relative sizes.

(2/3)L L/3

M

Mg

FL FR

What is the ratio FR/FL?

A: 2/3 B: 1/3 C: 1/2 D: 2

E: some other answer

Clicker question 1 Set frequency to BA

A mass M is placed on a very light board supported at the ends, as shown. The free-body diagram shows directions of the forces, but not their correct relative sizes.

(2/3)L L/3

M

Mg

FL FR

What is the ratio FR/FL?

A: 2/3 B: 1/3 C: 1/2 D: 2

E: some other answer

Sum of the Torques = 0, or - FL*(2/3L ) + FR*(L/3) = 0, so FR/FL = 2/1.

Newton’s Law of Gravitation can be written as

Newton’s Law of Gravity Newton and Einstein are generally thought to be the two greatest physicists ever.

Not only did Newton come up with the three laws of motion and invent calculus, he was the first to realize that the force associated with things falling was also responsible for astronomical phenomena.

Between any two masses (here m1 & m2) there is an attractive force proportional to the product of the masses and inversely proportional to the square of the distance between them.

is the force of gravity which is felt by each mass and directed towards the other mass.

Gravitational Force

Newton figured out the 1/r2 dependence assuming that the celestial objects and the Earth were point particles.

Therefore for any two spherically symmetric objects, the distance r that enters into the force of gravity is the distance between the centers of the spheres.

By inventing integral calculus he could prove that for a mass m2, outside a spherical mass m1, the force of gravity was as if all of the mass m1 was in the center of the sphere.

Newton’s Shell Theorem

A uniform spherical shell of matter attracts a particle that is outside the shell as if all of the shell’s mass were concentrated at its center.

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Let ρ =M

4πR2

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Find F =GMm /r2

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dA = (2πRsinθ)Rdθ

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dM = ρdA =12M sinθ dθ

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dF =GmdMs2

cosα =12GmMs2

cosα sinθdθ

Force rules is the force of gravity with

Newton’s 2nd law still works. The net force on an object determines the object’s acceleration:

Remarkably, the mass in Newton’s 2nd law (called the inertial mass) is the same as the mass in the law of gravitation (called the gravitational mass). Einstein figured out (230 years later) that this “coincidence” could be explained by assuming space and time were curved (in the theory of general relativity). Remember, force is still a vector and the law of superposition still works. To find the net gravitational force on an object, determine the magnitude and direction of the force from all other masses and then add these forces together.

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F 1,net =

F 12 +

F 13 +

F 14 + ......+

F 1n

Two asteroids in inter-galactic space are a distance r = 20 km apart. Asteroid 2 has 10 times the mass of asteroid 1. The magnitudes of the accelerations of asteroids 1 and 2 are a1 and a2, respectively. What is the ratio a1/a2?

Clicker question 2 Set frequency to BA

A. 1/100 B. 1/10 C. 1 D. 10 E. 100

m1 m2 r = 20 km

Two asteroids in inter-galactic space are a distance r = 20 km apart. Asteroid 2 has 10 times the mass of asteroid 1. The magnitudes of the accelerations of asteroids 1 and 2 are a1 and a2, respectively. What is the ratio a1/a2?

Clicker question 2 Set frequency to BA

A. 1/100 B. 1/10 C. 1 D. 10 E. 100

m1 m2 r = 20 km

The force on m1 is the same as the force on m2:

Acceleration is force divided by mass so

and which gives us

Comments about Earth The density of the Earth is higher than most of the other planets in our solar system. Sources vary when it comes to the density of the Earth. ~5.5 g/cm3

Inner Core: solid, ~13 g/cm3 mainly Fe + Ni

Outer Core: Thought to be mainly responsible for earth’s magnetic field.~11g/cm3

Lower Mantle: Silicon, magnesium, Oxygen ~3.5g/cm3

Earth is not a sphere + rotating! The equator radius is larger than the polar radius by 21 km! Means gravitational acceleration is larger at poles than at equator!

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F = ∑ FN −mag = m(−v 2 /R)

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FN = mag −mω2R

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g = ag −ω2R

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ω =dθdt

=2π radians

24 hr

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R = 6.37 ×106m

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g = 9.8m /s2 − 0.034m /s2

Force of gravity on Earth How does correspond to our new force ?

If we consider mass 2 to be the Earth (ME) and r to be the radius of the Earth (RE) then we can write

Using known values we can find that

So, on the surface of the Earth, the force of gravity between the Earth and an object m1 is

We can only use if the distance above the surface is very small compared to the radius.

Used to find near the Earth’s surface

Planet X has the same mass as the Earth, but ½ the radius due to its higher density. What is the acceleration of gravity on Planet X?

Clicker question 3 Set frequency to BA

A. ¼ g B. ½ g C. g D. 2 g E. 4 g

For Earth

Used to find near the Earth’s surface

Planet X has the same mass as the Earth, but ½ the radius due to its higher density. What is the acceleration of gravity on Planet X?

Clicker question 3 Set frequency to BA

A. ¼ g B. ½ g C. g D. 2 g E. 4 g

For Earth

For Planet X

The higher density of Krypton (being made of Kryptonite) makes the force of gravity at the surface stronger, meaning Superman must be stronger to do any old normal thing.

A rock is released from rest in space beyond the orbit of the Moon. The rock falls toward the Earth and crosses the orbit of the Moon. At this point, the acceleration of the rock is…

A. greater B. smaller C. the same as

the acceleration of the Moon.

Clicker question 4 Set frequency to BA

Earth

Moon

rock

A rock is released from rest in space beyond the orbit of the Moon. The rock falls toward the Earth and crosses the orbit of the Moon. At this point, the acceleration of the rock is…

A. greater B. smaller C. the same as

the acceleration of the Moon.

Clicker question 4 Set frequency to BA

Earth

Moon

rock

If the Moon and the rock are a distance r from the center of the Earth then the acceleration of either mass can be determined by

independent of whether it is the Moon or a rock

Note, the speeds are probably not the same but the accelerations are!

Gravitational potential energy When we used we found a potential energy of .

What is the potential energy associated with the force ?

A while ago we learned force is the derivative of potential energy.

The potential energy gives the force when you take the derivative with respect to r.

To make sense, potential energy should increase as the distance increases and be smallest when the objects are closest together.

Gravitational potential energy Potential energy increases (less negative) as the separation increases. This is what we wanted.

Maximum potential energy is 0 when r approaches infinity.

Since two objects cannot share the same space, r > 0. The minimum potential energy is when the objects are touching.

Earth’s gravitational potential energy

Distance from center of the Earth (km)

RE = radius of Earth = 6380 km

Potential energy due to Earth’s gravity is

where r is the distance from the center and h is the height above the surface of the Earth.

Suppose a rock is released from rest at r = 30000 km. Initially it only has potential energy. It will start falling, converting potential energy to kinetic energy. The total energy remains the same.

The rock cannot go past r = 30000 km because it would have negative kinetic energy at that point (which is impossible).

For total energy < 0 the object is bound by the gravitational field (and orbits are ellipses). Examples are planets around the sun.

Effect of total energy on trajectory

Distance from center of the Earth (km)

RE = radius of Earth = 6380 km

If the total energy were 0 then it is possible for the object to make it to r = ∞.

For total energy > 0, object is unbound with a hyperbolic orbit.

For total energy of 0 the object is barely unbound (parabolic orbit).

We can identify three basic scenarios for a total energy which is positive, negative, or 0.