Gravity

Unanswered questionsGalileo describes falling objects by rolling objects down a ramp.

But why does everything accelerate the same rate regardless of its mass?

Kepler describes planetary motions.

What force can account for the elliptical paths of the planets?

Two seemingly unrelated observations, but…

Newton unites the two in one master stroke at the age of 24.

Inverse Square Law

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ra c =

2π( )2R

T 2

T 2 = kR3

r a c =

2π( )2R

kR3 ~1

R2

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FG = maC

~ m1

R2

~ Mm1

R2

FG =GMm

R2

Circular motion result

Kepler’s observational law

Centripetal acceleration behaves like inverse square

Newton’s 2nd Law for centripetal direction

Constant of proportionality determined by Cavendish

Newton’s 3rd Law suggests both masses are important

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G = 6.67 ×10−11 Nm 2

kg2

The Force Law

Direction?Along a line connecting the center of the two masses.

Action at a distance. How does the force GET from the Earth to the moon?

“I feign no hypothesis regarding action at a distance.”

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FG =GMm

R2

m

M

R

Potential Energy Revisited

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FG = mg

Remember how we used to write the force of gravity.

From this, we derived an expression for gravitational potential energy:

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UG = mgh

This only applies near surface of earth! (When GME/RE2=9.8 m/s2)

More generally, we have

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UG = −GMm

R

As usual, we have a choice for where we set UG = 0. When using this equation, the choice is made for us.

Two ApplicationsTerrestrial (free fall near the surface of a planet or star)

Celestial (circular orbit around a planet or star)

This most general expression is always true, but sometimes the first expression is simpler to implement (it has limited application, however, so be careful!).

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FG = mg

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FG =GMm

R2

Terrestrial ApplicationExample:

The radius of the Earth is 6.4 x 106 m and the value of g is 9.8 m/sec2. What is the mass of the Earth?

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FG =GME m

R2

FG = gm

GME m

R2 = gm

ME =gR2

G

=9.8 6.4 ×106

( )2

6.67 ×10−11 = 6.0 ×1024 kg

Celestial Application

2

2

22

2411

2

6

or,

6 106.67 10

7680

6.79 10

c G

s es

e e

F F

m mvm G

R Rm m

v G R GR v

xR x

x m

Radius of the earth is about 6.38 x 106 m at the equator. That gives the altitude above the surface to be:6.79x106 – 6.38x106 = 0.41x 106 m, or 410 km.

“g” ~ 8.68 m/s2

The ISS orbits the earth with a speed of approximately 7680 m/s.What is the orbital radius of the station, and what is its altitude?

Torque

Dynamics

Which applied force results in the largest angular acceleration of the bolt?

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rF 1

€

rF 2

€

rF 3

Dynamics

€

rF 1

€

rF 2

€

rF 3

Which applied force results in the largest angular acceleration of the bolt?

Dynamics

€

rF 1

€

rF 2

Which applied force results in a clockwise angular acceleration? A counter-clockwise angular acceleration?

O

€

rr

€

rF Perp

€

rF Par

Dynamics quantified

Consider a force acting on a rigid body, some distance away from a fixed pivot point.

Only the perpendicular part contributes to rotation!Where does the parallel part go?

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rτ = r

r r F Perp

Can split the force into components.

Dynamics quantified

Which applied force results in the largest angular acceleration of the bolt?

€

rF 1

€

rF 2

€

rF 3

Dynamics quantified

€

rF 1

€

rF 2

€

rF 3

Which applied force results in the largest angular acceleration of the bolt?

Direction of torque

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rF 1

€

rF 2

But torque is a vector, and vectors only point in a single direction.

Direction of torque is given by right hand rule.Draw and tip to tailPoint fingers of right hand in the direction of Curl fingers in direction ofThumb points in direction of torque: either into (clockwise, negative) or out of

(counter-clockwise, positive) page

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rr

€

rF

€

rr

€

rF

€

rr

€

rr

Direction of torque is the direction the applied force tends to cause the object to rotate.

F1 provides a clockwise torque.F2 provides a counter-clockwise torque.

Newton’s LawsNewton’s 1st Law – If there is no net torque on an object, then the object rotates at a

constant angular velocity (could be zero angular velocity).Newton’s 2nd Law – If the net torque on an object about a point is not zero, then the

net torque produces an angular acceleration about that point.

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rF

€

rr

€

τ =rF

= r ma( )

= r m rα( )( )

= mr2α

The quantity mr2 is the rotational equivalent of mass, and is called moment of inertia.

Newton’s 3rd Law – For every action, there is an equal and opposite reaction.

O

Example

20m 40m

A B

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rτ net, A =

r τ T, A +

r τ B, A

0 = (±)FT rT + (±)FB rB

0 = −FT rT + FB rB

FB =FT rT

rB

=8000(20)

(60)= 2667 N

A truck crosses a massless bridge supported by two piers. What force much each pier exert when the truck is at the indicated position?

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rF A

€

rF B

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rF T = 8000 N

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rτ net, B =

r τ T, B +

r τ A, B

0 = (±)FT rT + (±)FA rA

0 = +FT rT − FA rA

FA =FT rT

rA

=8000(40)

(60)= 5333 N

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2667 + 5333 = 8000 N

Sample Torque Problems

X = __________

X = __________