nats 1311 - from the cosmos to earth examples: pulling a table cloth out from under a table setting...
TRANSCRIPT
NATS 1311 - From the Cosmos to Earth
Examples:
Pulling a table cloth out from under a table setting
The reaction of coffee in a cup when accelerating or decelerating in a car
Tightening of a hammerhead by banging hammer on the ground
Getting ketchup out of a bottle
Not wearing a seatbelt during a head-on car crash
Headrests in a car to prevent whiplash during a read-end collision
NATS 1311 - From the Cosmos to Earth
Velocity and Acceleration
Newton showed that acceleration (a) is the change of a body’s velocity (v) with time (t):
1. Acceleration in the conventional sense (i.e. increasing speed)
a = Dv/Dt
Differential calculus!
Different cases of acceleration:
Velocity and acceleration are vectors.
3. Change of the direction of motion (e.g., in circular motion)
2. Deceleration (i.e. decreasing speed)
a
v
NATS 1311 - From the Cosmos to Earth
Newton’s 2nd Law Explains the Feather and the Ball
1 kg on the Earth weighs 9.8 N or 2.2 lbs
F = W = mg
W = 1kg X 9.8 m/s = 9.8 kg m/s = 9.8 N
Take a 1 kg rock and a 10 kg rock and drop them from the same height
a1 = F1/m1 = W1/m1 = 9.8 N/1 kg = 9.8 m/s = g
a2 = F2/m2 = W2/m2 = 98 N/10 kg = 9.8 m/s = g
NATS 1311 - From the Cosmos to Earth
A body subjected to a force reacts with an equal counter force to the applied force:
That is, action and reaction are equal and oppositely directed, but never act on the same body.
Newton’s Third Law
For every action (force), there is an equal and opposite reaction (force)
NATS 1311 - From the Cosmos to Earth
Examples of Action/Reaction
Swimming - your hands and the water
Walking - your feet and the ground
Driving - a car’s tires and the road
A bug and a car’s windshield
A falling object - the object and the earth
A person pulling a spring
A deflating balloon - the air rushing out and the balloon
Pushing on the wall - your hand and the wall
Rocket ship - expelled fuel and rocket
NATS 1311 - From the Cosmos to Earth
apparent weight - weight force that we actually sense not the downward force of gravity, but the normal (upward) force exerted by the surface we stand on
- opposes gravity and prevents us falling to the center of the Earth - what is measured by a weighing scale.
For a body supported in a stationary position, normal force exactly balances earth's gravitational force
- apparent weight has the same magnitude as actual weight.
If no contact with any surface to provide such an opposing force - no sensation of weight (no apparent weight).
- free-fall - experienced by sky-divers and astronauts in orbit who feel "weightless" even though their bodies are still subject to the force of gravity - also known as microgravity.
A degree of reduction of apparent weight occurs, for example, in elevators. In an elevator, a spring scale will register a decrease in a person's (apparent) weight as the elevator starts to accelerate downwards. This is because the opposing force of the elevator's floor decreases as it accelerates away underneath one's feet.
Apparent Weight
NATS 1311 - From the Cosmos to Earth
The Earth is round - its surface drops about 5 m for every 8 km of distance. If you were standing at sea level, you would only see the top of a 5-meter mast on a ship 8000 m away - remember the story of Columbus and the orange.
Given h=1/2gt2, if t=1 s then h = 5 m. So if a projectile is fired horizontally at ~8 km/s, it will fall fast enough to keep “falling around” the Earth - becomes a satellite. So a spacecraft is in free fall around the Earth - free fall is not an absence of gravity. If a satellite is given a velocity greater than 8 km/s, it will overshoot a circular orbit and trace an elliptical path.
Escape velocity - velocity at which gravity can not stop outward motion - 40,000 km/hr for Earth
Cannonball Animation
Orbital Velocity
NATS 1311 - From the Cosmos to Earth
Momentum is mass times velocity, a vector quantity:Mom=mv
Law of Conservation of MomentumThe total momentum of an isolated system is conserved, I.e., it remains constant.
An outside or external force is required to change the momentum of an isolated system.
The Law of Conservation of Momentum is an alternate way of stating Newton’s laws:
1. An object’s momentum will not change if left alone2. A force can change an object’s momentum, but…3. Another equal and opposite force simultaneously changes some other object’s momentum by same amout
Momentum
NATS 1311 - From the Cosmos to Earth
A Rifle and a Bullet
When a bullet is fired from a rifle, the rifle recoils due to the interaction between the bullet and the rifle.
The force the rifle exerts on the bullet is equal and opposite to the force the bullet exerts on the rifle.
But the acceleration of the bullet is much larger that the acceleration of the rifle - due to Newton’s 2nd law: a = F/m
The acceleration due to a force is inversely proportional to the mass.
The force on the rifle and the bullet is the same but the mass of the rifle is much larger than the the mass of the bullet so the acceleration of the rifle is much less than the acceleration of the bullet.
NATS 1311 - From the Cosmos to Earth
Angular Momentum
Momentum associated with rotational or orbital motionangular momentum = mass x velocity x radius
NATS 1311 - From the Cosmos to Earth
Torque and Conservation of Angular Momentum
Conservation of angular momentum - like conservation of momentum -in the absence of a net torque (twisting force), the total angular momentum of a system remains constant
Torque - twisting force
NATS 1311 - From the Cosmos to Earth
A spinning skater speeds up as she brings her arms in and slows down as she spreads her arms because of conservation of angular momentum
NATS 1311 - From the Cosmos to Earth
The force on a body of mass m1 is:
(Newton’s Second Law)
If this force is due to gravity, then:
m1 cancels out, and:
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F =m1a
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m1a=Gm1m2
d2
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a=Gm2
d2
Newton’s 2nd Law and the Acceleration Due to Gravity
NATS 1311 - From the Cosmos to Earth
The acceleration due to the force of gravity is called g, so:
Mass of the Earth (m2) = 5.97 X 1024 kgRadius of Earth (d) = 6.378 X 106 mG= 6.67 x 10-11 Nm2/ kg2
g= (6.67 x 10-11 Nm2/ kg2) X (5.97 X 1024 kg)/(6.378 X 106 m)2
g= 9.79 m/s2
g does not depend on the mass of the body m1 - so the feather falls at the same speed as the steel ball - Galileo learned this by experimentation (the Leaning Tower of Pisa experiment) - Newton showed why.
Weight is the result of the force of gravity on a body of mass m1:
Therefore all objects on earth having the same mass have the same weight.
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g=Gm2
d2
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W =m1g
NATS 1311 - From the Cosmos to Earth
The acceleration of gravity and therefore a person’s weight is dependent on a planet’s mass and radius.
Planetary Mass, Radius and Weight
NATS 1311 - From the Cosmos to Earth
Newton’s Formulation of Kepler’s Laws
As a planet moves around its orbit, it sweeps out equal areas in equal times - a planet moves slower when it is farther from the Sun and faster when it is closer
Kepler’s Laws were based on observation (experimentation). Newton’s lawsexplained Kepler’s Laws
Kepler’s Second Law
NATS 1311 - From the Cosmos to Earth
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F =m1a
For a circular orbit: (r = radius of orbit)
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a=v2
r
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F =m1
v2
r=Gm1m2
r2
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v2 =Gm2
r
Substitute (2) into (1):F is the force of gravity:
Cancel m1and r; then
(1)
(2)
(3)
(4)
The smaller the radius, the greater the speed.The orbital speed is independent of the mass of the orbiting body (m1). As the radius (the distance to the orbiting body) increases, the orbital speed decreases.
When you swing a ball around, the string exerts a force that pulls the ball inward (gravity for orbiting body). The acceleration is also inward.
NATS 1311 - From the Cosmos to Earth
The square of any planet's period of orbital revolution, P, is proportional to the cube of its mean distance, r, from the sun.
Kepler’s 3rd Law
Orbital Period vs Distance Animation
NATS 1311 - From the Cosmos to Earth
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v2 =Gm2
r
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v=2πrP
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v2 =4π 2r2
P2
From Kepler’s Second Law (previous slide):
Speed around orbit:Circumference (2r)/ timeP=period, time of 1 orbit
(1)
(2)
(3)
(4)
(5)
Combine (1) and (3):
Rearrange terms:
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Gm2
r=
4π 2r2
P2
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P2 =4π2
Gm2
r3
Square both sides:
NATS 1311 - From the Cosmos to Earth
A more complex derivation of this equation yields:
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P2 =4π 2
G(m1 +m2)r3
From this equation, if one knows the mass of the orbiting body (m1), the mass of the central body (m2) may be calculated.
What is the mass of the Sun?
MSun (m1) >> MEarth (m2) so: m1 + m2 m1
M1 = 42r3/GP2
G = 6.67 x 10-11 Nm2/ kg2
r = 1.5 x 1011mP = 3.15 x 1011 s
So:
Msun = 2 x 1030 kg
NATS 1311 - From the Cosmos to Earth
Geosynchronous/Geostationary OrbitsA geosynchronous orbit has a period the same as the rotational speed of the Earth - e.g., it orbits in the same amount of time that the Earth rotates - 1 sidereal day. A geostationary orbit is a geosynchronous orbit at the equator - it always stays above the same place on the Earth - communications satellites, satellite TV, etc…
What is the altitude of a geostationary orbit?
From Newton’s formulation of Kepler’s 3rd Law:
MEarth (m1) >> MSatellite (m2) so:
r = (GMEarth P2/42)1/3
G = 6.67 x 10-11 Nm2/ kg2
P = 3.15 x 1011 s
MEarth = 5.97 X 1024 kgSo:
R = 42,000 km above the center of the Earth and the altitude is about 35,600 km
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P2 =4π 2
G(m1 +m2)r3