Download The Beginning of Modern Astronomy

The Beginning of Modern
Astronomy
Isaac Newton (1642-1727)
•
•
•
•
did research on optics (the properties of light)
invented calculus
discovered the three laws of Motion
discovered the law of universal gravitation
Motion Concepts
•
•
•
•
•
•
inertia: resistance to a change in motion.
mass: numerical measure of inertia, amount of material in the object.
speed: how fast something moves.
velocity: speed and direction.
acceleration: rate of change of velocity.
Acceleration in the direction of motion speeds up the object. Acceleration
perpendicular to the object’s path changes the direction of its motion.
momentum: mass × velocity.
Newton’s Laws of Motion
• An object moves with constant velocity unless acted on by an
unbalanced external force.
• When an object is acted on by an unbalanced force, it accelerates in the
direction of the unbalanced force. The magnitude of the acceleration is
related to the magnitude of the net (unbalanced) force by the equation
Fnet = ma, where m is the mass of the object and a is its acceleration.
• When two objects interact, they exert equal and opposite forces on
each other. FA on B = - FB on A.
Circular Motion, Centripetal Acceleration, and
Centripetal Force
• An object in uniform circular motion is accelerating even if its speed is
constant.
• In this case, there is no acceleration along the path of the object; the
acceleration, called centripetal acceleration , is perpendicular to the
path (toward the center of the circle).
• The force that causes a centripetal acceleration is called a centripetal
force.
v2
The centripetal acceleration is calculated using the formula a 
r
where a is the
centripetal acceleration, v is the speed, and r is the radius of the circle.
Example 1
The average distance from Earth to the Moon is 3.84×108 m, and the moon’s
average orbital speed is 1022 m/s. Calculate its centripetal acceleration.
2
r  3.84  108 m / s
v  1022m / s
m

1022


s 

a
3.84  108 m
a  2.73  103 m / s 2
Newton’s Law of Gravity
Any two particles in the universe attract each other with a force that is directly proportional
to the product of their masses and inversely proportional to the distance between them.
FG  
GMm
r2
The minus sign reminds us that
the force is attractive.
G = 6.673×10-11 Nm2/kg2
M and m are the masses of the two particles.
r is the distance between their centers.
M
r
m
A spherically symmetric object is one whose mass is distributed equally in all
directions. The force on a particle outside an object with spherical symmetry is
the same as if all of of the object’s mass were concentrated at its center. This
allows Newton’s law of gravity to be used for things like planets, which are
almost spherically symmetric.
Relation Between Weight and Mass
weight = the gravitational force on an object.
Consider the falling object shown at the right. If air resistance
is negligible, the only force acting on it is gravity.
Fnet = FG
Fnet = ma
GMm
F
R2
M = mass of Earth m = mass of object
R = radius of earth
ma 
GMm
GM

a

R2
R2
The acceleration due to gravity is directly proportional to the mass of the planet and
inversely proportional to the square of the distance from the planet’s center. It is usually
denoted by the symbol g.
g
GM
R2
W = weight = mg
Near earth’s surface, a = g = 9.8 m/s2.
When air resistance is negligible, the acceleration of a falling body does not depend on
its weight.
Newton’s second law of motion and his law of gravity enable us to determine the
masses of planets and stars.
Example 2
The acceleration due to gravity at the surface of the Earth is 9.8 m/s2, and the radius of the
Earth is 6380 km. What is the mass of Earth?
GM
g 2
R
gR  GM
2
g  9.8m / s
gR 2
M
G
2
G  6.673 10
11
Nm2
kg 2
2
m

6
9.8
6.38

10
m



s2 

M
2
11 Nm
6.673  10
kg 2
R  6380km
M?
M  5.98  1024 kg
Example 3
An astronaut whose weight is 150 lb on Earth is launched to an
altitude of twice earth’s radius. What is his weight at that altitude?
GM
W  mg  m 2
R
R 2  3 R1
W2 R12  R1 

 
W1 R 2 2  R 2 
W1  150 lb
2
R 
W2  W1 1 
 R2 
1
W2  150 
 3
2
2
W2 
150 lb
 16.7 lb
9
Example 4
From the moon’s orbital speed and its distance from the center of Earth, Newton knew that
the centripetal acceleration of the Moon is 0.0027 m/s/s; this is the experimental value of
the moon’s acceleration. Since the (average) distance from Earth to the Moon is about 60
times the radius of Earth, Newton’s law of gravity predicts that the acceleration should be
9.8
m
9.8
m
2
2
m
s
s
a

 0.0027
3600
602
s2
The agreement between this theoretical value and the
experimental value was an important confirmation of
Newton’s law of gravity.
Conservation of Angular Momentum
The angular momentum of a particle moving relative to some
fixed point O is defined by L  mv  r.
v  the velocity component perpendicular to OP.
P
v
O
When the net force on a particle is always directed toward a fixed point, its angular
momentum relative to that point does not change with time; i.e., its angular momentum
is conserved.
The gravitational force of the Sun on a planet is always directed toward the Sun, so the
angular momentum of the planet relative to the sun is conserved.
It can be shown that the conservation of the angular momentum of a planet is equivalent
to the statement that the line from the Sun to the planet sweeps out equal areas in equal
times, so Kepler’s second law of planetary motion is equivalent to the law of
conservation of angular momentum applied to a planet in orbit around the Sun.
v
Kinetic Energy, Radiative Energy, and Potential Energy
Energy is the ability to move an object while exerting a force on it.
The energy of an object due to its motion is called kinetic energy. It is defined by the equation
1
K  mass speed2
2
Potential Energy is the energy that a group of objects has because of their relative positions.
There is no single formula for potential energy.
When you exert a force on an object and cause it to move, you put energy into it. The
process of putting mechanical energy into an object is called doing work on the object.
Radiative Energy is the energy of the electric and magnetic fields in electromagnetic
radiation.
4 π 2a 3
P 
G m M 
The law of conservation of energy can be used to prove Kepler’s third
law of planetary motion and add some detail to it.
2
P = the sidereal period of the planet (or satellite).
m = the mass of the planet (or satellite).
M = the mass of the Sun (or planet).
G = 6.673×10-11 Nm2/kg2.
Example 5
The sidereal period of the Moon is 27.32 days, and its average distance from Earth is
384,000 km. Calculate the mass of Earth.
Assume that the mass (m) of the Moon is negligible compared to that of Earth (M).
4 π 2a 3
2
P 
G m M 
If m  M
4 π 2a 3
2
P 
GM
Solving for M
P = 27.32 days = 2.732×101×8.64×104 s = 2.360×106 s
a = 3.84×105 km = 3.84×105×103 m= 3.84×108 m
G = 6.673×10-11Nm2/kg2
M
4 π 2a 3
M
GP 2
4 π 2  3.84  108 
3
 6.673  10  2.360 10 
11
6 2
M = 6.02×1024 kg
Circular Velocity
The orbital speed of a satellite in a circular orbit of radius r around a planet of mass M is
V 
c
GM
where G = 6.67310-11Nm2/kg2.
r
Example 6
Calculate the orbital speed of a satellite in a circular orbit 150
km above the surface of Earth. Assume that the radius of earth
is 6380 km and its mass is 5.98×1024 kg.
r = 150 km + 6380 km = 6530 km
r = 6.53×106 m
r = 6.53×103 x 103 m
M = 5.98×1024 kg
 6.673 10 5.98 10 
11
Vc 
24
6.53 106
Vc = 28,500 km/hr = 17,500 mi./hr
Vc  7.82  103 m/s  7.82km/s
r
M
Escape Velocity
If an object is at a distance r from the center of a planet of mass M, it can escape from
the planet if its speed is at least equal to Ve where
Ve 
2GM
r
Ve is called the escape velocity.
Example 7
Calculate the escape velocity from the surface of Earth.
r = 6.38×106 m
Ve 
M = 5.98×1024 kg
 2   6.673 1011  5.98 1024 
6.38 10
6
Ve  1.12  104 m/s
Ve  11.2km/s
Tides
The length of a day increases by about 0.0023 seconds per century, and the Moon
moves farther from Earth by about 3.8 cm per year. Why?
900 million years ago, earth’s day was 18 hours long.