Transcript Newton and Kepler - Physics Presentations

Newton and Kepler
Newton’s Law of Gravitation
The Law of Gravity
Isaac Newton deduced that two particles of masses m1 and
m2, separated by a distance r, will attract each other with a
force
This is the universal law of gravity. If two spheres were
treated as if all the mass were concentrated at the center of the
sphere, the same law applies. The universal gravity constant
G is
.
The force of gravity on a person of mass m due to the earth’s
mass ME is called the person’s weight. Thus
so
At the earth’s surface, r ' RE and g = 9.80 m/s2.
Gravitational Potential Energy
Gravity is a conservative force, and we may define a potential energy
associated with it. The gain in potential energy is the work you must do to
lift a mass from one point to another. Work is done against gravity only
when the displacement is up or down. Going sideways to r doesn’t require
any work. The work done is the gain in potential energy U(r)
For objects a small distance h above the surface of the earth where h <<
RE., we set r1 = RE and r2 = RE + h and the equation becomes
We neglected hR << R2.. If we choose U = 0 at r = RE, then
U = mgh
Tides
The tidal force arises because of the gravitational force between
the two bodies acts on an axis connecting the two centers of
gravity. The earth is subject to the moon's gravitational forces
causing the water in the oceans to change their shape, forming
bulges on the sides near the moon and far from the moon. The
bulges lie on the axis connecting the two centers of gravity of
the bodies.
When the Moon and Sun are in a line with Earth, the tides
are higher. We call those spring tides. When they're at right
angles in outer space, the tides are lower. We call those
neap tides.
The moon, which is approximately 386,240 km from the
earth, exerts a greater influence on the tides then does the
sun, which sits 150 million km from the earth. Although
the mass of the moon is only about a hundred millionths of
the sun's mass, it still exerts a greater force on the ocean
than the sun does because it is about 400 times closer to the
earth than the sun is. The sun's tidal influence is thus about
half (46 percent) of the moon's.
Problems
1.) Determine the minimum speed that a rocket at the surface of the
earth must have to escape completely the earth’s gravity field.
Solution
The rockets energy is conserved. We want the rocket to go far away
(r = 1), and barely reach there (v = 0). Thus U1 + KE1 = 0 + 0.
so
vesc is the escape velocity. If the values for RE = 6371 km and ME =
5.9736 £ 1024 kg are inserted in the equation, then the escape
velocity from the earth is 11.2 km/s. For the values of Rsun=6.955 £
105 km and Msun= 1.99 £ 1030 kg, the escape velocity from the
solar system is 618 km/s.
Kepler’s First Law
Every planet moves in an elliptical orbit with the sun at one focus.
Kepler’s Second Law
As a planet moves in an orbit, a line drawn from the sun to the planet
sweeps out equal areas in equal time intervals.
Kepler’s Third Law
It T is the period and a is the length of the semimajor axis of a planet’s
orbit, then the ratio
is the same for all of the planets.
Kepler’s Second Law is derived by using the conservation of
angular momentrum.
because the lever arm is 0 and v is parallel to v, the equation
equals zero. Thus the angular momentum is a constant and is
conserved.
In time dt the planet has displacement
out an area dA. But the magnitude of
the area of the parallelogram formed by
half of this so
and sweeps
is
,
and
. dA is just
or
Kepler’s third law is easy to prove using the force of gravity.
The gravitational force is equal and opposite to the centripetal
force needed to move the planet in a circle. Thus
The orbital speed is v = 2¼ r/T where T is the period of the
motion.
or
Problems
1. ) Geosynchronous satellites are placed in an orbit in the
equatorial plane such that they are always above a given point
on the earth’s surface, that is, their period is 24 hours. How far
from the center of earth is such a satellite?
Solution
From the derivation of Kepler’s third law we have the equation
Substituting T = (24 h)(3600 s/h) and ME = 5.9736 £ 1024 kg
and solving for r results in r = 42,300 km.
2.) The moon goes around the earth every 27.3 days. The mass
of the earth is 5.98 £ 1024 kg. Determine the earth to moon
distance
Solution
Solve
for r. r = 3.83 £ 108 m.
3.) An artificial satellite of mass m travels at a constant speed
in a circular orbit of radius R around the earth (mass ME).
What is the speed of the satellite?
Solution
The centripetal force on the satellite is provided by Earth’s
gravitational pull. Therefore
Solving this equation for v yields
Note that the satellite’s speed doesn’t depend on its mass, even if
it were a baseball, if its orbit radius were R, then it would still be
.