Prof.
Clements Notes/Hints for Physics Courses
Chapter
6 Part 2 Open Stax College Physics or Most Introductory Physics
Courses
Gravity,
Orbits, Escape Velocity, "Weightlessness", Cavendish
Experiment
Terms
you should know at the end of this unit: gravity, Cavendish,
“weightlessness,” geosynchronous
Newton’s Universal
Law of Gravitation
The force of gravity
can be calculated by this equation:
G M1
M2 Where F is the force, G is a constant,
F = -------------- M1 and M2 are the masses that are
attracting,
R2 R is the distance between the centers of the two
(spherical) objects
It takes two objects to
produce a gravitational force.
The gravitational force
always attracts, never repels.
If the mass is greater
the force will be greater.
If the distance between
the centers of the objects is greater the force will be less.
The effect of R
in this case is called an inverse square relation.
The force gets smaller as the distance becomes larger at a rate
given by the square of R.
G
= 6.673×10−11N
⋅
m2
/ kg2
This value was not known until 1798.
e.g. An astronaut
weighs 180 pounds on the Earth. How much will the astronaut weigh
when the astronaut is a distance of two times the radius of the Earth
above the Earth? You don’t need to do a detailed calculation. You
can use the principle of proportions to solve this.
Answer: The 180 pounds
is the force value when R = 1 REarth . The mass of the
Earth and the mass of the astronaut do not change when the astronaut
is in space compared to their values on the Earth. We do not have to
calculate using the full gravity equation, we may use the concept of
proportions. In space the distance to the astronaut from the center
of the Earth is 3 REarth . The force (weight) will be
less because R is larger. The force will be less by a factor of R
squared. R is larger by a factor or 3 so the force is smaller by a
factor of 9…. (32). The new weight of the astronaut is
180 lbs / 9 or 20 lbs.
You should read the
section in your textbook that discusses how the force of gravity of
the Earth on the Moon creates the observed centripetal acceleration
of the Moon. Newton did not do this calculation since he did not
know the value for G. But he did observe that the centripetal
acceleration for the Moon is about 1/3600 of “g.” Newton also
knew that the distance to the Moon is about 60 * RE.
How would this have
helped Newton confirm he was on the right track for his law of
gravity?
Answer: Q2. Note that
3600 is 602 . Newton reasoned that the force became
weaker proportional to the square of the distance between the centers
of the objects.
Does Jupiter exerts a
gravitational force on you?
Answer: Yes Every
piece of mass in the universe attracts every other piece of mass in
the universe. That is why this gravity law is called the “universal”
law of gravitation. But, the force of gravity on you due to Jupiter
is very small compared to the force of the gravity on you due to the
Earth.
Some scientists who
learned of Newton’s ideas thought that gravity had a limited range
(e.g. some said the Earth’s gravity did not extend beyond our
atmosphere). The study of comets by Edmund Halley (of Halley’s
comet fame) showed that the Sun’s gravity extended throughout the
solar system. As far as physicists know, the force of gravity works
the same way on Earth as it does throughout most of the universe.
The exceptions are situations in which the density value is high
(e.g. White Dwarf stars, neutron stars, black holes).
Mass of the Earth
The acceleration due to
gravity on an object can be measured. The radius of the Earth is
known from surveying on the Earth (and using proportions). After the
value for G, the constant in the Law of Gravitation, was measured
scientists could calculate the mass of the Earth. You should
convince yourself that this is possible.
Hint:
G
MEarth mperson
mpersong
= -------------------- Now, cancel the mass of the person
R2 and solve for Mass of the Earth.
The Earth exerts
gravitational force on objects in space. How could this fact be used
to determine the mass of the Earth?
Answer: Occasionally
small asteroids pass near the Earth. As they fly by the Earth the
force of gravity due to the Earth causes their flight path to change.
Astronomers measure this change and then calculate the mass of the
Earth.
You should substitute
Newton’s expression for the force of gravity into the equation for
centripetal force, Fc = mV2 /R to eliminate
Fc. Then simplify.
mV2 G M1 m
------------ =
--------------
R R2
V2 G M1 so M = V2
R / G or V2 = G M / R
------------ =
-------------- V is the speed of an object in a circular
orbit.
1 R
Masses of other
planets and stars
Think of a way
astronomers might have found a value for the mass of Pluto before the
New Horizon spacecraft flew by Pluto in July 2015.
Answer: If the size of
the orbit and the time for one orbit for a moon of Pluto is measured,
then the mass can be calculated. The force of gravity from the
planet provides the centripetal force for the orbiting moon. The
only unknown is the mass of the planet. This method was used in 1978
when the moon Charon was discovered to calculate the mass of Pluto.
This same technique is used to find the mass for other planets,
asteroids with moons, stars, and galaxies.
The mass of the Sun can
be computed from data for the Earth's orbit.
The mass of an asteroid
can be approximately calculated from data on the orbit of a moon
around the asteroid.
Orbital Speed
You may have noticed
that a rocket headed to orbit does not leave the launch pad in a
completely vertical path. The initial velocity but soon the rocket
starts heading “down range.” i.e. From the Florida launch
facility the rocket starts moving eastward over the Atlantic Ocean.
The vertical motion during launch is needed to get the rocket above
the Earth’s atmosphere to reduce air resistance to the motion. The
horizontal (down range) part of the motion is needed to give the
satellite enough speed to go into orbit around the Earth.
Satellites in orbit
around the Earth are always pulled towards the Earth and accelerate
towards the Earth. In order to not hit the Earth, the satellite has
to have a forward speed parallel to the land underneath the
satellite. As the satellite moves forward and falls towards the
Earth, the land of the spherical Earth also “falls away” from the
satellite. For the Earth, a satellite in circular “low Earth orbit”
has an orbital speed of about 17,500 miles/hour. A low Earth orbit
is an orbit that is only a few hundred miles above the surface of the
Earth. A satellite in low Earth orbit takes about 90 minutes to
complete one orbit.
Objects in orbit have a
certain speed based on the mass of the planet and the size of the
orbit.
1) For a given orbit
size, the speed of the orbit is greater if the mass of the planet is
greater.
2) For a given planet
mass, the speed of the orbit is smaller if the orbit size is larger.
You should consider why
both of these are reasonable statements given what you know about the
calculation for the force of gravity.
Calculate the orbital
period and speed for a satellite that orbits 236 miles above the
Earth. (This is "low" earth orbit and is approximately the
orbital position of the International Space Station.)
Geosynchronous
Satellites
You have probably
noticed that residential satellite TV dishes on different homes are
pointed in the same direction. These dishes are receiving signals
from a geosynchronous satellite.
Why doesn’t the dish
antenna have to continuously move to track the satellite across the
sky during the day? ...and how is it possible to receive a signal
from the satellite at night??
Answer: The satellites
that supply signals to these dish antennas orbit the Earth's center
once every 24 hours. This is the same time taken for the satellite
dish to complete one rotation around the Earth's rotation axis. So,
the two objects have the same rotation rate (e.g. 15 degrees per
hour). From the point of view of the satellite dish antenna the
satellite is not moving across the sky.
A geosynchronous
satellite is in an orbit relatively far from the Earth (22, 236 miles
above the Earth). Reference:
http://en.wikipedia.org/wiki/Geosynchronous_satellite
At this distance the
orbital speed (slow) and orbit size (large) combine to produce one
orbit every 24 hours. The Earth also spins once in 24 hours. The
position of the geosynchronous satellite in the sky is constant so
the dish antenna does not need to move.
Calculate the size of
the orbit for a geosynchronous satellite in orbit around the Earth.
GPS
The GPS network of
satellites are in a medium Earth orbit (not geosynchronous). Your
GPS device continuously receives signals from at least four of these
satellites and can then calculate your position because it knows the
positions of these satellites and the time they sent your device a
signal. Reference:
http://en.wikipedia.org/wiki/Global_Positioning_System
Escape Velocity
An object that is
launched from a planet with “escape velocity” leaves the planet
or star permanently. Its speed is large enough that the force of
gravity cannot stop its motion and have it return to the planet or
star. The escape velocity is equal to the square root of 2 times the
circular velocity. The Earth’s orbital velocity around the Sun is
about 30 kilometers / second. (You are carried along with the Earth
18 miles every second.)
What is the approximate
escape velocity from the Sun for an object starting from the position
of Earth’s orbit?
Answer: The circular
velocity for a circular orbit at the Earth's orbital location is is
about 30 km/sec. The square root of 2 is about 1.414. 1.414 times
30 kilometers/second is 42.4 kilometers/second. This is about 95
miles/second.
There are five man-made
satellites that are leaving our solar system, Pioneer 10 and 11,
Voyager 1 and 2, and New Horizons. They were given the necessary
escape velocity to leave the Sun from rockets and gravitational
assist when flying by planets
Earth-Moon Orbit
The Earth and Moon both
orbit the Sun. Their mutual gravitational forces cause both objects
to wobble back and forth around the ellipse that is the path of the
center of mass of the Earth –Moon system.
You should calculate
the force of gravity between the Sun and the Earth, between the Moon
and the Earth, and between the Sun and the Moon. You will discover
that the Sun exerts a larger force on the Moon than the Earth exerts
on the Moon.
“Weightlessness”
The force of gravity
gives an object weight. The force of gravity does not go to zero
right above the surface of the Earth or just above the surface of any
body. Astronauts in orbit are not truly weightless. They are
apparently weightless. Both the astronaut and the spacecraft are
accelerating towards the Earth. Both the astronaut and the
spacecraft are in orbit around the Earth. No contact force is needed
between the astronaut and the spacecraft. If an astronaut steps on a
bathroom scale in orbit the scale reads zero. Imagine placing a pen
on top of a whiteboard eraser. If they are both allowed to fall
towards the floor the eraser will not apply any force to the pen.
The pen is apparently weightless since there is no contact force
between the objects.
You should view some
YouTube videos that discuss Mass and Weight in Orbit.
The Cavendish
Experiment: Determining "G"
In 1797 and 1798
Cavendish performed an experiment to determine the value of "G,"
the universal constant of gravitation. You should read discussion of
this experiment in your textbook and watch a video.
Copyright©
2017 by Greg Clements Permission is granted to reproduce this
document as long as 1) this copyright notice is included, 2) no
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