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Type: Double
Objective:
Gravitation II
Gravitation III
Homework: Do PROBLEMS #(22, 27) Ch. 4
Do PROBLEM #(29) Ch. 5
Satellites in orbit
http://www.youtube.com/watch?v=eYUh-hMaNiM
Mt. Everest
http://en.wikipedia.org/wiki/Image:Everest_kalapatthar_crop.jpg
Panoramic view from top of Mt. Everest
http://www.panoramas.dk/fullscreen2/full22.html
The Hubble Space Telescope
http://nicmosis.as.arizona.edu:8000/GRAPHIC_HST.GIF
http://www.youtube.com/watch?v=upXTZE57Z5U
Pluto
Pluto the dog
Pluto the dog 2
Pluto and Charon
From the Earth to the Moon
JV - Rocket
Date:________
AP Physics “B”
Mr. Mirro
Date: ________
Gravity at a Distance
Gravity and altitude
We have defined the weight of a body as the attractive gravitational force exerted on it by the earth.
Now, we can broaden our definition to be more general as follows:
The weight of a body is the “total” gravitational force exerted on the body by ALL other
bodies in the Universe.
When the body is near a planet or other celestial body, the weight is a measure of the force of attraction
with respect to that planet or celestial body. For instance:
‰
At the surface of the earth, we consider our weight to be the gravitational attraction of the earth.
‰
At the surface of the moon, we consider a body’s weight to be the gravitational attraction of the
moon, and so on…
We can neglect all other gravitational forces and consider the weight as just the planet or celestial
body’s gravitational attraction.
But, we also know that the weight of a body (W) is the force that causes the acceleration g of free-fall.
So by relating Newton’s Second Law and Newton’s Law of Gravitation, we obtain the expression:
g = GM
R2
Where the mass M and distance R depend on the “body”
The weight of an object with respect to the earth’s surface “decreases” inversely with the square of its
distance from the surface of the earth.
For example, consider the gravitational force of attraction of a person who weighs 600 N at the
surface of the earth as they are moved through various distances from earth.
600 N
r = RE = 6.38 x 106 m
500 N
400 N
300 N
200 N
100 N
5
10
15
20
25
30
R[r – RE] ⇒ ( x 106 m)
Orbital velocity
Today there are many satellites in orbit about the earth. The ones in circular orbits are examples of
Uniform Circular Motion.
Like a model airplane on a guideline, each satellite is kept on its circular path by a centripetal force.
The gravitational pull of the earth provides the centripetal force and acts like an invisible
guideline for the satellite.
There is only one “speed” that a satellite can have if the satellite is to remain in an orbit with a fixed “radius.”
If we relate the centripetal force to the gravitational force of attraction as:
Fc ⇒ m v2 = G m ME
r
r2
We can develop an expression for the orbital velocity v of the satellite at this fixed position r.
Fc ⇒ (m) v2 = G (m) ME
r
r(2) → 1
canceling mass (m) and
reducing radius (r)
v 2 = G ME
r
“un”squaring velocity (v)
Yeilds,
V =
√
G ME
r
If the satellite is to remain in orbit of radius r, the speed v must have precisely this value.
Note: Since the radius r of the orbit is in the denominator, the closer the satellite is to the
earth, the smaller is the value for r and the greater the orbital velocity must be.
AP Physics “B”
Mr. Mirro
Date: ________
Gravity at a Distance
Ex 1: Estimate the effective value of gravity g on the top of Mt. Everest, 8848 m (29,028 ft) above the
Earth’s surface. That is, what is the acceleration due to gravity of objects allowed to fall freely at
this altitude ? [Giancoli5.14]
Ex 2: The Hubble Space Telescope orbits approximately 375 miles above Earth, working around the clock
to unlock the secrets of the Universe. It uses excellent pointing precision, powerful optics, and stateof-the-art instruments to provide stunning views of the Universe that cannot be made using ground
-based telescopes or other satellites. Hubble was originally designed in the 1970s and launched in
1990. [Cutnell4.6+5.9mod]
Hubble is the first scientific mission of any kind that is specifically designed for routine servicing by space walking
astronauts. It has a visionary, modular design which allows the astronauts to take it apart, replace worn out
equipment and upgrade instruments. These periodic service calls make sure that Hubble produces first-class
science using cutting-edge technology. Each time a science instrument in Hubble is replaced, it increases Hubble
scientific power by a factor of 10 or greater!
a. If the mass of the Hubble Space Telescope is 11,600 kg determine the weight of the telescope
when it was resting on the surface of the earth.
b. What is the value of the acceleration due to gravity at a point 598 km above the earth’s surface
where the telescope now orbits ?
c. Determine the approximate force of gravitational attraction the Hubble experiences while in its
orbit.
d. Compute the orbital velocity in (m) of the Hubble Space Telescope while it is in its orbit and then
convert it to (MPH).
Ex 3: In Roman mythology, Pluto is the god of the underworld. The planet received this name perhaps
because it's so far from the Sun (5900 x 106 Km) that it is in perpetual darkness and perhaps because
"PL" are the initials of astronomer Percival Lowell. [HallidayTbl15.3]
Pluto was discovered in 1930 by a fortunate accident. Calculations which later turned out to be in error had
predicted a planet beyond Neptune, based on the motions of Uranus and Neptune. Not knowing of the error,
Clyde W. Tombaugh at Lowell Observatory in Arizona did a very careful sky survey which turned up Pluto
anyway…Planet “X” as it was referred to in its early days.
Pluto is the only planet that has not been visited by a spacecraft. Even the Hubble Space Telescope can
resolve only the largest features on its surface. There is a planned mission called New Horizons that will
launch in 2006 if it gets funded. Use Kepler’s Third Law to determine the period of Pluto in Earth years ?
Recall,
Ksun = 3.36 x 1018 m3/s2
Gravitational Potential Energy - Revisited
When we first introduced the concept of gravitational potential energy, we assumed that the gravitational
force on a body is constant in magnitude and direction. This led to the basic expression for the
potential energy of an object relative to the surface of the earth, which originated by considering the
work done against gravity.
Wgravity = (Fgravity) d
PEgravity = (mg) h
where h is the height of the object above
(or below) some reference level.
The equation is only valid when the object is near the earth’s surface. For objects high above the
earth’s surface, such as a satellite, moon, etc, an alternative expression must be used to analyze and
compute the gravitational potential energy between the bodies.
We know that the earth’s gravitational force on a body of mass m at any point outside the earth is given
more generally by the Universal Law of Gravitation.
Fgravity = G MEm
RE2
However, if an object is fired straight upward from the earth’s surface, there is a radial component of the
gravitational force called Fradial which is negative because it is opposite in direction WRT Fgrav.
v
That is,
Fgravity = - Fradial
h
OR
Fr
Fradial = - G MEm
r2
Fg
Therefore, the work done (WGravity) done by the gravitational force when the body moves directly away from
or toward the center of the earth from a distance r, can be expressed as:
Wgravity = (Fradial) d
PEgravity =
- G MEm
r2
r
FTBO ⇒
Thus, the general expression for the gravitational potential energy (PEg = Ug) for an object of mass m at
a distance r from the earth’s surface can be expressed as:
Ugravity = - G MEm
r
U
Considering the figure, we can see how the gravitational potential energy
depends on the distance r between the body of mass m and the center of
the earth.
0
RE
r
When the body moves away from the earth, r increases, the gravitational
gravitational force does negative work, and U “increases”
- becomes less negative.
When the body “falls” toward earth, r decreases, the gravitational
force does positive work, and U “decreases”
- becomes more negative.
-GMEm
RE
You may be troubled by the equation for Ug because it states that gravitational potential energy is always
negative. But in fact, we have seen negative work - as done by the weight lifter on the bench press.
‰
Positive work was done when the weightlifter lowered (with g) the barbells downward to his chest
WDown = F * [cos 0° = +1] * S
‰
Negative work was done when the weightlifter raised (against g) the barbells upward from his chest
WUp = F * [cos 180° = -1] * S
Note: We have chosen U to be zero when the body of mass m is infinitely far away from the earth (r →∝)
If you fire a projectile upward, usually it will slow, stop momentarily and return to earth. There is, however,
a certain initial speed that will cause it to move upward forever, theoretically coming to rest only at infinity.
This initial speed is known as the “ESCAPE SPEED.”
If we consider the conservation of energy theorem, we can develop such a speed as follows:
Einitial = Efinal
KE initial + Ug initial = KE final + Ug final
½ mv2 +
- GMEm
R
= ½ m (0)2 - G MEm
(∝)
½ (m) v2 = G ME (m)
R
Therefore,
2 G ME
VESCAPE =
√
R
⇒
AP Physics “B”
Mr. Mirro
Date: ________
Gravitational Potential Energy - Revisited
Ex 1: In Jules Verne’s 1865 story titled “From the Earth to the Moon”, three men were shot to the moon in
a shell fired from a giant cannon sunken into the earth in Florida. [Sears12.5]
a. Draw a diagram and derive the equation for the muzzle velocity needed to shoot the shell
straight up to a height above the earth equal to the earth’s radius in terms of only mass, radius
and appropriate constant(s).
b. Compute the muzzle velocity needed to shoot the shell straight up to a height above the earth
equal to the earth’s radius.
c. Find the escape-speed, that is, the muzzle speed that would allow the shell to escape from the
earth completely. [Neglect air resistance, rotation of the earth and gravitational pull of moon]
Ex 2: Below you will find several astronomical facts pertaining to the planets in our solar system. Note that
escape velocities (ve) are given in meters per second. [SerwayTbl7.2]
a. Convert the escape velocities for the missing celestial bodies below from (m / s) to (mi / hr).
Mass
Radius
Escape Velocity
Planet
(kg)
(m)
(m/s)
Mercury
0.33 x 1024
2.4 x 106
4 300
Venus
4.87 x 1024
6.1 x 106
10 300
23 031
Earth
5.98 x 1024
6.4 x 106
11 200
25 043
Moon
0.0736 x 1024
1.7 x 106
2 300
Mars
0.65 x 1024
3.4 x 106
5 000
Jupiter
1900 x 1024
72 x 106
60 000
Saturn
570 x 1024
60 x 106
35 600
79 602
Uranus
87 x 1024
26 x 106
21 200
47 403
Neptune
100 x 1024
2.3 x 106
23 600
52 770
Pluto
0.011 x 1024
1.2 x 106
1 100
Escape Velocity
x 2.236 = (MPH)
b. Determine the escape speed for the SUN from the data below, then compare and contrast.
Sun
1.989 x 1030
6.96 x 108
c. Combine the mass of all of the planets in our solar system and compare your result to our the mass of
our Sun. For the various satellites of the planets (including our Moon) add in another 10 x 1024 kg.
The rule of thumb is general taken in terms of volume, Jupiter is about 1000 times bigger than Earth, and the
Sun is about 300 times bigger than Jupiter. The Sun is pretty big!
So all the planets add up to only just over 1/1000 th the mass of the Sun.
Ex 2:
Pluto The Dog
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Pluto The Dog 2
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Pluto and Charon
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Jules Verne Rocket
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