F. Gravitation and Oscillation HW I

** **

Things you need to know for these problems: 1 mile = 1609 meters

**M _{earth} = 5.98 x
10^{24} kg R_{earth}
= 6.378 x 10^{6} m (equator)
**

**M _{sun} = 2.0 x 10^{30
}kg R_{sun} = 7
x 10^{8} m (equator)**

**M _{moon} = 7.4 x 10^{22}
kg R_{moon} = 1.75 x 10^{6}
m (equator)**

**R _{earth-sun }= 1.5
x 10^{11} m (average distance) Known as 1 A.U. (astronomical unit)**

**R _{earth-moon} = 3.8
x 10^{8} m (average distance)**

** **

**7.5 Newton’s Law of
Gravitation**

**70, 71, 72, 73, 74, 76, 77,
78, 79, 82, 83, 85**

72: Try it!

73: This question is looking for a
re-arrangement of Newton’s Universal law of Gravitation set equal to M_{earth}. Assume you know G, R_{earth} and a_{g}(9.8m/s^{2})

74: b)
Earth is wider at the equator than at the poles because it bulges due to
rotation.

77: R_{sun-moon }varies as the moon
goes around the earth. When there is a
solar eclipse, the moon is between the sun and the Earth. When there is a lunar eclipse, the Earth is
between the sun and the moon (so the moon is 1 A.U. **PLUS **1 R_{earth-moon }away from the sun at this point)

78: If r is tripled, the result of 1/r^{2}
will be 1/9)

79: FIRST, DRAW IT OUT!!!!!!! Next, choose one mass as your reference
point. Note that this mass is being
pulled in three different directions at once:
up, across and diagonal. We will
call these F_{up}, F_{across} and F_{diagonal}.

Note that F_{up} and F_{across}
are ^ to each other and are equal in magnitude
since the distances between the masses are the same and since the masses are
all equal in mass.

Note that Fup and Facross form the y and
x components of some resultant force (R_{Fup/Facross}) which must act
45^{o} above the horizontal since the distance of both legs (Fup and
Facross) is 1 m. This makes a 45-45-90
right triangle.

You can find the Resultant force by _{} (the 2’s are
“squares”)

Note that F_{diagonal} is 45^{o}
above the horizontal and that the distance between the masses diagonally across
from each other is _{} since the length of
the legs are 1 m each.

Since the Resultant force and the F_{diagonal
}are each 45o above the horizontal, their forces can be added such that SF_{45odirection} = Resultant force +
F_{diagonal}

82: F_{earth} = G M_{earth} m_{astronaut}/
(r_{earth-astronaut})^{2} = F_{moon} = G M_{moon }m_{astronaut}/(r_{moon-astronaut})^{2}

Notice
that when you set F_{earth} equal to F_{moon}, the G’s and the
m_{astronauts} cancel.

To
determine the different values of r, let us assume that the astronaut is some
distance **r**
away from the earth and that he is 3.8 x 10^{8}m – **r** away
from the moon. Since the R_{earth-moon
}is 3.8 x 10^{8}m, we can assume that he is r away from one and
3.8 x 10^{8}m - r away from the other.

83: To find the ratio, divide F_{earth-sun}/F_{earth-moon}.

F_{earth-sun
}= G M_{earth} M_{sun}/ (r_{sun-earth})^{2}

F_{earth-moon
}= G M_{earth} M_{moon}/ (r_{moon-earth})^{2}

WOW,
look at that! When you divide
these, G’s and M_{earth }cancel! Oh, Happy Days!

85: Just like example 7.14 on p. 233 where
Utotal = -U_{12}-U_{13}-U_{23}.

**7.6
Kepler’s Law and Earth Satellites**

**86, 87, 88, 89, 92, 95, 96,
104**

92: T^{2}_{earth} = K_{earth}
r^{3} where K = 4p^{2}/G M_{earth}. T_{earth} = # of seconds/1 rotation
of earth on its axis. Therefore, you
will have to convert 1 day into seconds.
DON’T FORGET TO THEN SQUARE THE VALUE FOR T THAT YOU GET.

Don’t forget that the value for r that you will find is a
distance FROM THE CENTER OF THE EARTH to the satellite in space. Therefore, you must subtract R_{earth}
from your answer. ALSO, PLEASE CONVERT THESE METERS INTO MILES SO WE KNOW HOW HIGH
IT MUST BE IN MILES.

95: Realize that the angular velocity of the sun
(w) must be exactly the same as that of the
earth. WHY? Because the sun has no angular velocity in the sky since it is
not the one turning. If it weren’t for
the rotation of the earth on its axis, the sun would have no apparent value for
w.
Therefore, we can say w_{earth} = w_{sun(apparent motion).}

And, since we know that w_{earth} = Dq/Dt and that Dq for the earth in one day is 2p (Because 2p = 360^{o} of
rotation). You can then solve for w_{earth}. (Dt = 24 hours x 60 min x 60
sec = 86,400 seconds)

Once you have w_{earth}, you can then use that same
equation to figure out Dt since the problem tells
you that the angular width of the sun is 0.50^{o}. First, convert 0.50^{o} into radians
by multiplying by p/180.

Then, Dt = Dq/w_{earth}.

104: F_{c} = ma_{c} = mv^{2}/r so that the m’s cancel. Therefore, a_{c} = v^{2}/r where v = 2.7 m/s

F_{tangent} = ma_{t}
= m (gsinq) since g is not acting
directly down on the perspective we are interested in. At the tangential, g is acting with sinq. THINK ABOUT IT.
What if you held the bob completely perpendicularly to the ceiling. Then, it would be at an angle 90^{o}
to the vertical. Sin 90 = 1 so that all
of g would be acting on it in that direction – no duh! So, as q approaches 0^{o}
away from vertical, the force of g in that direction also approaches 0. (of course m’s cancel and you can solve for
a_{t}).

**Even Answers:**

70. d 72. discuss in class.

74. a.
no. Spring scale works with gravity. It is only calibrated to give you kg which
are a mass, not a weight. b) less because earth is wider at equator than
at poles due to rotation.

76. 2.0 x 10^{20} N 78. 0.10 N

82. 3.4 x 108 m from Earth. Not completely without gravity since gravity decreases with distance but never goes away. How about the gravity on the astronaut from the sun?

86: c

88: a)
none b) no because they are still falling with the
elevator at “g” for the time period that they are falling. By pushing off the floor the moment before
impact, they are reducing their rate by only a small amount and therefore still
hit the ground at a velocity equal to g times the amount of time they were in
free-fall.

92. 3.6 x 10^{7} m. This works out to be 22,374 miles (realize
that the circumference of the earth at the equator is about 24,000 miles. So, geosynchronous satellites are about as
high off the earth as the earth is around.

96. discuss in class.

104. a) a_{t}
= 2.5 m/s^{2} and a_{c} = 9.7 m/s^{2}. B)
at its lowest point, all of the acceleration is dependent upon a_{c}
and none on a_{t}.