F. Gravitation and Oscillation HW I
Things you need to know for these problems: 1 mile = 1609 meters
Mearth = 5.98 x
1024 kg Rearth
= 6.378 x 106 m (equator)
Msun = 2.0 x 1030
kg Rsun = 7
x 108 m (equator)
Mmoon = 7.4 x 1022
kg Rmoon = 1.75 x 106
m (equator)
Rearth-sun = 1.5
x 1011 m (average distance) Known as 1 A.U. (astronomical unit)
Rearth-moon = 3.8
x 108 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 Mearth. Assume you know G, Rearth and ag(9.8m/s2)
74: b)
Earth is wider at the equator than at the poles because it bulges due to
rotation.
77: Rsun-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 Rearth-moon away from the sun at this point)
78: If r is tripled, the result of 1/r2
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 Fup, Facross and Fdiagonal.
Note that Fup and Facross
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 (RFup/Facross) which must act
45o 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 Fdiagonal is 45o
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 Fdiagonal
are each 45o above the horizontal, their forces can be added such that SF45odirection = Resultant force +
Fdiagonal
82: Fearth = G Mearth mastronaut/
(rearth-astronaut)2 = Fmoon = G Mmoon mastronaut/(rmoon-astronaut)2
Notice
that when you set Fearth equal to Fmoon, the G’s and the
mastronauts 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 108m – r away
from the moon. Since the Rearth-moon
is 3.8 x 108m, we can assume that he is r away from one and
3.8 x 108m - r away from the other.
83: To find the ratio, divide Fearth-sun/Fearth-moon.
Fearth-sun
= G Mearth Msun/ (rsun-earth)2
Fearth-moon
= G Mearth Mmoon/ (rmoon-earth)2
WOW,
look at that! When you divide
these, G’s and Mearth cancel! Oh, Happy Days!
85: Just like example 7.14 on p. 233 where
Utotal = -U12-U13-U23.
7.6
Kepler’s Law and Earth Satellites
86, 87, 88, 89, 92, 95, 96,
104
92: T2earth = Kearth
r3 where K = 4p2/G Mearth. Tearth = # 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 Rearth
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 wearth = wsun(apparent motion).
And, since we know that wearth = Dq/Dt and that Dq for the earth in one day is 2p (Because 2p = 360o of
rotation). You can then solve for wearth. (Dt = 24 hours x 60 min x 60
sec = 86,400 seconds)
Once you have wearth, you can then use that same
equation to figure out Dt since the problem tells
you that the angular width of the sun is 0.50o. First, convert 0.50o into radians
by multiplying by p/180.
Then, Dt = Dq/wearth.
104: Fc = mac = mv2/r so that the m’s cancel. Therefore, ac = v2/r where v = 2.7 m/s
Ftangent = mat
= 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 90o
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 0o
away from vertical, the force of g in that direction also approaches 0. (of course m’s cancel and you can solve for
at).
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 1020 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 107 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) at
= 2.5 m/s2 and ac = 9.7 m/s2. B)
at its lowest point, all of the acceleration is dependent upon ac
and none on at.