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

71:   In the immoral words of Tom Petty:  “I’m free…..free fallin’…”

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.