ANSWER TO CHAPTER 6 HOMEWORK

1.  The acceleration due to gravity at a distance r from the center of the Earth is

                  g = F/M = GmEarth/r2.

          If we form the ratio of the two accelerations for the different distances, we have

                 g/gsurface = [(rEarth/(rEarth + h)]2;

          (a)  g = (9.80m/s2)[(6400km)/(6400km + 3.2km)]2 =  9.80m/s2.

          (b)  g = (9.80m/s2)[(6400km)/(6400km + 3200km)]2 =  4.36m/s2.

2.  For the magnitudes of the forces on the mass m from the other masses we have

                   F2 = Gm(2m)/xo2 = 2Gm2/xo2;

                   F3 = Gm(3m)/(xo2 + yo2) = 3Gm2/(xo2 + yo2);

                   F4 = Gm(4m)/yo2 = 4Gm2/yo2.

          The force F3 is at an angle q above the x-axis, with 

                   sinq = yo/(xo2 + yo2)1/2, and cosq = xo/(xo2 + yo2)1/2.

           Thus the resultant force is

                    F = (F2 + F3cosq)i + ( F3sinq + F4)j

                        = Gm2{(2/xo2) + [3/(xo2 + yo2)][xo/(xo2 + yo2)1/2]}i  +

                                Gm2{[(3/xo2 + yo2)][yo/(xo2 + yo2)1/2] + (4/yo2)}j

                        =  Gm2{(2/xo2) + [3xo/(xo2 + yo2)3/2]}i  + Gm2{[3yo/(xo2 + yo2)3/2] + (4/yo2)}j.

 

  
3. We take the positive direction upward. The spring scale reads the normal force expressed as an effective mass: FN/g.

        We write SF = ma from the force diagram:

                       FN - mg = ma,  or  meffective  = FN/g = m(1 + a/g).

         (a)  For a constant speed, there is no acceleration, so we have

                       meffective = m(1+ a/g) = m =   56kg.

          (b)  For a constant speed, there is no acceleration, so we have

                       meffective = m(! + a/g) = m =   56kg.

          (c)  For the upward (positive) acceleration, we have

                       meffective = m(1 + a/g) = m(1 + 0.33g/g) 1.33(56kg) =   75kg      

          (c)  For the downward (negative) acceleration, we have

                        meffective = m(1 + a/g) = m(1 - 0.33g/g) = 0.67(56kg) =   38kg.

           (d)  In free fall the acceleration is -g, so we have

                        meffective = m(1 + a/g) = m(1 - g/g) =   0.

  

4.  we use Kepler's third law, T2 = 4p2r3/GmE, for the motion of the Sun:

                         T2 = 4p2r3/GmGalaxy;

                              = 4p2[(3 x 104ly)(3 x 108m/s)(3.16 x 107s/yr)]3/(6.67 x 10-11N.m2/kg2)(4 x 1041kg),

          which gives

                         T = 6 x 1015s =  2 x 108yr.

5.  We call D the separation of the Earth and Moon and take the positive direction toward the Earth. Because the gravitational field are in opposite directions, we have

                         g = gE - gM = [GmE/(D/2)2] - [GmM/(D/2)2] = (4G/D2)(mE - mM)

                            = 4[(6.67 x 10-11N.m2/kg2)/(3.84 x 108m)2](5.98 x 1024kg  - 7.4 x 1022kg)

                            =  1.07 x 10-2N/kg toward Earth.    

*6. (a) The acceleration due to gravity at a distance r = rE + Dr from the center is

                        g' = Gm/r2 = Gm/(rE + Dr)2.

If we use the binomial expansion and keep only the first two terms, we get

                        g' = (Gm/rE2)[1 + (Dr/rE)]-2 = g[1 - 2(Dr/rE) + 3(Dr/rE)3 + . . .] » g - 2g(Dr/rE).

Thus we have

                       g' - g = Dg » - 2g(Dr/rE).

Note that this could also be obtained by treating the changes as differentials:

                      dg = -2g(Gm/r3)dr = -2gdr/r.

(b) The negative sign means that g decreases with an increase in height.

(c) At a height of 100km we get

                      Dg = -2g(Dr/rE) = 2(9.80m/s2)(100km)/(6.38 x 103km) = - 0.307m/s2.

Thus we have

                      g' = g + Dg = 9.800m/s2 - 0.307m/s2 = 9.493m/s2.

If we use Eq. 6-1, we get

                     g' = GmE/(rE + Dr)2

                         = (6.67 x 10-11N·m2/kg2)(5.98 x 1024kg)/(6.38 x 106m + 100 x 103m)2 = 9.499m/s2.

*7. If we take the upward acceleration of the elevator as positive, the effective value of gravity in the elevator is

                    geff = g + aelev.

The acceleration of the mass relative to the plane must be along the surface of the plane. From the diagram we see that

                   mgeffsinq = marel, so arel = (g + aelev)sinq .

 (a) For an upward acceleration of the elevator, we get

                   arel = (g + aelev)sinq  = (g + 0.50g)sin30°

                         = 0.75g (down the plane).

(b) For a down acceleration of the elevator, we get

                   arel = (g + aelev)sinq  = (g - 0.50g)sin30°

                         = 0.25g (down the plane).

(c) For free fall arel = -g, so we have aelev = 0, and we get

                   arel = (g + aelev)sinq  = 0.

(d) For constant speed aelev = 0, so we have geff = g, and we get

                   arel = (g + aelev)sinq  = gsin30° = 0.50g (down the plane).

 

 

HOME  SEND QUESTIONS