SOLUTIONS TO EXERCISE 5


(a) The potential energy the falling chunk of satellite has before it falls is

                      - (6.67 * 10^-11) * (6.0 * 10^24) * (2)
 E(start) = -GMm/r  = --------------------------------------- = -0.2085 * 10^8 J.
                         (32000 * 10^3 m) + (6.38 * 10^6 m)

The potential energy of the chunk after it reaches Earth's surface is

                    - (6.67 * 10^-11) * (6.0 * 10^24) * (2)
 E(end) = -GMm/r  = ---------------------------------------  = -1.2545 * 10^8 J.
                         (0 m) + (6.38 * 10^6 m)

So the amount of gravitational potential energy released is

  dE = E(start) - E(end) = (-0.2085 + 1.2545) * 10^8 = 1.05 * 10^8 J.

Now, if this amount of energy were released all at once upon impact, it would
be equivalent to an explosion of about 200 pounds of dynamite.  However, since
Earth has an atmosphere, most of the gravitational potential energy would be
released as heat produced by the friction of re-entry; in fact, if the chunk
of satellite got too hot, it would burn up before making it to the ground.  In
that case, the amount of potential energy released would be smaller, since the
distance it fell would be shorter.

(b) The potential energy the brick has before it falls is

                      - (6.67 * 10^-11) * (6.0 * 10^24) * (2)
 E(start) = -GMm/r  = --------------------------------------- = -1.2482 * 10^8 J.
                         (32 * 10^3 m) + (6.38 * 10^6 m)

The potential energy of the brick after it reaches Earth's surface is

                    - (6.67 * 10^-11) * (6.0 * 10^24) * (2)
 E(end) = -GMm/r  = ---------------------------------------  = -1.2545 * 10^8 J.
                         (0 m) + (6.38 * 10^6 m)

So the amount of gravitational potential energy released is

  dE = E(start) - E(end) = (-1.2482 + 1.2545) * 10^8 = 6.3 * 10^5 J.

(c) Since g = 9.8 m/s^2 at Earth's surface, then the simplified formula gives

  dE = mgh = (2 kg) * (9.8 m/s^2) * (32 * 10^6 m) = 6.3 * 10^8 J

for the chunk of satellite

  dE = mgh = (2 kg) * (9.8 m/s^2) * (32 * 10^3 m) = 6.3 * 10^5 J

for the chunk of robotic plane.  Part (b) is identical, to the accuracy
of our calculations, but Part (a) misses by a factor of about six.  The
easiest way to explain the difference is that, at an altitude of 32 km,
the gravitational acceleration, g, is just about the same as it is at Earth's
surface; but at an altitude of 32,000 km, the gravitational acceleration is
much different than at Earth's surface - in fact, about 1/36 as large.  So the
approximation breaks down, and the simplified equation is invalid for Part (a).