SOLUTIONS TO EXERCISE 29

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(a) Orbital velocity is given by v = sqrt(GM/r). In this case, M is the mass of
Earth, 6 * 10^24 kg; and r = the distance between the Harley-Davidson and the
center of our planet.  Thus, to go into orbit, the Harley must go horizontally at

6.67 * 10^(-11) m^3 kg^(-1) s^(-2) * 7.36 * 10^22 kg
v = sqrt( ------------------------------------------------------- ) = 7900 m/s
6,378,000 m + 1200 m

This number is equal to about 17,000 miles per hour.

(b) Escape velocity is the square root of two times the orbital velocity; so
for this situation, the Harley must go horizontally at

v = sqrt(2) * 7900 m/s = 11,200 m/s.

This number is equal to about 25,000 miles per hour.

(c) No matter how far the Harley falls, one inch off the ground the force on
the Harley is still F = ma = 450 kg * 10 m/s^2 = 4500 Newtons.

(d) The force in both cases are the same.  But we intuitively "know" that a Harley
dropped from an altitude of 80 m will suffer less than damage than one dropped
from a much higher altitude of 1200 m.  That's because the higher Harley has a
longer falling time, so it gets accelerated for a longer period of time.  Thus,
it's moving faster when it hits the ground, so its momentum (P = mv) is greater.
The damage suffered does not depend on the force experienced by an object at any
given time.  Rather, it depends on the amount of momentum change over any given
period of time.

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