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Transcript
because the speed υ is proportional to the square root of the mass of the planet, the
speed increases as the mass of the planet increases.
Example 13.6 The Change in Potential Energy
A particle of mass m is displaced through a small vertical distance Δy near the Earth’s
surface. Show that in this situation the general expression for the change in
gravitational potential energy given by Equation 13.13 reduces to the familiar
relationship ΔU = mg Δy.
SOLUTION
Conceptualize Compare the two different situations for which we have developed
expressions for gravitational potential energy: (1) a planet and an object that are far
apart for which the energy expression is Equation 13.14 and (2) a small object at the
surface of a planet for which the energy expression is Equation 7.19. We wish to
show that these two expressions are equivalent.
Categorize This example is a substitution problem.
Combine the fractions in Equation
13.13:
 1 1
 r r
(1) U  GM E m     GM E m  f i
r r 
 rr
i 
 f
 i f



Evaluate rf − ri and rirf if both the rf − ri = Δy rirf ≈ RE2
initial and final positions of the
particle are close to the Earth’s
surface:
Substitute these expressions into
Equation (1):
U 
GM E m
y  mg y
RE 2
where g = GME/RE2 (Eq. 13.5).
WHAT IF? Suppose you are performing upper-atmosphere studies and are asked by
your supervisor to find the height in the Earth’s atmosphere at which the “surface