Math 1321
Week 14 Lab Worksheet
Due Thursday 04/18
Many forces in our known universe can be modeled with conservative force fields. In
particular the gravitational field and the electric field due to a static point charge can be
modeled in such a way. This is nice because it allows us to make computations such as how
much interaction an object and the force field will undergo as the object passes through the
field. The following problems guide you through the process of finding the amount of work
a force field will do to an object moving through it.
1. Suppose that F is an inverse square force field, that is
F(r) =
cr
|r|3
for some constant c and r = xi + yj + zk. Determine a formula for the work done by
moving an object from a point P1 to the point P2 in terms of the distances d1 and d2 of
these points from the origin. Hint: If F is conservative what does that mean?
Solution: Here F(r) = cr/|r|3 and r = xi + yj + zk. Then f (r) = −c/|r| is a
potential function for F, that is, ∇f = F. Hence F is a conservative field and its
line integral is independent of path. Let P1 = (x1 , y1 , z1 ) and P2 = (x2 , y2 , z2 ). Then
the amount of work in moving an object from P1 to P2 is given by
Z
c
1
1
c
−
=c
−
W =
F·dr = f (P2 )−f (P1 ) = − 2
(x2 + y22 + z22 )1/2 (x21 + y12 + z12 )1/2
d1 d2
C
2. The gravitational force field is modeled by
F(r) =
−mM Gr
|r|3
Using your answer to part (1), determine the work done by the gravitational field when
the earth moves from aphelion to perihelion. These correspond to the earth’s maximum
distance of 1.52 × 108 km from the sun, and it’s minimum distance of 1.47 × 108 km
from the sun respectively. Use the values of m = 5.97 × 1024 kg, M = 1.99 × 1030 kg,
and G = 6.67 × 10−11 N · m2 /kg2 .
Solution: In this case, c = −(mM G) =⇒
1
1
−
W = −mM G
1.52 × 108 1.47 × 108
= −(5.97 × 1024 )(1.99 × 1030 )(6.67 × 10−11 )(−2.2377 × 10−10 ) ≈ 1.77 × 1035 J
3. Another application of inverse square force fields involves the electric field induced by
a static point charge. Using Coulomb’s Law, the electric field due to a static charge
located at the origin is given by
qQr
F(r) =
|r|3
Suppose that an electron with a charge of −1.6 × 10− 19 C is located at the origin.
A positive unit charge is positioned a distance 10−12 from the electron and moves to
a position half that distance from the electron. Again using the formula you derived
in part (1), find the work done by the electric field in moving the particle toward the
electron. (Use the value of = 8.985 × 109 .)
Solution: In this case, c = qQ =⇒
1
1
−
W = qQ
10−12 5 × 10−13
= (8.985 × 109 )(1)(−1.6 × 10−19 )(−1012 ) ≈ 1400 J
p
4. Assume r is the distance from the center of our “perfectly circular” earth, r = x2 + y 2 + z 2
where x and y describe distances from the axis of rotation (assume perfect symmetry),
and z describes the distance above the plane which contains the equator (see image).
Suppose air resistance of a moving object can be modeled by the field AR(x, y, z) =
hr, r, zi. This assumes that the moving object undergoes an additional force in the direction of AR of magnitude |AR| at the point (x, y, z) due to air resistance and drag
forces. Can you apply the same formula you derived in part (1) to compute the work
done by air resistance and drag forces in moving an object from a point P1 to a point
P2 with given distances d1 and d2 from the center of the earth? Why or why not? Hint:
Your answer should line up with your intuition about frictional forces and choice of path,
and if not see Theorem 3 and Example 2 in Section 13.5.
Solution: Intuition should tell us that if a moving object takes a longer path through
our atmosphere, the frictional forces such as air resistance and drag will do more work
to the object. Thus we should expect to not be able to apply the same formula which
is independent of path. This is due to the idea that the frictional force field should
be non-conservative. In fact, if we explicitly compute the curl of AR, we find
z
z
x−y
∇ × AR = − i + j +
k 6= 0 for all x, y, z
r
r
r
But we know from Theorem 3, the curl of a potential field is zero (that is ∇×(∇f ) =
0), thus we cannot represent F as a potential function. That is there is no f such that
F = ∇f . Therefore, by definition F is not conservative which means the integral of
it will be path dependent. Thus we cannot apply the previous formula because F is
not a conservative field.