Problem#1
(a) What will an object weigh on the Moon’s surface if it weighs 100 N on Earth’s surface? (b) How many Earth radius must this same object be from the center of Earth if it is to weigh the same as it does on the Moon?Answer:
(a) the gravitational acceleration at the surface of the Moon is
gEarth/gmoon = WEarth/Wmoon
with, gEarth = 9.81 m/s2 and gmoon = 1.67 m/s2
then,
[9.81 m/s2/1.67 m/s2] = 100 N/Wmoon
Wmoon = 17.02 N
(b) If the force f of 17.02 N above is the gravitational force between the object and the earth, we can express that using Newton's universal law of gravity.
F = GmM/r2
With, r = distance from the center of the earth to the object, M = mass of the earth
So, the distance r is
17.02 N = (6.67 × 10−11 Nm2/kg2)(10.19 kg)( 5.98 × 1024 kg)/r2
r = 1.545 x 107 m
The radius of the earth is 6378 km, so the number of the earth's radius is
N = 1.545 × 107 m/6378000 m = 2.42
Problem#2
A large mountain can slightly affect the direction of “down” as determined by a plumb line. Assume that we can model a mountain as a sphere of radius R = 2.00 km and density (mass per unit volume) 2.6 x 103 kg/m3. Assume also that we hang a 0.50 m plumb line at a distance of 3R from the sphere’s center and such that the sphere pulls horizontally on the lower end. How far would the lower end move toward the sphere?
Answer:
Known:
sphere of radius R = 2.00 km = 2.00 x 103 m
density (mass per unit volume) ρ = 2.6 x 103 kg/m3
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| Fig.1 |
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| Fig.2 |
Fg = T sin θ and T cos θ = mg
Then,
Fg/mg = tan θ
From the Figure 2, tan θ = d/L, because sin θ <<, so that
d = FgL/mg = GML/gr2
where d is distance moved from lower end.
M = ρV = ρ(4πR3/3)
M = (2.6 x 103 kg/m3)[4π x (2.00 x 103 m)3/3] = 8.71 x 1013 kg
So,
d = (6.67 × 10−11 Nm2/kg2)(8.71 x 1013 kg)(0.50 m)/[9.81 m/s2 x (3 x 2.00 x 103 m)2]
d = 7.4 x 10-6 m = 7.4 μm
Problem#3
At what altitude above Earth’s surface would the gravitational acceleration be 4.9 m/s2?
Answer:
Let a body of mass m be placed on the surface of the Earth:
g = GM/R2
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| Fig.3 |
Let the acceleration due to gravity at that position be g’. Then,
g' = GM/(R + h)2
For comparison, the ratio between g’ and g is taken
g'/g = [R/(R + h)]2
4.9 m/s2/9.81 m/s2 = [6.4×106 m/(6.4 × 106 m + h)]2
0.71 (6.4 × 106 m + h) = 6.4 × 106 m
h = 2.655 x 106 m = 2655 km
Problem#4
In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constructed. Ignoring Earth’s rotation, find the change in your weight if you were to ride an elevator from the street level, where you weigh 600 N, to the top of the building.
Answer:
Let a body of mass m be placed on the surface of the Earth:
g = GME/R2
Difference in acceleration at head and feet is
dag/dr = ─2GME/r3
which implies that the change in weight is
Wto ─ Wbottom ≈ m(dag)
However. Since Wbottom = GmME/R2, we have
m(dag) = ─2GmME (dr/R3)
m(dag) = ─2Wbottom(dr/R) = ─2 x 600 N x 1.61 x 103 m/6.37 x 106 m
m(dag) = ─0.303 N
the minus sign indicating that it is a decrease in W
Problem#5
Certain neutron stars (extremely dense stars) are believed to be rotating at about 1 rev/s. If such a star has a radius of 20 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation?
Answer:
Known:
the angular velocity, ω = 1 rev/s = 2π rar/s
radius, r = 20 km
The gravitational pull is the centripetal force. If gravity is not strong enough the objects will get thrown off the surface, so they must be at least equal to each other.
GmM/r2 = mv2/r or
GM/r = v2 Where v = ωr
We want to solve for M:
M = ω2r3/G
M = (2π rad/s)2(20 x 103 m)3/(6.67 × 10−11 Nm2/kg2) = 4.74 x 1024 kg
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