PHY 2053, Section 3794, Fall 2009, Quiz 9
G = 6.67428 × 10−11 m3 /kg/s2 . MEarth = 5.9742 × 1024 kg.
1. An artificial satellite circles Earth in a circular orbit at a location where the acceleration
due to gravity is 9.30 m/s2 .
a) What is the orbital period of the satellite?
√
First, we compute r from a = GM/r2 . We find r = GM/a. It is also possible to compute v
√
from v 2 /r = a, which gives v = ar = (GM a)1/4 . Now we can compute T from v = 2πr/T ,
which gives:
2πr
T =
= 2π ×
v
√
GM
1
= 2π ×
×
a
(GM a)1/4
(
)
GM 1/4
= 87.9 min = 1.46 hr.
a3
2. A 4.25 m, 235 N uniform ladder leans against a smooth wall. The coefficient of static
friction between the ladder and the ground is 0.425, and the ladder makes an angle of 53.13◦
with the ground. A 645 N person begins to climb the ladder.
a) How far can the person climb up the ladder before it begins to slip?
We have Fx :
FR = NW , where FR is the force of friction and NW is the normal force
exerted by the wall, and Fy :
NG = WP + WL , where NG is the normal force exerted by
the ground, WP is the weight of the person and WL is the weight of the ladder. We can put
these together: FR = µNG = µ(WP + WL ) = NW . Now we calculate torque about the base
of the ladder.
τ : L NW sin(θ) − L/2 WL cos(θ) − x WP cos(θ) = 0.
Thus:
x=
Lµ(WP + WL ) sin(θ) − L/2 WL cos(θ)
= 2.51 m.
WP cos(θ)