University Physics
Quiz 6 Solution
Fri Oct 11, 2012
Your name:
Please show your work, write neatly, and put a box or circle around each of your answers.
This week the quiz is also being given in MasteringPhysics with the same problems.
1. (4 points total) What is the approximate value of the gravitational force between a
75 kg astronaut and a 7.7×104 kg spacecraft when they’re 90 m apart? Express your
answer using two significant figures, in Newtons.
Solution: The gravitational force is Fg = (Gm1 m2 )/r2 , so we have
FG =
(6.67 × 10−11 N m2 /kg2 )(75 kg)(7.7 × 104 kg)
= 4.8 × 10−8 N = FG
(90 m)2
2. (3 points total) A sensitive gravimeter (a device for measuring the acceleration of gravity) is carried to the top of Chicago’s Willis (formerly Sears) Tower, where its reading
for the acceleration of gravity is 1.36 mm/s2 lower than at street level. Find the building’s height in meters.
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Solution: The acceleration due to gravity is a = GMEarth /rEarth
where MEarth is the
mass of the Earth, and rEarth is the distance from the center of the earth. If the height
of the building is h and the magnitude of the acceleration due to Earth’s gravity at the
surface of the Earth is g, then we have
g=
GMEarth
2
rEarth
(g − 0.00136 m/s2 ) =
GMEarth
(rEarth + h)2
Subtracting the second equation from the first gives
GMEarth
GMEarth
−
2
rEarth
(rEarth + h)2
GMEarth
GMEarth
=
− 0.00136 m/s2
2
(rEarth + h)2
rEarth
GMEarth
(rEarth + h)2 =
2
(GMEarth /rEarth − 0.00136 m/s2 )
s
GMEarth
h =
− rEarth
2
(GMEarth /rEarth − 0.00136 m/s2 )
0.00136 m/s2 =
and using G = 6.67 × 10−11 N m2 /kg2 , MEarth = 5.97 × 1024 kg, and rEarth = 6.371 ×
106 m, we have h = 441 m .
3. (3 points total) Mars’s orbit has a diameter 1.52 times that of Earth’s orbit. How long
does it take Mars to orbit the Sun? Express your answer in Earth years.
Solution: The period T of a planet’s orbit around the sun is related to its radius r by
T 2 = 4π 2 r3 /GM where M = Msun + Mplanet is the total mass of both planet and sun.
Taking TE , rE , and ME for the period, radius and mass of the Earth, and TM , RM , and
MM as the period, radius, and mass of Mars, we have
TE2
4π 2 rE3
=
G(Msun + ME )
2
TM
3
4π 2 rM
=
G(Msun + MM )
Taking the ratio of the second over the first, we then have
2 3 TM
rM
Msun + ME
=
TE
rE
Msun + MM
1
or
s
TM
3 Msun + ME
rM
= TE
rE
Msun + MM
p
≈ (1 year) (1.52)3
≈
1.87 years = TM
This is very close to the Wikipedia value of 1.88 years.
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