Download UCM/CNF Worksheet 3: Universal Gravitation

School of the Future
Physics II
Name:
Date:
UCM/CNF Worksheet 3: Universal Gravitation
Adapted from AMTA 2013 Modeling Instruction Materials (U7 CNF Model – WS4, V3.1)
•
•
Body
Mass (kg)
Sun
Earth
Moon
Mars
Jupiter
Saturn
Pluto
1.989 x 1030
5.972 x 1024
7.348 x 1022
6.419 x 1023
1.898 x 1027
5.685 x 1026
1.305 x 1022
CELESTIAL REFERENCE TABLE
Distance from
Radius (km)
Sun’s Center (km)
696,300
N/A
6,371
1.496 x 108
1,737
N/A
3,396
2.279 x 108
69,911
7.785 x 108
60,268
1.433 x 109
1,153
5.874 x 109
Surface Gravitational
Field Strength (in g’s)
????
1
???
0.376
???
1.065
???
Reminder: 1.00lbs = 4.45N (approximately)
Reminder: ON THE SURFACE OF THE EARTH: 1.0kg = 2.2lbs (approximately)
1. Two students sitting next to one another are roughly 50cm away from one another. If both
students have the same mass of 70kg,
a. What gravitational force does each apply to the other?
b. Why is it that, when drawing FBDs and analyzing systems, we ignore the
gravitational forces applied by any else except Earth?
2. Mr. Z weighs 158lbs on Earth. What would his weight be on:
a. Mars?
b. Saturn?
c. In deep space (very, very far away from any large bodies)?
3. Mr. Z weighs 158lbs on Earth. How would this change if:
a. His mass doubled?
b. Earth’s mass doubled?
c. The distance between him and Earth doubled?
d. His Earth’s mass doubled AND the distance between hi and Earth doubled?
4. Planes travel at roughly 35,000ft above sea level. Determine how much your weight
changes as the plane goes from the ground to this altitude.
5. Mr. Z (158lbs) travels to Pluto.
a. What is the magnitude of the gravitational force he experiences while on the
surface of this planet?
b. What is the gravitational field strength at the surface of this planet?
c. What does your answer to (b) mean?
6. What is the gravitational field strength at the surface of:
a. The Moon
b. Jupiter
c. The Sun
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School of the Future
Physics II
Name:
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7. Suppose you are at mission control on the moon, in charge of launching a moon-orbiting
communications satellite. The satellite has a mass of 1500kg.
a. Draw an FBD for the satellite.
The only force that acts on the satellite is a gravitational force applied by the moon in the radial
direction (towards the Moon’s center).
b. The satellite is to have an altitude of 100 km above the moon's surface. When the
satellite is in orbit, how big will the gravitational force be?
R = 1,737km + 100km = 1837km = 1,837,000m
mm
FG = G 1 2 2
R
m 2 (1500kg)(7.348x10 22 kg)
FG = (6.67x10 −11 N ⋅ 2 )
kg
(1,837, 000m)2
FG = 2178.6N
c. What will the net centripetal force on the satellite be? Why?
ΣF = FG = 2178.6N
Since the gravitational force applied by the moon is the only force on the satellite, that
represents the entire net force on the satellite.
d. What is the required orbital velocity for the satellite?
ΣF = m
v2
R
2178.6N = (1500kg)
v = 1633.4 m
v2
(1,837, 000m)
s
e. How long will it take the satellite to orbit the moon? (This time is called the orbital
period.) 2π R
2π R
===> T =
T
v
2π (1,837, 000m)
T=
1633.4 m
s
T = 7066.4s = 1.96hrs
v=
f. Is this satellite accelerating while in orbit? If so, what is the direction and
magnitude of the acceleration?
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School of the Future
Physics II
Name:
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Yes, since the satellite is changing direction during its orbit, it is experiencing centripetal
acceleration (towards the center of the Moon).
2
v 2 (1633.4 m s )
aC = =
= 1.42 m 2
s
R (1,873, 000m)
8. Consider astronauts aboard the ISS.
a. Are astronauts in orbit really "weightless"? Why or why not?
Astronauts aboard the ISS are not actually “weightless”. They must experience a gravitational
force from Earth, which happens to be the centripetal force that keeps them moving in UCM. If
they didn’t, they would remain in equilibrium and continue in a straight-line path instead of
orbiting the Earth.
b. Why do astronauts float aboard the international space station?
Astronauts float aboard the ISS because they are falling just as the ISS is. Recall, orbiting
satellites are technically falling, but just moving fast enough tangentially to maintain a constant
orbit. The same can be said about the passengers inside the ISS. For this reason, since they
experience the same falling motion as the ISS, they appear to “float” inside the ISS. For this
reason, they experience “apparent weightlessness” – meaning they don’t experience a normal
force.
9. The earth's orbit around the sun is very nearly circular.
a. The Earth completes one revolution around the sun in 365 days. Determine the
average speed of the earth in its orbit around the sun. R = 1.496 x 108 km = 1.496 x 1011m
T = 365days = 31536000s
v=
2π R 2π (1.496x1011 m)
=
= 29806 m
s
T
31536000s
b. What is the magnitude of the earth's average acceleration in its orbit around the
sun?
2
v 2 (29806 m s )
aC = =
= 0.00594 m 2
11
s
R (1.496x10 m)
c. Determine the gravitational force on the earth by the sun. How does the force on
the earth by the sun compare to the force on the sun by the earth?
NEVER DOUBT THAT A SMALL GROUP OF COMMITTED INDIVIDUALS CAN CHANGE THE WORLD, INDEED IT IS THE ONLY THING THAT EVER HAS. MEAD
School of the Future
Physics II
FG = G
ΣF = m ⋅ a
ΣF = (5.972x10 24 kg)(0.00594 m
s2
ΣF = 3.55x10 22 N
)
Name:
Date:
m1m2
R2
OR FG = (6.67x10 −11 N ⋅
m 2 (5.972x10 24 kg)(1.989x10 30 kg)
)
kg 2
(1.496x1011 m)2
Fg = 3.55x10 22 N
Fg = 3.55x10 22 N
10. Determine the length of a “year” on Jupiter (the number of days it takes for it to complete
a full orbit around the Sun).
FG = G
m1m2
R2
FG = (6.67x10 −11 N ⋅
m 2 (1.898x10 27 kg)(1.989x10 30 kg)
)
kg 2
(7.785x1011 m)2
Fg = 4.155x10 23 N
ΣF = m
v2
R
4.155x10 23 N = (1.898x10 27 kg)
v = 13054 m
v2
(7.785x1011 m)
s
2π R
2π R
===> T =
T
v
11
2π (7.785x10 m)
T=
13054 m
s
8
T = 3.747x10 s = 4, 336.8 Earth Days
v=
11. CHALLENGE: Back in Galileo's day, one of the
objections to the heliocentric model of the solar
system is that if the earth is spinning, we should
all be "thrown off the earth." In reality, a scale
would say that you weigh a bit less on the
equator than you would at the poles. Calculate
how much less you would weigh. (Hint: Start by
making FBDs for a person at the equator and at
the North Pole.)
NEVER DOUBT THAT A SMALL GROUP OF COMMITTED INDIVIDUALS CAN CHANGE THE WORLD, INDEED IT IS THE ONLY THING THAT EVER HAS. MEAD
School of the Future
Physics II
Name:
Date:
v2
2π R
ΣF
=
m
v=
R
T
ΣF = FN + FG
(463.3 m )2
2π (6.371x10 6 m)
s
v=
THEN ΣF = (75kg)
THEN 2.53N = FN + (75kg)(9.81 N )
kg
(6.371x10 6 m)
86, 400s
FN = −733.22N (upward)
ΣF = 2.53N (downward)
v = 463.3 m
s
A 75kg person standing on a scale on the North Pole will read 735.75N (165.3lbs). On the
Equator, the scale will read 733.22N (164.8 lbs). They’ll be 0.5lbs lighter!
NEVER DOUBT THAT A SMALL GROUP OF COMMITTED INDIVIDUALS CAN CHANGE THE WORLD, INDEED IT IS THE ONLY THING THAT EVER HAS. MEAD