Download Radial Net Force Wkst 4

gravitational constant:G
 6.67  1011 Nm
kg 2
2
Name
Date
Pd
Central Net Force Model Worksheet 4:
Orbital Motion
1. Suppose you are at mission control on the moon, in charge of launching a moon-orbiting
communications satellite.
a. First, how much would a 1500 kg satellite weigh at the surface of the moon? (The
gravitational field strength on Earth's moon is 1.6 m/s2, or 1.6 N/kg.)
N
Fg  mg  1500kg (1.6 kg
)  2400N
b. The satellite is to have an altitude of 100 km above the moon's surface. What is the radius
of the orbit of the satellite? (The radius of the moon is1.74×106 m and the mass of the
moon is 7.36×1022 kg.)
6
6
6
m
1.74  106 m  100km( 1000
1km )  1.74  10 m  .1  10 m  1.84  10 m
c. When the satellite is in orbit, how big will the centripetal force be? Explain.
The centripetal force will be the force of the moons gravity on the satellite. We use
Newton's Law of Universal Gravitation to calculate this.
22
mmoonmsat
11 Nm2 7.36  10 kg(1500kg )
Fg  G
 6.67  10 kg2
 2175N
r2
(1.84  106 m)2
d. Find the required orbital velocity for the satellite.
Fnet  Fg 
F r 2175N (1.84  106 m)
mv 2
 v2  g 
r
m
1500kg
v 2  2,670,000 ms 2  v  1600 ms
2
e. How long will it take the satellite to orbit the moon? (This time is called the orbital period.)
v  1600
m
s
2 r
2 (1.84  106 m)


 7200s  2.0hours

1600 ms
f. Is this satellite accelerating while in orbit? If so, what is the direction and magnitude of
the acceleration?
2175 N
The satellite is always accelerating with a magnitude of 1.45 sm2 ( 1500
kg ) in the direction of
the moon.
©Modeling Instruction – AMTA 2013
1
U7 Central Force Model - ws4 v3.1
2. a. Why do astronauts float aboard the international space station? What sensation does an
astronaut feel while in orbit?
The astronauts are in a space ship that is falling around the earth. They are falling at the
same rate as the space ship. The feel the same sensation you would feel if you were in a
free falling elevator. (Until it hits the ground.)
b. Are astronauts in orbit really "weightless"? What might be a better description?
The astronauts are not weightless. There is a pull of gravity on them. They are
"normalless". There is no force pushing them away from the center of the earth.
3. The space shuttle aims for an orbit about 250 km above the surface of the earth. In orbit, the mass
of the space shuttle is about 95,000 kg. The radius of the earth is approximately 6.38 x 103 km and
the mass of the earth is 5.98×1024 kg.
a. Calculate the orbital speed of the space shuttle.
mems msv 2
m
Fnet  Fg  G 2 
 v2  G e
r
r
r
24
24
5.98  10 kg
2
11 Nm2
11 Nm2 5.98  10 kg
v  6.67  10 kg 2
 6.67  10 kg 2
m
6.38  103 km  250km( 1000
6.63  106 m
1km )
v 2  60,200,000 ms 2  v  7800 ms
2
b. Calculate the orbital period of the space shuttle.
v  7800 ms 
2 r
2 (6.63  106 m)

 5300s  89min

7800 ms
©Modeling Instruction – AMTA 2013
2
U7 Central Force Model - ws4 v3.1
4. 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." Actually, you do weigh a bit
less on the equator than you would at the poles. Calculate how much. (Hint: Construct force
diagrams for a 100 kg person standing on a bathroom scale at the equator and at the pole, and
do the Fnet calculations. Also, the radius of the earth is approximately 6.38 x 103 km.)
Fnet  Fg  FN 
FN
FN  G
G
Fg
me ms

r2
mpv 2
r
mpv 2
 FN  Fg 
mpv 2
r
r
me ms

r2
5.98  1024 kg(100kg)
 980N
6.38  106 m
mpv 2 mp ( 2 r )2 mp 4 2 r



r
r
2
100kg(6.38  106 m)4 2
 3.4N
2
s
24
hr
(3600
)

hr 
6.67  1011 Nm
kg 2
2
At the North pole where the person would not be moving the normal force equals the force of
gravity so the scale reads 980N. At the equator the normal force would be 3.4 N less.
5. The earth's orbit around the sun is very nearly circular, with an average radius of 1.5 x 108 km.
Assume the mass of the earth is 5.98 x 1024 kg.
a. Determine the average speed of the earth in its orbit around the sun.
m
2 (1.5  108 km( 1000
2 r
1km ))
v
v 
 29900 ms
hr

365.25days(24 day
)3600 hrs
b. What is the magnitude of the earth's average acceleration in its orbit around the sun?
ac 
v 2 (29900 ms )2

 0.0060 sm2
11
r
1.5  10 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?
Fc  mac  5.98  1024 kg(0.0060 sm2 )  3.6  1022 N
The force on the sun by the earth is also 3.6×1022 N but in the opposite direction.
©Modeling Instruction – AMTA 2013
3
U7 Central Force Model - ws4 v3.1