Download PHYS 2210 — Fall 2015 GA8 Solutions

PHYS 2210 — Fall 2015
GA8 Solutions
Problem 1
Work and Kinetic Energy: Apollo 11 moon landing
During the Apollo 11 moon landing on July 20, 1969, a lunar module containing American astronauts Neil Armstrong and Buzz Aldrin descended from a lunar orbit onto the surface of the moon.
The module’s rocket thrusters were used to extract the module from its stable circular orbit and
control its descent onto the moon’s surface. During its descent, the lunar module followed a spiral
path down to the moon’s surface (see figure). This path had a length roughly equivalent to half the
circumference of one lunar orbit. The lunar orbit was approximately circular at a height of about
110 km above the surface of the moon. The moon’s mass is 7.3 × 1022 kg, and its average radius is
1737 km. The mass of the lunar module was 15 × 103 kg.
(a) What was the kinetic energy of the lunar module while in orbit?
(b) What was the net work done on the lunar module during its descent to the moon’s surface?
(c) What was the work done by gravity during the lunar module’s descent to the moon’s surface?
(d) What was the work done by the rocket engines during the lunar module’s descent?
(e) Calculate the average force exerted by the rocket thrusters on the lunar module itself during
its descent to the moon’s surface.
(f) The lunar module’s rocket thrusters work by ejecting mass (hot gas) at very high velocity out
d(mv)
dv
dm
of a nozzle. We can write Newton’s 2nd Law in the form, F = dp
dt = dt = m dt + v dt , where
F refers to the force on the ejected gas, m and v are the mass and speed of the ejected gas.
If we assume that v ∼ 2500 m/s is constant over time, then the force on the gas is F = v dm
dt .
Using this information, calculate the average “burn rate” for the rocket fuel in kg/s.
1
2
Solution
I. Understand the Problem: As the lunar module descends from orbit to land on the surface
of the moon, it must use its rocket thrusters to control its trajectory and speed so that it doesn’t
crash land. We are going to approximate the work done by the rocket thrusters to achieve a safe
landing, the average force (thrust) exerted by the engines on the module, and the fuel “burn rate”.
II. Describe the Physics: When in orbit, the lunar module has a particular speed, vorbit , to
maintain a circular orbit with radius, Rorbit = Rmoon + H, where H = 110 km. When the module
lands on the surface, it must land gently, so its final speed just before touchdown should be vfinal ≈ 0.
Thus the kinetic energy of module decreases from its initial orbital value to approximately zero
at touchdown. While descending to the moon’s surface, the gravitational force from the moon
does positive work on the module since the module is getting closer to the center of the moon. If
this were the only force doing work on the module, its kinetic energy should therefore increase by
virtue of the work-kinetic energy theorem: ∆K = Wnet . Since we know the module’s kinetic energy
actually decreases during the descent, we conclude that the rocket thrusters do negative work on
the module, that is they exert a force in the opposite direction to its motion.
A few other key assumptions to consider:
(1) We can neglect the effect of drag/air resistance. In this case, this is perfectly valid because the
moon doesn’t have an atmosphere.
(2) The height of the module’s orbit above the surface of the moon is H = 110 km, which is ≈ 6%
of the radius of the moon. This is sort of borderline for using the simpler expression for work
done by gravity, Wgrav = −mg∆y and to be more precise,
one should probably use the more
general form, Wgrav = GMmoon Mmodule 1/r22 − 1/r12 , but since we are making estimates, we’ll
use the simpler form to make the math slightly easier.
(3) Since we aren’t near the surface of the earth, we need to recalculate the value of g corresponding
to the moon:
gmoon = G
Mmoon
2
Rmoon
= (6.67 × 10−11 m3 /kg/s2 )
= 1.61 m/s2
(1)
7.3 × 1022 kg
(1.74 × 106 m)2
(2)
(3)
III. Plan the Solution: The kinetic energy of the module while in orbit is easily found by
equating the gravitational force on the module at heigh H with the product of the module’s mass
and its centripetal acceleration (Newton’s second law). This is numerically equal to the net work
done on the module during its descent (with a “minus” sign) so since the kinetic energy of the
module upon touchdown is zero. The work done by gravity only depends on the modules change in
height, as per the approximate equation above. Subtracting the work done by gravity (positive),
from the net work (negative) will give an even larger negative value for the work done by the rocket
thrusters. Once we have the work done by the rocket thrusters, we can use the given path length
D = π(Rmoon + H) to find the average force exerted by the thrusters on the module and combining
this with the form of Newton’s second law provided in the problem, we can find the fuel burn rate.
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IV. Execute the Solution:
(a) What was the kinetic energy of the lunar module while in orbit?
First find the orbital speed:
r
vorbit =
s
=
GMmoon
Rmoon + H
(4)
(6.67 × 10−11 )(7.3 × 1022 )
= 1.62 km/s
(1.74 × 106 ) + (110 × 103 )
(5)
(6)
Now find the kinetic energy:
1
2
Korbit = Mmodule vorbit
2
1
= (1.5 × 104 kg)(1.62 × 103 m/s)2
2
= 1.97 × 1010 Joules
(7)
(8)
(9)
(b) What was the net work done on the lunar module during its descent to the moon’s surface?
Use the work-kinetic energy theorem:
Wnet = ∆K = 0 − Korbit
= −1.97 × 10
10
(10)
Joules
(11)
(c) What was the work done by gravity during the lunar module’s descent to the moon’s surface?
Using the approximate formula for objects close to the moon’s surface (see justification above):
Wgrav = −Mmodule gmoon ∆y = −Mmodule gmoon (−H)
= (1.5 × 104 kg)(1.61 m/s2 )(110 × 103 m)
10
= +2.66 × 10
(12)
(13)
Joules
(14)
(d) What was the work done by the rocket engines during the lunar module’s descent?
We have computed the net work done on the lunar module and the work done by gravity. We
now simply subtract the work done by gravity from the net work. What’s left over is the work
done by the rocket thrusters:
Wnet = Wgrav + Wthrust
(15)
10
=⇒ Wthrust = Wnet − Wgrav = (−1.97 × 10
= −4.63 × 1010 Joules
10
− 2.66 × 10 ) Joules
(16)
(17)
(e) Calculate the average force exerted by the rocket thrusters on the lunar module itself during
its descent to the moon’s surface.
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The problem states that the lunar module descends along a spiral path roughly equivalent to
half an orbital circumference:
1
D = [2π(Rmoon + H)] = π(Rmoon + H)
2
= π (1.74 × 106 ) + (110 × 103 )
= 5.81 × 106 m
(18)
(19)
(20)
Assuming the rocket thrust is always directed opposite the module’s direction of travel, then
we have:
~ = −Favg D
Wthrust = F~avg · D
Wthrust
4.63 × 1010 Joules
=
D
5.81 × 106 m
= 7.97 kN
=⇒ Favg = −
(21)
(22)
(23)
(f) Calculate the average “burn rate” for the rocket fuel in kg/s.
Using the provided information, we have:
F = vfuel
Favg
dMfuel
dMfuel
=⇒
=
dt
dt
vfuel
7.97 × 103 N
=
2.5 × 103 m/s
= 3.2 kg/s
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(24)
(25)
(26)