Problem 1.
In what year were the first iron-cased rockets used against the British in the Anglo-Mysore
Wars?
Solution.
1792
Problem 2.
What are the units for specific impulse?
Solution.
The units for specific impulse are seconds.
Problem 3.
Which of the following are rockets?
• Spaceship
• Missile
• Airplane
• Jet
Solution.
The correct answers are Spaceship and Missile. Recall from lecture that ”A rocket is any type of vehicle that uses a rocket engine to produce thrust”. Additionally, a rocket, unlike a jet or airplane, carries its own fuel and oxidizer for combustion.
Problem 4.
What is the force generated between two objects, mass
1 by 0.5 meters?
= 10kg and mass
2
= 20kg separated
Solution.
m
2
The equation to calculate force between two objects was given as F =
Gm
R
1
2 m
2 where m
1 and are given as 10kg and 20kg, respectively.
The gravitational constant G is given as G = 6 .
67408 x 10 − 11 m 3 kg − 1 s − 2 and the distance between the objects is R = 0 .
5m. Evaluating the equation, F =
6 .
67408 × 10
− 11
· 10 kg · 20 kg
0 .
5 m 2
→ F = 5 .
338 × 10
− 8 N.
Problem 5.
Calculate the centripetal force of a 50kg object moving at 100m/s within a 10m radius, enter the number without units.
Solution.
The equation to calculate centripetal force is given as F c
= − mv
2 r
ˆ . In this equation, the force is a vector in terms of the radius unit vector, ˆ , which is pointing away from the center of the circle. The centripetal force is pointing towards the center of the circle, which is opposite of ˆ , therefore we include a negative sign. Identifying the given information, m=50kg, v=100m/s and r=10m. Plugging these into the equation we get, F c
= − 50 × 100
2
10
→ F c
= − 50 , 0000 N.
Problem 6.
Using the atmospheric scale height for Earth derived in class, calculate the density at 60 km above sea level. Now convert kg/m3 to g/m3. Assume H is constant at 8 km and n
0
, the density at sea level is 1 .
225 kg · m
− 3
.
1

Solution.
The equation of atmospheric density as a function of height is given as n ( r ) = n
0 e
− r/H where
H = k
B
T /mg . The scale height, H, is given as 8km and remains constant with height in the problem.
The other components are n
0
, the density of the atmosphere at sea level which is 1 .
225 kg · m − 3 and the height above the Earth’s surface which is 60km. Plugging this information into the equation yields n ( r ) = 1 .
225 e
− 60 / 8 → n ( r ) = 6 .
775 × 10
− 4 kg · m
− 3 .
The last step is to convert 6 .
775 × 10
− 4 kg · m
− 3 to g · m
− 3
. We do this as,
6 .
775 × 10
− 4 kg m 3 x
1 , 000 g
1 kg
→ n ( r ) = 0 .
6775 g · m
− 3
Problem 7.
Which are advantages of liquid rockets (can be more than one correct answer)
• Higher effective exhaust velocities
• You can restart them
• Pressure in combustion chamber is low compared to solid rocket fuel
• Can only reuse liquid fuel rockets
• Fuel tanks do not have to handle high pressures
Solution.
The correct answers are: Higher effective exhaust velocities, this is because they have a higher specific impulse. You can restart them. Fuel tanks do not have to handle high pressures. This is due to the high density of the liquids which does not require the tanks to be pressurized.
Problem 8.
Above Armstrong’s line at 18 km above the surface of the Earth
• Supplemental oxygen allows you to breath even through the air is thin
• You would need to wear a pressure suit to prevent the water in your lungs from boiling
• A spacecraft would travel in a line because air resistance is low
• If you tried to skydive you would fall faster than the speed of sound
Solution.
The correct answers is: You would need to wear a pressure suit to prevent the water in your lungs from boiling. This is because the vapor pressure at this height allows for water to boil at body temperature.
Problem 9.
Use 6,378 km for the radius of the Earth. How far from the center of the Earth (not from the surface of the Earth!) would a spacecraft in a circular orbit have to be so it appears to hover over the same spot on Earth? Hint: this is called a geosynchronous orbit because the spacecraft takes exactly a day to orbit Earth and the Earth rotates once in a day. Use 24 hours as the period of the orbit. Write your answer in units of Earth radii.
Solution.
Using the equation from lecture ’In Space’, we know v = ( G · m
E where G is the gravitational constant, m where T
E is the mass of the Earth, r is the radius of our orbit. Also, T =
= 24 h for geosynchronous orbit. We can rearrange the equation to, velocity equation mentioned above to solve for r. We get r = (
Gm
E
T
(2 π ) 2
2
) 1 / 3
· /r) 1 / 2 v for a stable circular orbit
=
2 πr
T
2 πr v
, and equate to the
, and plugging in the gravitational constant G that is given as G = 6 we get the answer to be r = 4.184
.
67408 x 10
× 10
7
− 11 m m = 6.6 R
E
3 kg
− 1 s
− 2
, given R
, m
E
E
= 5 .
972 ×
= 6 .
378 × 10
6
10 m.
24 kg, T = 24 h = 86,400 s,
Problem 10.
Match the rockets.
• R-7
• Redstone
• Vanguard
2

Solution.
• R-7 - First Soviet ICBM refurbished to launch Sputnik
• Redstone - US rocket descended from V-2
• Vanguard - Unclassified US rocket meant to be used for scientific missions
Problem 11.
Remember the thrust produced by a solid fuel rocket is proportional to the surface area of the fuel that is burning at any moment in time, so we can adjust the thrust of a solid rocket by changing the shape of the burning surface and considering how the burning fuel will change shape with time. Match the solid fuel rocket grain shape to the resulting performance.
• Rod and tube
• Tubular
• Multi-fin
• Star
Solution.
These can be found in the plots under ’Playing Games with the Grain’ slide in the ’Solid Fuel
Nozzle’ lecture.
• Rod and tube - Thurst is constant with time.
• Tubular - Thurst increases with time.
• Multi-fin - A burst of high thrust and then steady low thrust.
• Star - roughly constant thrust with time.
Problem 12.
I would like to design an upper stage rocket that produces a Delta V three times larger than the effective escape velocity of the exhaust from its combustion chamber. What is the ratio of the initial to final mass of the rocket after the fuel is burned, including the payload..
Solution.
Here you have to use the rocket equation, ∆ the exhaust velocity, and m
◦ and m f v = v e ln m
◦ m f where ∆ v , is your change in velocity, v e is are the initial and final masses of the rocket, respectively. To determine the initial to final mass ratio that would produce a Delta V = 3 v e that ∆ v v e
= ln m
◦ m f
, we need to rearrange the equation such then raise both sides of the equation to the power of e, where we can use e ln ( x ) = x to simplify. Finally, we get m
◦ m f initial to final mass is 20:1.
= e
∆ v ve . We can substitute ∆ v = 3 v e into the equation, m
◦ m f
= e
3 ve ve = 20. So the
Problem 13.
The Space Shuttle solid rocket boosters were filled with APCP-PBAN (about 70% ammonium perchlorate, 16% aluminum power, and about 12% PBAN). What was the primary source of fuel within this mixture?
Solution.
The answer is Aluminum powder.
Problem 14.
What is the generally accepted date for the start of the Space Race?
Solution.
The answer is August 2, 1955 with both countries announcing intent to fly a science mission for the IGY.
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