This set of Spaceflight Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Two-Body Mechanics”.

1. What is the force of attraction between earth and moon? Given, distance between earth and moon is equal to 384,400 km, standard gravitational parameter of earth is equal to 398,600 km^{3}/s^{2} and mass of the moon is equal to 7.348 x 10^{22} kg.

a) 1.982 x 10^{17} N

b) 1.771 x 10^{26} N

c) 2.042 x 10^{12} N

d) 6.192 x 10^{17} N

View Answer

Explanation: Given,

Standard gravitational parameter of earth (μ) = 398,600 km

^{3}/s

^{2}

Mass of moon (m) = 7.348 x 10

^{22}kg

Distance between earth and moon (r) = 384,400 km

Force of attraction (F) = (μ.m) / (r)

^{2}

= (398,600 x 7.348 x 10

^{22}) / (384,400)

^{2}

= 1.982 x 10

^{17}N

2. Earth revolves slower around the sun when it is farther from the sun compared to when it is closer from the sun.

a) True

b) False

View Answer

Explanation: True because this phenomenon can be described by Kepler’s second law which states that a line joining a planet and the sun sweep equal areas in equal times. Kepler’s laws were major breakthroughs in orbital mechanics, and they are astonishingly accurate to this day.

3. What is the acceleration due to gravity at a height of 10 km from the surface of earth? The acceleration at the MSL is 9.81 m/s^{2} and the radius of earth is 6378 km.

a) 9.68 m/s^{2}

b) 9.78 m/s^{2}

c) 9.99 m/s^{2}

d) 10 m/s^{2}

View Answer

Explanation: Given,

Acceleration due to gravity at MSL (g

_{0}) = 9.81 m/s

^{2}

Radius of Earth (R

_{E}) = 6378 km

Height above the surface (h) = 10 km

Acceleration due to gravity at 10 km (g) = g

_{0}/ (1 + h / R

_{E})

^{2}

= 9.81/(1+10/6378)

^{2}

= 9.78 m/s

^{2}

4. What is the resultant instantaneous acceleration of a rocket with a combined mass of 1.98 x 10^{6} kg at that instant. Total thrust produced by the rocket is 31 x 10^{6} N.

a) 4.2 m/s^{2}

b) 5.85 m/s^{2}

c) 7.1 m/s^{2}

d) 3.25 m/s^{2}

View Answer

Explanation: Given,

Thrust (T) = 31 x 10

^{6}N

Mass of rocket (m) = 1.98 x 10

^{6}kg

Resultant Force (F) = T-mg

= 31*10

^{6}-(1.98*10

^{6}*9.81)

= 11.57 x 10

^{6}N

From Newton’s second law,

Resultant instantaneous acceleration (a) = F/m

=(11.57*10

^{6})/(1.98*10

^{6})

= 5.85 m/s

^{2}

5. The mean distance from the Sun and Mercury is 0.39 AU. What is the mean distance from the Sun and Venus? The sidereal periods for Mercury and Venus are 0.24 hours and 0.62 hours respectively.

a) 2.91 x 10^{8} km

b) 1.08 x 10^{4} km

c) 1.08 x 10^{8} km

d) 3.41 x 10^{11} km

View Answer

Explanation: Given,

Sidereal period of Mercury (T

_{1}) = 0.24 hours

Mean distance of Mercury from the Sun (d

_{1}) = 0.39 AU

Sidereal period of Venus (T

_{2}) = 0.62 hours

From Kepler’s third law: T

^{2}α d

^{3}

Therefore, Mean distance of Venus from the Sun (d

_{2}) = [(T

_{2}

^{2}d

_{1}

^{3})/(T

_{1})

^{2}]

^{1/3}

= [(0.62

^{2}*0.39

^{3})/(0.24)

^{2}]

^{1/3}

= 0.73 AU

= 1.08 x 10

^{8}km

6. It takes 2 months for a planet to travel from point A to point B and from point C to point D. Given A_{1} = 5.625 x 10^{9} km^{2}, what is A_{2}?

a) 100% of A_{1}

b) 75% of A_{1}

c) 125% of A_{1}

d) 119% of A_{1}

View Answer

Explanation: Kepler’s second law states that a line joining a planet and the sun sweep equal areas in equal times. Areas A

_{1}and A

_{2}are swept in the same sidereal period of 2 months. This means that both the areas must be equal. Therefore, A

_{2}= 100% of A

_{1}.

7. The standard parameter of Earth is 398,600 km^{3}/s^{2}, distance between Earth and Mars is 225 x 10^{6} km, force of gravitational attraction between Earth and Mars is 6.39 x 10^{23} N. What is the standard gravitational parameter of Mars? Given, universal gravitational constant is 6.67408 × 10^{-11} m^{3} kg^{-1} s^{-2}.

a) 78,901 km^{3}/s^{2}

b) 12,890 km^{3}/s^{2}

c) 34,567 km^{3}/s^{2}

d) 42,647 km^{3}/s^{2}

View Answer

Explanation: Given,

Force of attraction between Earth and Mars (F) = 6.39 x 10

^{23}N

Distance between Earth and Mars (r) = 225 x 10

^{6}km

Standard Gravitational Parameter of Earth (μ

_{E}) = 398,600 km

^{3}/s

^{2}

Universal gravitational constant (G) = 6.67408 × 10

^{-11}m

^{3}/(kg.s

^{2})

From Newton’s law of gravitation,

Mass of Mars (m) = (Fr

^{2})/μ

_{E}

= (6.39*10

^{23}*(225*10

^{6})

^{2})/(398,600*10

^{9})

= 6.39 x 10

^{23}kg

Standard Gravitational Parameter of Earth (μ

_{M}) = G.m

= 6.67408*10

^{-11}*6.39*10

^{23}

= 42,647 km

^{3}/s

^{2}

8. Which of these is NOT an application of satellites in orbit?

a) TV Broadcasting

b) Weather Forecasting

c) Global Positioning System

d) Fixed Wireless

View Answer

Explanation: Fixed Wireless is a type of communication that occurs between two fixed devices or between two fixed locations. An example of fixed wireless is communication between a computer and a Wi-Fi router. All other options are well implemented through satellites.

9. Kepler’s laws CANNOT be derived from Newton’s laws of motion and gravitation.

a) True

b) False

View Answer

Explanation: False. All three Kepler’s laws can be derived from Newton’s laws of motion and gravitation. Sir Isaac Newton was the first to prove Kepler’s laws mathematically. He also proved that an orbit can exist in 4 different shapes: circular, elliptical, parabolic and hyperbolic.

10. What is the tangential velocity of a satellite moving in a circular orbit at a height of 5250 km above the surface of Earth? The standard gravitational parameter of Earth is 398,600 km^{3}/s^{2}. The radius of Earth is 6378 km.

a) 11.1 km/s

b) 6.01 km/s

c) 5.85 km/s

d) 7.91 km/s

View Answer

Explanation: Given,

Distance between Earth and satellite (r) = 5250 + 6378

= 11,628 km

Standard gravitational parameter of Earth (μ) = 398,600 km

^{3}/s

^{2}

Since, gravitational attraction between Earth and satellite = centrifugal force of the satellite, we can derive,

Tangential velocity of satellite (V) = (μ/r)

^{1/2}

= (398,600/11,628)

^{1/2}

= 5.85 km/s

**More MCQs on Two-Body Mechanics:**

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