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 km3/s2 and mass of the moon is equal to 7.348 x 1022 kg.
a) 1.982 x 1017 N
b) 1.771 x 1026 N
c) 2.042 x 1012 N
d) 6.192 x 1017 N
View Answer
Explanation: Given,
Standard gravitational parameter of earth (μ) = 398,600 km3/s2
Mass of moon (m) = 7.348 x 1022 kg
Distance between earth and moon (r) = 384,400 km
Force of attraction (F) = (μ.m) / (r)2
= (398,600 x 7.348 x 1022) / (384,400)2
= 1.982 x 1017 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/s2 and the radius of earth is 6378 km.
a) 9.68 m/s2
b) 9.78 m/s2
c) 9.99 m/s2
d) 10 m/s2
View Answer
Explanation: Given,
Acceleration due to gravity at MSL (g0) = 9.81 m/s2
Radius of Earth (RE) = 6378 km
Height above the surface (h) = 10 km
Acceleration due to gravity at 10 km (g) = g0 / (1 + h / RE)2
= 9.81/(1+10/6378)2
= 9.78 m/s2
4. What is the resultant instantaneous acceleration of a rocket with a combined mass of 1.98 x 106 kg at that instant. Total thrust produced by the rocket is 31 x 106 N.
a) 4.2 m/s2
b) 5.85 m/s2
c) 7.1 m/s2
d) 3.25 m/s2
View Answer
Explanation: Given,
Thrust (T) = 31 x 106 N
Mass of rocket (m) = 1.98 x 106 kg
Resultant Force (F) = T-mg
= 31*106-(1.98*106*9.81)
= 11.57 x 106 N
From Newton’s second law,
Resultant instantaneous acceleration (a) = F/m
=(11.57*106)/(1.98*106)
= 5.85 m/s2
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 108 km
b) 1.08 x 104 km
c) 1.08 x 108 km
d) 3.41 x 1011 km
View Answer
Explanation: Given,
Sidereal period of Mercury (T1) = 0.24 hours
Mean distance of Mercury from the Sun (d1) = 0.39 AU
Sidereal period of Venus (T2) = 0.62 hours
From Kepler’s third law: T2 α d3
Therefore, Mean distance of Venus from the Sun (d2) = [(T22 d13)/(T1)2]1/3
= [(0.622*0.393)/(0.24)2]1/3
= 0.73 AU
= 1.08 x 108 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 A1 = 5.625 x 109 km2, what is A2?
a) 100% of A1
b) 75% of A1
c) 125% of A1
d) 119% of A1
View Answer
Explanation: Kepler’s second law states that a line joining a planet and the sun sweep equal areas in equal times. Areas A1 and A2 are swept in the same sidereal period of 2 months. This means that both the areas must be equal. Therefore, A2 = 100% of A1.
7. The standard parameter of Earth is 398,600 km3/s2, distance between Earth and Mars is 225 x 106 km, force of gravitational attraction between Earth and Mars is 6.39 x 1023 N. What is the standard gravitational parameter of Mars? Given, universal gravitational constant is 6.67408 × 10-11 m3 kg-1 s-2.
a) 78,901 km3/s2
b) 12,890 km3/s2
c) 34,567 km3/s2
d) 42,647 km3/s2
View Answer
Explanation: Given,
Force of attraction between Earth and Mars (F) = 6.39 x 1023 N
Distance between Earth and Mars (r) = 225 x 106 km
Standard Gravitational Parameter of Earth (μE) = 398,600 km3/s2
Universal gravitational constant (G) = 6.67408 × 10-11 m3/(kg.s2)
From Newton’s law of gravitation,
Mass of Mars (m) = (Fr2)/μE
= (6.39*1023*(225*106)2)/(398,600*109)
= 6.39 x 1023 kg
Standard Gravitational Parameter of Earth (μM) = G.m
= 6.67408*10-11*6.39*1023
= 42,647 km3/s2
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 km3/s2. 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 km3/s2
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:
Sanfoundry Global Education & Learning Series – Spaceflight Mechanics.
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