This set of Orbital Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Orbits”.

1. Given, the semi-major axis and semi-minor axis are 6580 km and 5910 km respectively, what is the eccentricity of the orbit?

a) 0.89

b) 0.44

c) 0.64

d) 0.25

View Answer

Explanation: Given,

Semi-major axis (a) = 6580 km

Semi-minor axis (b) = 5910 km

Eccentricity (e) = (1 – (b/a)

^{2})

^{1/2}

= (1-(5910/6580)

^{2})

^{1/2}

= 0.44

2. What is the eccentricity of a spacecraft around earth? Given, specific energy is -10.29 km^{2}/s^{2}, specific angular momentum is 10,112 km^{2}/s. The standard gravitational parameter of earth is 398,600 km^{3}/s^{2}.

a) 0.478

b) 1.45

c) 0.691

d) 0.456

View Answer

Explanation: Given,

Specific energy (ε) = -10.29 km

^{2}/s

^{2}

Specific angular momentum (h) = 10,112 km

^{2}/s

Standard gravitational parameter (μ) = 398,600 km

^{3}/s

^{2}

Using the e-h-ε relation

e = (1+(2εh)/μ)

^{1/2}

= (1+(2*-10.29*10,112)/398,600)

^{1/2}

= 0.691

3. Which of the following orbit shapes will be formed if a satellite is travelling at exactly the escape velocity of earth?

a) Circular

b) Parabolic

c) Hyperbolic

d) Elliptical

View Answer

Explanation: Parabolic orbits are formed at the exact escape velocity of a planet. Elliptical and circular orbits are formed when the satellite’s velocity is less than the escape velocity. Whereas, at velocities higher than the escape velocity, hyperbolic orbits are formed.

4. If the standard gravitational parameter of earth, μ is 398,600 km^{3}/s^{2} and the radius of earth is 6378 km, what is the period of a satellite revolving around earth in a circular orbit 500 kms above the surface of the earth?

a) 2,012.2 s

b) 5,801.06 s

c) 1,156.21 s

d) 937.267 s

View Answer

Explanation: The Time Period of a circular orbit can be derived as:

T = (2πR

^{3/2}/√(μ)

R = radius of earth + distance from the surface of earth to satellite

R = 6378 + 500

R = 6978 km

Therefore, T = (2π(6978)

^{3/2}/√(398600)

T = 5,801.06 s

5. The eccentricity of an asteroid is 0.7. What is the shape of its orbit around the sun?

a) Circular

b) Hyperbolic

c) Parabolic

d) Elliptical

View Answer

Explanation: Elliptical orbits have an eccentricity greater than zero and less than one. Most planets and orbits orbiting around the sun are elliptical in shape.

6. What is the eccentricity of a satellite around a planet if the periapsis and apoapsis are at 6,778 km and 10,378 km respectively from the center of the planet?

a) 0.7051

b) 0.5

c) 0.2098

d) 0.901

View Answer

Explanation: The eccentricity of an orbit can be determined as follows:

e = (Apoapsis distance – Periapsis distance) / (Apoapsis distance + Periapsis distance)

e = (10,378 – 6778) / (10,378 + 6778)

e = 0.2098

7. What is the semi-major axis of an elliptical orbit with perigee and apogee distances of 2500 km and 8000 km respectively?

a) 5250 km

b) 5000 km

c) 4950 km

d) 5150 km

View Answer

Explanation: Semi-major axis of an ellipse is half of the major axis. Major axis can be represented as a sum of perigee and apogee distances. Therefore,

Semi-major axis = (2500 + 8000) / 2 = 5,250 km

8. Which of the following describes the angle measured between the ascending node and the periapsis?

a) Argument of periapsis

b) Inclination

c) Longitude of the ascending node

d) True anomaly

View Answer

Explanation: Argument of periapsis is defined as the angle between ascending node and periapsis. The angle between vernal equinox and ascending node is longitude of the ascending node. The angle between earth’s equatorial plane unit vector and satellite’s angular momentum vector is inclination. True anomaly is the angle between perigee and spacecraft position relative to center of the earth.

9. What is the specific angular momentum, h of a satellite in orbit as shown in figure below? Given, radius of earth is 6378 km and standard gravitational parameter is 398,600 km^{3}/s^{2}.

a) 44,342 km^{2}/s

b) 59,642 km^{2}/s

c) 57,112 km^{2}/s

d) 12,012 km^{2}/s

View Answer

Explanation: Perigee distance, r

_{p}= Radius of earth + distance from surface

= 6378 + 400

= 6978 km

Similarly, Apogee distance, r

_{a}= 6378 + 6000

=12,378 km

Eccentricity, e = (r

_{a}– r

_{p}) / (r

_{a}+ r

_{p})

= (12,378 – 6,978) / (12,378 + 6,978)

= 0.2789

From Keplerian solution to two-body problems, Angular momentum, h can be found:

r = (h

^{2}/ μ) / (1 + e.cosθ)

For perigee, θ = 0 degrees

Therefore, r

_{p}= (h

^{2}/ μ) / (1 + e)

h = √(r

_{p}(1 + e) μ)

h = √(6978 x (1 + 0.2789) x 398,600)

h = 59,642 km

^{2}/s

10. Which of the following coordinate systems has sun as its center?

a) Geocentric-Equatorial System

b) Geocentric-Ecliptic System

c) Heliocentric-Ecliptic System

d) Right Ascension-declination System

View Answer

Explanation: Heliocentric-Ecliptic system is a system which has sun as its center or origin. Fundamental plane of operation of this coordinate system is ecliptic plane of our solar system. Most of the planets along with the sun lie on this plane.

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