This set of Orbital Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Circular Orbits – Set 2”.

1. How much is the delta-v required to put a satellite in circular orbit of altitude 3,000 km over Mars surface. The satellite is travelling initially at 27 km/s. Radius and gravitational parameter of Mars are 3,389.5 km and 42,828 km^{3}/s^{2}.

a) 24.41 km/s

b) 21.11 km/s

c) 15.8 km/s

d) 22.67 km/s

View Answer

Explanation: Given,

Initial velocity (v

_{i}) = 27 km/s

Gravitational Parameter (μ) = 42,828 km

^{3}/s

^{2}

Radius of Satellite (r) = 3,389.5 + 3,000

= 6,389.5 km

Orbit velocity (v) = (μ/r)

^{1/2}

= (42,828/6,389.5)

^{1/2}

= 2.589 km/s

Delta-v = v

_{i}– v

= 27 – 2.589

= 24.41 km/s

2. What is the escape velocity of a satellite around Venus if its specific angular momentum is 47,862.73 km^{2}/s? The orbit of satellite is circular and gravitational parameter of Venus is 324,859 km^{3}/s^{2}.

a) 10.36 km/s

b) 11.11 km/s

c) 9.599 km/s

d) 9.45 km/s

View Answer

Explanation: Given,

Specific angular momentum (h) = 47,862.73 km

^{2}/s

Gravitational parameter (μ) = 324,859 km

^{3}/s

^{2}

Escape velocity (v

_{esc}) = 2

^{1/2}(μ/h)

= 2

^{1/2}*(324,859/47,862.73)

= 9.599 km/s

3. What is the time period of revolution of a satellite around Venus if its specific angular momentum is 18,919 mi^{2}/s. The orbit of the satellite is circular. Gravitational parameter of Venus is 77,937.75 mi^{3}/s.

a) 1.809 hours

b) 1.946 hours

c) 2.01 hours

d) 22 hours

View Answer

Explanation: Given,

Specific angular momentum (h) = 18,919 mi

^{2}/s

= 18,919 x 1.60934

^{2}km

^{2}/s

= 49,000 km

^{2}/s

Gravitational parameter (μ) = 77,937.75 mi

^{3}/s

= 77,937.75 x 1.60934

^{3}km

^{3}/s

= 324,856 km

^{3}/s

Satellite radius (r) = h

^{2}/μ

= 49,000

^{2}/324,856

= 7,390.967 km

Time period (T) = 2πr

^{3/2}μ

^{-1/2}

= 2π*7,390.967

^{3/2}*324,856

^{-1/2}

= 7,004.65 s

= 1.9457 hours

4. Velocity of a satellite in a circular orbit is dependent on radius of orbit, mass of the earth and true anomaly.

a) True

b) False

View Answer

Explanation: False, because the velocity of a satellite in a circular orbit is independent of true anomaly. Unlike, elliptical, parabolic, hyperbolic trajectories where the velocity is dependent on the true anomaly. Mass of the earth/planet and radius of orbit are dependent on the velocity for all cases of orbits.

5. Velocity of a satellite in circular orbit is _____________________ the radius of orbit.

a) directly proportional to

b) inversely proportional to

c) inversely proportional to square root of

d) inversely proportional to square of

View Answer

Explanation: Only for circular orbits is the velocity of a satellite inversely proportional to square root of the radius of orbit. In all the other cases, the relationship is not as straightforward due to the tangential and radial velocity components. It cannot be inversely proportional either, though v = h/r. Since h, is dependent on r. Only equation that best describes the relationship is v = (μ/r)

^{1/2}and μ is a constant.

6. What is the specific energy of an asteroid in a circular orbit around the Sun of radius 4.8 AU. The gravitational parameter of Sun is 1.327124 x 10^{11} km^{3}/s^{2}.

a) -2.705 m^{2}/s^{2}

b) 92.405 km^{2}/s^{2}

c) -2.705 x 10^{-3} m^{2}/s^{2}

d) -2.705 x 10^{-3} km^{2}/s^{2}

View Answer

Explanation: Given,

Radius of orbit (r) = 4.8 AU

= 4.8 x 1.496 x 10

^{8}km

= 7.181 x 10

^{8}km

Gravitational parameter (μ) = 1.327124 x 10

^{11}km

^{3}/s

^{2}

Specific energy (ε) = -μ/(2r)

= -(1.327124 x 10

^{11})/(2 x 7.181 x 10

^{8})

= 92.405 km

^{2}/s

^{2}

7. Describe the equation of the circular orbit in cartesian coordinates. Given, a point is P(6,054.78,6492.95) along the orbit in cartesian coordinates. Assume the centre of larger body to be the origin O(0,0).

a) (x – 6,054.78)^{2} + (y – 6,492.95)^{2} – 8,878 = 0

b) x^{2} + y^{2} – 12,548 = 0

c) x^{2} + y^{2} – 8,878 = 0

d) x^{2} + y^{2} – 78,818,884 = 0

View Answer

Explanation: Given, origin O(0,0) at larger body centre.

Therefore, point P(6,054.78,6492.95) can be used to determine the orbit radius.

Orbit radius (r) = (6,054.78

^{2}+ 6492.95

^{2})

^{1/2}

= 8,878 km

Equation of a circle can be written as x

^{2}+ y

^{2}= r

^{2}, if origin point is (0,0)

Therefore, equation of circle is

x

^{2}+ y

^{2}– 78,818,884 = 0

8. A satellite is in an orbit with a velocity of 3.156 km/s, radius of 3,000 km and eccentricity of 0. What is the specific angular momentum of the satellite?

a) 126,269.5 km^{2}/s

b) 16,269.4 km^{2}/s

c) 9,468 km^{2}/s

d) 14,121.4 km^{2}/s

View Answer

Explanation: For eccentricity e = 0,

Satellite velocity (v) = Tangential velocity = 3.156 km/s

Satellite radius (r) = 3,000 km

Specific angular momentum (h) = rv

= 3,000*3.156

= 9,468 km

^{2}/s

9. What is the percentage of visible surface area (shaded area) from the GEO satellite as shown? The surface area can be calculated as (surface area of hemisphere) x (1 – cos(latitude angle)).

a) 100%

b) 70.76%

c) 69.42%

d) 42.04%

View Answer

Explanation: Given,

Latitude angle (Φ) = 73°

Percentage of visible surface = (Visible surface area) / (Surface area of hemisphere) x 100%

= (Surface area of hemisphere x (1 – cos Φ)) / (Surface area of hemisphere)

x 100%

= (1 – cos Φ) x 100%

= (1 – cos 73°) x 100%

= 70.76%

10. What is the specific energy of a LEO satellite of specific angular momentum of 10,000 km^{2}/s?

a) -149.21 km^{2}/s^{2}

b) -912.01 km^{2}/s^{2}

c) -794.41 km^{2}/s^{2}

d) -49.25 km^{2}/s^{2}

View Answer

Explanation: Given,

The satellite is orbiting Earth, therefore,

Gravitational parameter (μ) = 398,600 km

^{3}/s

^{2}

Specific angular momentum (h) = 10,000 km

^{2}/s

Also, Low Earth Orbits (LEO) are circular, therefore

Specific energy (ε) = -(1/2)(μ/h)

^{2}

= -(1/2)(398,600/10,000)

^{2}

= -794.41 km

^{2}/s

^{2}

**Sanfoundry Global Education & Learning Series – Orbital Mechanics**.

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