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

1. The two-body problem assumes both bodies as point masses.

a) True

b) False

View Answer

Explanation: True. Kinematic properties of the moving body relative to the stationary body can be calculated by assuming the bodies as point masses concentrated at their respective center of masses. Despite the bodies being rigid bodies, knowledge of point mass dynamics is all that is required. Rigid body dynamics is applied to calculate more complex data which is beyond the purpose of a two-body problem.

2. What is the area swept by a satellite in an elliptical orbit around earth at t = T/5, where T is the time period of the orbit? Given, semi-major axis and semi-minor axis of the orbit are 9,000 km and 4,000 km respectively.

a) 22.62 x 10^{6} km^{2}

b) 29.89 x 10^{6} km^{2}

c) 120.5 x 10^{3} km^{2}

d) 117.9 x 10^{4} km^{2}

View Answer

Explanation: Given,

Semi-major axis (a) = 9,000 km

Semi-minor axis (b) = 4,000 km

Area swept (at t = T/5) (A)=?

Due to similarity,

A/(T/3) = πab/T || Area of ellipse = πab

A = πab/5

= π*9,000*4,000/5

= 22.62 x 10

^{6}km

^{2}

3. What is the velocity of a satellite orbiting around moon in a circular orbit of altitude 100 km from moon’s surface. Given, standard gravitational parameter of moon is 4903 km^{3}/s^{2} and radius of moon is 1738 km.

a) 2.089 km/s

b) 3.211 km/s

c) 1.633 km/s

d) 1.015 km/s

View Answer

Explanation: Given,

Standard gravitational parameter (μ) = 4903 km

^{3}/s

^{2}

Radius of moon (R

_{M}) = 1738 km

Altitude from surface of moon (z) = 100 km

Velocity of satellite (v) = (μ/(R

_{M}+ z))

^{1/2}

= (4903/(1738+100))

^{1/2}

= 1.633 km/s

4. How long does it take for a satellite to travel between perigee and apogee? Apogee and perigee altitude from surface of the earth is 900 km and 400 km respectively. Earth’s radius and standard gravitational parameter is 6378 km and 398,600 km^{3}/s^{2} respectively.

a) 7983.13 s

b) 2931.8 s

c) 4211.11 s

d) 5863.53 s

View Answer

Explanation: Given,

Perigee distance (r

_{p}) = 6378 + 400 = 6778 km

Apogee distance (r

_{a}) = 6378 + 900 = 7278 km

Semi-major axis (a) = (r

_{p}+ r

_{a})/2

= (6778 + 7278)/2

= 7028 km

Time period of the orbit (T) = 2πa

^{3/2}/μ

^{1/2}

= 2π*7028

^{3/2}/398,600

^{1/2}

= 5863.53 s

Time taken to travel from perigee to apogee = T/2 = 5863.53/2 = 2931.8 s

5. What is the specific angular momentum of an elliptical orbit, if the flight path angle of the satellite in orbit at an instant is 15°? The satellite is travelling at a velocity of 10.7 km/s at the same instant and is parked at an altitude of 7048 km measured from earth’s center.

a) 70,234.12 km^{2}/s

b) 72,876.32 km^{2}/s

c) 61,456.23 km^{2}/s

d) 4,536.32 km^{2}/s

View Answer

Explanation: Given,

Velocity of satellite (v) = 10.7 km/s

Flight path angle (γ) =15°

Satellite distance (r) = 7048 km

Tangential velocity (v

_{t}) = vcos(γ) = 10.7*cos(15°) = 10.34 km/s

Specific angular momentum (h) = rv

_{t}= 7048*10.34 = 72,876.32 km

^{2}/s

6. What is the eccentricity of an orbit if tangential and radial velocity are 9.08 km/s and 2.01 km/s respectively. Specific angular momentum is 63,585 km^{2}/s. Standard gravitational parameter is 398,600 km^{3}/s^{2}.

a) 0.423

b) 0.127

c) 0.934

d) 0.551

View Answer

Explanation: Given,

Radial velocity (v

_{r})= 2.01 km/s

(v

_{t}) = 9.08 km/s

Specific angular momentum (h) = 63,585 km

^{2}/s

Satellite distance from earth’s center (r) = h/v

_{t}

= 63,585/9.08

= 7,002.75 km

From orbit equation,

ecos(θ) = h

^{2}/(μr) – 1

= 63,585

^{2}/(398,600*7002.75) – 1

= 0.448

esin(θ) = v

_{r}h/μ

= 2.01*63,585/398,600

= 0.321

esin(θ)/ecos(θ) = 0.321/0.448

θ = atan(0.716)

= 35.59°

Eccentricity (e) = 0.448/cos(θ)

= 0.551

7. Position, Velocity and Acceleration of a celestial body is as follows:

\(\vec{r}\) = 290\(\hat i\) + 640\(\hat j\) + 410\(\hat k\) (m)

\(\vec{v}\) = 90\(\hat i\) + 125\(\hat j\) 200\(\hat k\) (m/s)

\(\vec{a}\) = 14\(\hat i\) + 125\(\hat j\) 27\(\hat k\) (m⁄s^{2})

What is its radius of curvature?

a) 129.20 m

b) 158.12 m

c) 680.44 m

d) 456.89 m

View Answer

Explanation:

v = \(\sqrt{90^2 + 125^2 + 200^2}\) = 252.44 m/s

\(\hat u_t = \frac{\vec v}{v} = \frac{90\hat i + 125\hat j + 200\hat k}{252.44} \)= 0.357\(\hat i\) + 0.495\(\hat j\) + 0.79\(\hat k\)

a

_{t}= \(\vec{a}\).\(\hat u_t\) = (14\(\hat i\) + 125\(\hat j\) 27\(\hat k\)) . (0.357\(\hat i\) + 0.495\(\hat j\) + 0.79\(\hat k\)) = 88.203 m/s

^{2}

a = \(\sqrt{14^2 + 125^2 + 27^2}\) = 128.65 m⁄s

^{2}

\(\vec{a}\) = a

_{t}\(\hat u_t\) + a

_{n}\(\hat u_n\)

a

_{n}= \(\sqrt{a^2 – a_t^2} = \sqrt{128.65^2 – 88.203^2}\) = 93.65 m⁄s

^{2}

Radius of curvature can then be found as,

ρ = \(\frac{v^2}{a_n} = \frac{252.44^2}{93.65}\) = 680.44 m

8. Two asteroids with a mass of 6.687 x 10^{15} kg each are only acted upon by each other’s gravitational pull. If they are separated by a constant distance of 2000 km. What is the angular velocity of the line joining them? Gravitational constant is 6.67 x 10^{-11} m^{3} kg^{-1} s^{-2}.

a) 0.1221 rad/hr

b) 0.001202 rad/hr

c) 0.0027 rad/hr

d) 0.0195 rad/hr

View Answer

Explanation: Given,

Standard gravitational parameter (μ) = 398,600 km

^{3}/s

^{2}

Mass of asteroid (m) = 6.687 x 10

^{15}kg

Distance between two asteroids (d) = 2000 km

Radius of rotation (r) = d/2 || Since both the masses are equal, the point of rotation must be halfway

Gravitational Force (F

_{g}) = Gmm/d

^{2}

= (Gm

^{2})/(4r

^{2})

Centripetal Force (F

_{c}) = mω

^{2}r

We know, F

_{g}= F

_{c}

(Gm

^{2})/(4r

^{2}) = mω

^{2}r

Upon simplification,

Angular velocity (ω) = ((2Gm)/d

^{3})

^{1/2}

= ((2*6.67*10

^{-11}*6.687*10

^{15})/2000

^{3})

^{1/2}

= 3.339 x 10

^{-7}rad/s

= 0.001202 rad/hr

9. For two-body problems, satellite motion is represented as __________ motion.

a) Curvilinear

b) Rectilinear

c) Rotary

d) Brownian

View Answer

Explanation: Curvilinear motion because all orbits/trajectories have a radius of curvature. Or in other words, all orbits are conic section and therefore curved. Rectilinear motion could be useful if a satellite moved in a straight line. Rotary motion also rotates, but it rotates along its own axis. Though most celestial bodies have rotary motion, it also does not apply to point masses (which is how bodies are assumed in two body problems). Brownian motion is a random motion of particles and has application in quantum world.

10. In the following equation, what are orbital constants? r = [h^{2}/μ][1/(1 + ecosθ)].

a) h, μ and θ

b) Only μ

c) h and e

d) h, μ and e

View Answer

Explanation: h and e (specific angular momentum and eccentricity) are the orbital constants among all the letters in the orbit equation. μ (standard gravitational parameter) is a constant that is independent of the orbit and depends on the mass of the larger body. Whereas, true anomaly (θ) is not constant and is dependent on the spacecraft/satellite position relative to perigee line.

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