This set of Spaceflight Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Third Body Perturbations”.
1. Perturbing force due to third bodies is a non-conservative force.
a) True
b) False
View Answer
Explanation: When a celestial object is under the influence of gravitational force from a third body such as sun, moon then that perturbing force is conservative. The total energy is constant.
2. How does the presence of third-body perturb the orbit of satellites orbiting the Earth?
a) Alter the orbit due to gravitational force
b) Alter the orbit due to change in energy
c) Alter the orbit due to J2 effect
d) Alter the orbit due to J3 effect
View Answer
Explanation: The presence of third body such as sun, moon perturb the orbit of the Earth due to its gravitational force.
3. Which of these is the formula for perturbing acceleration due to lunar gravity?
a) \(\vec{a}_m\) = μm \(\Big(\frac{\vec{r}_{m-sc}}{r_{m-sc}^3} – \frac{\vec{r}_{m-E}}{r_{m-E}^3}\Big)\)
b) \(\vec{a}_m\) = -μm \(\Big(\frac{\vec{r}_{m-sc}}{r_{m-sc}^3} – \frac{\vec{r}_{m-E}}{r_{m-E}^3}\Big)\)
c) \(\vec{a}_m\) = \(\Big(\frac{\vec{r}_{m-sc}}{r_{m-sc}^3} + \frac{\vec{r}_{m-E}}{r_{m-E}^3}\Big)\)
d) \(\vec{a}_m\) = -μm \(\Big(\frac{\vec{r}_{m-sc}}{r_{m-sc}^3} + \frac{\vec{r}_{m-E}}{r_{m-E}^3}\Big)\)
View Answer
Explanation: The acceleration of spacecraft relative to Earth is given by:
\({\ddot{\vec{r}}}_{E-sc} = -\frac{G(M_E+m)}{r_{E-sc}^3}\) rE-sc – GMm \(\Big(\frac{\vec{r}_{m-sc}}{r_{m-sc}^3} – \frac{\vec{r}_{m-E}}{r_{m-E}^3}\Big)\)
Where, It is approximated that G(ME + m) ≅ GME = μ
The relative position vectors of Earth to spacecraft is given by: \(\vec{r}_{E-sc} = \vec{r}_{sc} – \vec{r}_E\)
Relative position vectors of moon to earth is given by: \(\vec{r}_{m-E} = \vec{r}_E – \vec{r}_m\)
Relative position vectors of moon to spacecraft is given by: \(\vec{r}_{m-sc} = \vec{r}_{sc} – \vec{r}_m\)
μm is the moon’s gravitational parameter
In the above formula the second term is considered as the perturbing acceleration due to lunar gravity which is the difference between lunar gravity acting on satellite and Earth.
4. Which of these is the formula for perturbing acceleration due to sun’s gravity?
a) \(\vec{a}_s\) = μs \(\Big(\frac{\vec{r}_{s-sc}}{r_{s-sc}^3} – \frac{\vec{r}_{s-E}}{r_{s-E}^3}\Big)\)
b) \(\vec{a}_s\) = -μs \(\Big(\frac{\vec{r}_{s-sc}}{r_{s-sc}^3} – \frac{\vec{r}_{s-E}}{r_{s-E}^3}\Big)\)
c) \(\vec{a}_s\) = \(\Big(\frac{\vec{r}_{s-sc}}{r_{s-sc}^3} + \frac{\vec{r}_{s-E}}{r_{s-E}^3}\Big)\)
d) \(\vec{a}_s\) = -μs \(\Big(\frac{\vec{r}_{s-sc}}{r_{s-sc}^3} + \frac{\vec{r}_{s-E}}{r_{s-E}^3}\Big)\)
View Answer
Explanation: Similar to the lunar gravitational perturbing force, sun also results in perturbation acceleration on the satellite. The formula to compute that is given by:
\(\vec{a}_s\) = -μs \(\Big(\frac{\vec{r}_{s-sc}}{r_{s-sc}^3} – \frac{\vec{r}_{s-E}}{r_{s-E}^3}\Big)\)
Where, \(\vec{r}_{s-sc}\) is the sun to spacecraft relative position vector
\(\vec{r}_{s-E}\) is sun to Earth position vector
μs is the sun’s gravitational parameter.
5. What is the total perturbing acceleration if the satellite is being perturbed by moon and sun?
a) \(\vec{a}_p = \vec{a}_m + \vec{a}_s\)
b) \(\vec{a}_p = \vec{a}_m – \vec{a}_s\)
c) \(\vec{a}_p = \vec{a}_m \vec{a}_s\)
d) \(\vec{a}_p = \frac{\vec{a}_m}{\vec{a}_s}\)
View Answer
Explanation: When a satellite orbiting Earth is under the gravitational force of sun and moon, the total perturbing acceleration \(\vec{a}_p\) is the sum of lunar and solar gravitational acceleration.
6. Moon’s perturbing acceleration is more than perturbing acceleration due to the Sun.
a) True
b) False
View Answer
Explanation: Moon’s perturbing acceleration is more than sun’s perturbing acceleration for a satellite orbiting the Earth. For LEO orbit, lunar gravitational acceleration is 1.18 × 10-6 m/s2 and solar gravitational acceleration is 5.29 × 10-7 m/s2. Similarly for GEO orbit, lunar gravitational acceleration is 8.65 × 10-6 m/s2 and solar gravitational acceleration is 3.43 × 10-7 m/s2.
7. What is the secular drift in Ω due to moon’s gravity for circular orbits?
a) \(\dot{\bar{\Omega}}_m = \frac{-0.0012}{N_{rev}}\) cos i \((\frac{deg}{day})\)
b) \(\dot{\bar{\Omega}}_m = \frac{-0.00338}{N_{rev}}\) cos i \((\frac{deg}{day})\)
c) \(\dot{\bar{\Omega}}_m = \frac{-0.12}{N_{rev}}\) sin i \((\frac{deg}{day})\)
d) \(\dot{\bar{\Omega}}_m = \frac{-0.645}{N_{rev}}\) sin i \((\frac{deg}{day})\)
View Answer
Explanation: Secular drift in Ω due to moon’s gravity for circular orbits is given by:
\(\dot{\bar{\Omega}}_m = \frac{-0.00338}{N_{rev}}\) cos i \((\frac{deg}{day})\)
Where, Nrev is the number of revolutions per day.
8. What is the secular drift in Ω due to sun’s gravity for circular orbits?
a) \(\dot{\bar{\Omega}}_s = \frac{-0.0012}{N_{rev}}\) cos i \((\frac{deg}{day})\)
b) \(\dot{\bar{\Omega}}_s = \frac{-0.00154}{N_{rev}}\) cos i \((\frac{deg}{day})\)
c) \(\dot{\bar{\Omega}}_s = \frac{-0.00384}{N_{rev}}\) sin i \((\frac{deg}{day})\)
d) \(\dot{\bar{\Omega}}_s = \frac{-0.645}{N_{rev}}\) sin i \((\frac{deg}{day})\)
View Answer
Explanation: Secular drift in Ω due to sun’s gravity for circular orbits is given by:
\(\dot{\bar{\Omega}}_s = \frac{-0.00154}{N_{rev}}\) cos i \((\frac{deg}{day})\)
Where, Nrev is the number of revolutions per day
9. For which of these orbits is the argument of perigee unaltered by third body gravity?
a) Molniya
b) LEO
c) GEO
d) Polar
View Answer
Explanation: When the inclination is 63.4 deg, the argument of perigee is unaltered by the third body gravity. For Molniya and Tundra orbits, the inclination is 63.4 deg thus it is unaffected by third body perturbations.
10. What is the secular drift in ω due to moon’s gravity for circular orbits?
a) \(\dot{\bar{\omega}}_m = \frac{0.00169}{N_{rev}}\) (4 – 5 sin2 i) \((\frac{deg}{day})\)
b) \(\dot{\bar{\omega}}_m = \frac{-0.00154}{N_{rev}}\) (4 – 5 cos2 i) \((\frac{deg}{day})\)
c) \(\dot{\bar{\omega}}_m = \frac{-0.00384}{N_{rev}}\) sin2 i \((\frac{deg}{day})\)
d) \(\dot{\bar{\omega}}_m = \frac{-0.645}{N_{rev}}\) sin i \((\frac{deg}{day})\)
View Answer
Explanation: Secular drift in ω due to moon’s gravity for circular orbits is given by:
\(\dot{\bar{\omega}}_s = \frac{0.00169}{N_{rev}}\) (4 – 5 sin2 i) \((\frac{deg}{day})\)
Where, Nrev is the number of revolutions per day.
11. What is the secular drift in ω due to sun’s gravity for circular orbits?
a) \(\dot{\bar{\omega}}_s = \frac{0.00169}{N_{rev}}\) (4 – 5 sin2 i) \((\frac{deg}{day})\)
b) \(\dot{\bar{\omega}}_s = \frac{0.00077}{N_{rev}}\) (4 – 5 sin2 i) \((\frac{deg}{day})\)
c) \(\dot{\bar{\omega}}_s = \frac{-0.00384}{N_{rev}}\) sin2 i \((\frac{deg}{day})\)
d) \(\dot{\bar{\omega}}_s = \frac{-0.645}{N_{rev}}\) cos i \((\frac{deg}{day})\)
View Answer
Explanation: Secular drift in ω due to sun’s gravity for circular orbits is given by:
\(\dot{\bar{\omega}}_s = \frac{0.00077}{N_{rev}}\) (4 – 5 sin2 i) \((\frac{deg}{day})\)
Where, Nrev is the number of revolutions per day.
12. What is the gravity-induced secular drift is in Ω caused by lunar gravity for a satellite orbiting Earth with semi-major axis of 7,500 km, eccentricity of 0.25 and inclination of 35 degrees?
a) -0.001245 deg/day
b) -0.0000154 deg/day
c) -0.005412 deg/day
d) -0.0002077 deg/day
View Answer
Explanation: Given, a = 7,500km, e = 0.25, i = 35 deg
In order to compute the number of revolutions per day, we first need to find the orbital period.
Tperiod = \(\frac{2\pi}{\sqrt{\mu}} a^{3/2} = \frac{2\pi}{\sqrt{398,600}}\) 7,5003/2 = 6,464 sec ≅ 1.8 hours
Nrev = \(\frac{24}{1.8}\) = 13.33
Nodal drift rates due to lunar gravity is:
\(\dot{\bar{\Omega}}_m = \frac{-0.00338}{N_{rev}}\) cos i \((\frac{deg}{day})\) = \(\frac{-0.00338}{13.33}\) cos 35 = -0.0002077 deg/day.
Sanfoundry Global Education & Learning Series – Spaceflight Mechanics.
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