This set of Spaceflight Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Perturbations due to Earth’s Oblateness”.

1. How much is the Earth’s equatorial radiusgreater than the polar radius?

a) 20 km

b) 15 km

c) 50 km

d) 21 km

View Answer

Explanation: In reality, the Earth is not an ideal spherical body, rather it is a flattened sphere with the equatorial radius of 21 km greater than the polar radius. This leads to the perturbations as there is uneven distribution of mass and density.

2. Which functions are used to represent the oblate ellipsoidal shape of the Earth?

a) Cubic Harmonics

b) Spinor Spherical Harmonics

c) Spherical Harmonics

d) Cylindrical Harmonics

View Answer

Explanation: The Earth’s mass is unevenly distributed and in order to model the planet’s surface, sphericalharmonics is used. It helps in modeling the surface of the earth and the varied gravitational field along the latitude and longitude.

3. Which of these does not make up the spherical harmonics in the disturbing potential function R(r, λ, Φ)?

a) Tesseral harmonics

b) Sectoral harmonics

c) Zonal harmonics

d) Orthogonal harmonics

View Answer

Explanation: The disturbing potential function R(r, λ, Φ) is expressed in the terms of spherical harmonics which consists of zonal harmonics (sections of latitudes), sectoral harmonics (section of longitudes) and tesseral harmonics (depend on both latitude and longitude).

4. Which of the second order harmonics functions is used to model the buldge of the Earth?

a) Tesseral coefficient

b) Sectoral coefficient

c) Zonal coefficient J_{2}

d) Orthogonal coefficient

View Answer

Explanation: The equatorial buldge of the Earth is modelled using the second order spherical harmonic function with zonal coefficient J

_{2}. This is about 1000 times larger than the other J

_{k}coefficients.

5. What is the formula of the geopotential function that is used to model the oblate Earth?

a) U(r, Φ’) = \(\frac{\mu}{r} \Big[\)1 – J_{k} \((\frac{R_E}{r})^2\) P_{k} (sin Φ’)\(\Big]\)

b) U(r, Φ’) = \(\frac{\mu}{r} \Big[\)1 – J_{2} \((\frac{R_E}{r})^2\) P_{2} (sin Φ’)\(\Big]\)

c) U(r, Φ’) = \(\frac{\mu}{r} \Big[\)1 – J_{2} \((\frac{R_E}{r})^2\) P_{2} (cos Φ’)\(\Big]\)

d) U(r, Φ’) = \(\frac{\mu}{r} \Big[\)1 – J_{k} \((\frac{R_E}{r})^2\) P_{k} (tan Φ’)\(\Big]\)

View Answer

Explanation: The general geopotential function which depends on the radius and the altitude is given by: U(r, Φ’) = \(\frac{\mu}{r} \Big[\)1 – \(\Sigma_{k=2}^∞ \)J

_{k}\((\frac{R_E}{r})^k\) P

_{k}(sin Φ’)\(\Big]\) where J

_{k}represents zonal harmonics, R

_{E}is the radius of the Earth and P

_{k}is the Legendre polynomial.

J

_{k}shows the buldges and dips on the surface of the Earth. But for the flattened earth, we use the J

_{2}model as it is 1000 times larger than the other coefficients thus making the formulas as: U(r, Φ’) = \(\frac{\mu}{r} \Big[\)1 – J

_{2}\((\frac{R_E}{r})^2\) P

_{2}(sin Φ’)\(\Big]\).

6. What is the mean nodal regression of the satellite orbiting Earth with the eccentricity of the orbit as 0.0012, inclination = 45 deg and semi-major axis = 6,500 km?

a) -2.825 × 10^{-6} rad/s

b) -5.162 × 10^{-6} rad/s

c) -1.01 × 10^{-5} rad/s

d) -1.327 × 10^{-6} rad/s

View Answer

Explanation: Given, i = 45 deg, e = 0.0012, a = 6,500 km

Mean motion n is given by: n = \(\sqrt{\frac{\mu}{a^3}} = \sqrt{\frac{398,60}{6,500^3}}\) = 0.0012 rad/s

J

_{2}= 0.0010826267 and R

_{E}= 6,378 km

\(\frac{d\bar{\Omega}}{dt} = \frac{-3nJ_2}{2(1 – e^2)^2} \Big(\frac{R_E}{a}\Big)^2\) cos i = \(\frac{-3×0.0012×0.0010826267}{2(1-0.0012^2)^2} \Big(\frac{6,378}{6,500}\Big)^2\) cos 45

\(\frac{d\bar{\Omega}}{dt}\) = -1.327 × 10

^{-6}rad/s.

7. The zonal harmonic affects the orbital plane for a polar orbit.

a) True

b) False

View Answer

Explanation: When the satellite is in a polar orbit, i.e. i = 90 deg, the zonal harmonic J

_{2}has no effect on the orientation of the orbital plane. \(\frac{d\bar{\Omega}}{dt}\) becomes zero and thus there is no change in the orbital plane. The satellite in the polar orbit faces all sunlight for the initial conditions, and after 3 months the satellite is encompassed in the Earth’s shadow.

8. For which angle of inclination does the orbit have zero secular change in ω?

a) 0 deg

b) 90 deg

c) 63.4 deg

d) 118.5 deg

View Answer

Explanation: The formula of the secular drift for ω is given by:

\(\dot{\overline{\omega_{J_2}}}\) = \(\frac{3nJ_2}{4(1 – e^2)^2} \Big(\frac{R_E}{a}\Big)^2\) (4 – 5 sin

^{2}i)

Thus, for angles of i = 63.4 deg and 116.6 deg, the secular change in ω is zero as sine component becomes 0.

9. How many times is the nodal regression for a satellite in LEO orbit greater than the satellite in a lunar orbit?

a) 6-7 times

b) 10 times

c) 12-15 times

d) 1-2 times

View Answer

Explanation: The moon is less oblate compared to the Earth with its zonal coefficient J

_{2}= 0.0002027. A satellite in a 100 km altitude orbit around the moon at an inclination of 10 deg will exhibit the nodal regression rate of \(\dot{\overline{\Omega_{J_2}}}\) = -1.18 deg/day which is about 6-7 times larger than the nodal regression for a satellite in the low-lunar orbit.

10. Which of these orbital perturbations is not a result of the Earth’s oblateness?

a) Torque at the center of Earth

b) Rotation of line of apsides

c) Change in object’s attitude dynamics

d) Nodes moves westward for direct orbits

View Answer

Explanation: The Earth is not a perfectly spherical body, rather it is oblate with poles that are flattened. Due to this there is a density variation inside the volume. The excess mass near the equator causes slight torque on the satellite about the center of the Earth. The torque causes the orbit plane to precess, and the line of the nodes moves westward for direct orbits and eastward for retrograde satellites. Earth’s oblateness also causes a rotation of the line of apsides.

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