# Astronautics Questions and Answers – Real Orbits – Third-Body Forces – Set 2

This set of Astronautics Multiple Choice Questions & Answers (MCQs) focuses on “Real Orbits – Third-Body Forces – Set 2”.

1. What is the energy of a satellite of mass 1 ton orbiting at a distance of 36,000 kilometers away from Earth’s center, assuming only Earth’s gravitational influence (Earth’s mass = 5.972 × 1024 kg)?
a) -5.5 Giga Joules
b) -5.5 Mega Joules
c) 5.5 Giga Joules
d) 5.5 Mega Joules

Explanation: Given, r = 36,000 kilometers = 36 x 106 m; M = 5.972 × 1024 kg; m = 1000 kg. Now, energy of the satellite is given by the following equation:
E = $$\frac{1}{2}mv^2 – \frac{GMm}{r}$$
we know that
v2 = $$\frac{GM}{r}$$
E = $$\frac{1}{2}m (\frac{GM}{r}) – \frac{GMm}{r}$$
E = $$\frac{1}{2}\frac{GMm}{r} – \frac{GMm}{r}$$
E = –$$\frac{1}{2}\frac{GMm}{r}$$
substituting the given data, we have
E = –$$\frac{1}{2}\frac{(6.67 * 10^{-11})(5.972 * 10^{24})(1000)}{36 * 10^6}$$ = -5.5359 * 109 Joules = -5.5 Giga Joules

2. What is the energy of the above satellite if we also consider the Moon’s gravity? Assume the Moon lies 384,400 km from Earth and has a mass of 7.348 × 1022 kilograms).
a) -5.55 x 109 Joules
b) -6.55 x 109 Joules
c) -5.55 x 1010 Joules
d) 5.55 x 109Joules

Explanation: The energy of a satellite taking only Earth’s influence into consideration is
E = $$-\frac{1}{2} \frac{GM_Em}{r}$$
where ME is the mass of Earth. But here, we need to consider the Moon’s gravitational potential energy too. So the formula becomes
E = $$-\frac{1}{2} \frac{GM_Em}{r_E} – \frac{GM_Mm}{r_E}$$
where rE is the distance of the satellite from Earth’s center, while rM is the distance of the satellite from the Moon’s center. We have
ME = 5.972 × 1024 kg; rE = 36,000 kilometers = 3.6 x 107 meters; MM = 7.348 × 1022 kg; rM = 384,400 – 40,000 = 344400 kilometers = 3.444 x 108 meters. Substituting these values, we get
E = -5.55 x 109 Joules. Clearly, the change in energy of the satellite by taking the Moon’s influence into consideration is not that impressive.

3. Gravitational perturbations from the planets are noticeable during __________
a) close encounters with the sun
b) close encounters with Earth
c) close encounters between each external object
d) solar maximum

Explanation: Third body forces on a high altitude satellite from the planets are significant during close encounters with Earth. During such encounters, the gravity from the nearby planets is more strongly felt by Earth-orbiting satellites.

4. The close encounters of planets like Venus, Mars and Jupiter with Earth occur when _______________
a) their orbits change shape
b) their orbits change size
c) their orbits intersect with that of Earth’s
d) their angular separation with respect to Earth is minimum

Explanation: The orbits of all the planets are fixed. They come close to Earth when the angular distance from our planet is minimum. The angular separation is measured in terms of the angle between the Earth-sun line (line joining the centers of the Earth and the sun) and the planet-sun line (line joining the centers of the concerned planet and the sun).

5. Assume that at a given instant, the distance of Earth from both Mars and Venus is the same. In such a scenario, the perturbations on a high-altitude satellite are more from Mars than from Venus.
a) True
b) False

Explanation: The perturbations from Venus are more due to the fact that Venus has a greater mass than the red planet, resulting in a stronger perturbing effect from Venus than from Mars.

6. At the closest point of approach, the perturbations on a geostationary satellite of mass 1 ton due to Jupiter are more than those due to Mars (assuming Jupiter is 588 million kilometers from Earth at its nearest, while Mars is 54.6 million kilometers from Earth at its closest; also, mass of Jupiter 1.898 × 1027 kg, and mass of Mars = 6.39 × 1023 kg).
a) True
b) False

Explanation: In order to find out the magnitude of the perturbations, we calculate the gravitational force on the geostationary satellite due to Jupiter and Mars separately. Let us assume that the satellite is situated along the Earth-Jupiter (as well as Earth-Mars) line and lies in between Earth and Jupiter (or Mars). Now, mass of the satellite, m = 1000 kg.
Because the satellite is a geostationary satellite, the distance of the satellite from Jupiter, rJ = (588 million kilometers) – (radius of the satellite’s orbit) = (588 x 109 m) – (36 x 106 m) = 5.87964 x 1011 m. Also, we have the mass of Jupiter, MJ = 1.898 × 1027 kg. So the force on the satellite due to Jupiter’s gravity is given by
$$F_J = \frac{GM_Jm}{(r_J)^2}$$
Substituting all the values, we get FJ = 3.664 x 10-4 N.
Now let’s consider the case of Mars. Given, mass of the red planet, MM = 6.39 x 1023 kg. Also, distance of the satellite from Mars, rM = (54.6 million km) – (36000 km) = 5.4564 x 1010 meters. So, the force on the satellite due to Mars is given by
$$F_M = \frac{GM_Mm}{(r_M)^2}$$
Substituting the given values, we get FM = 1.4325 x 10-5 N.
Clearly, FJ>FM. The perturbing effects from Jupiter are thus greater.

7. For nearly circular orbits, the equation for the rate of change in the argument of perigee of a satellite’s orbit due to the Moon’s influence is given by
$$\dot{w} = 0.00169 \frac{[4 – 5(sin i)^2]}{n}$$
where ‘i’ is the orbital inclination and n is the number of orbit revolutions per day. Here, the unit of ω ̇ is ‘degrees per day’, and is clearly independent of the _____________
a) inclination
b) orbital altitude
c) number of revolutions
d) orbital orientation

Explanation: Here, the rate of change of the argument of perigee is dependent on the inclination and orbital altitude (the number of revolutions per day depends on the height of the satellite above the Earth). The orientation of the orbit, however, does not affect the value of ω ̇ (given a fixed inclination and orbital height).

8. For nearly circular orbits, the equation for the rate of change in the argument of perigee of a satellite’s orbit due to the sun’s influence is given by
$$\dot{w}_{sun} = 0.00077 \frac{[4 – 5(sin i)^2]}{n}$$
where ‘i’ is the orbital inclination and n is the number of orbit revolutions per day. For what value of the inclination does the precession rate of the argument of perigee due to the sun become zero?
a) 56 degrees
b) 34 degrees
c) 63 degrees
d) 89 degrees

Explanation: Observing the equation, we see that $$\dot{w}$$ sun becomes zero when the numerator becomes zero. So, [4-5(sin⁡i )2 ]=0, or [5(sin⁡i)2 ]=4, implying (sin⁡i )2=4/5, or sin⁡i=0.89, or i=sin(-1)⁡0.89, which gives the value of ‘i’ as 63.43 degrees, which is approximately 63 degrees.

9. For a circular orbit, the secular rate of change of the right ascension of the ascending node caused by the Moon is given by ___________
$$\dot{Ω} = -0.00338 \frac{cos i}{n}$$
where ‘i’ is the orbital inclination and n is the number of orbit revolutions per day. Here, the unit of $$\dot{Ω}$$ is ‘degrees per day’. What happens to the rate of change of the right ascension for a polar orbit?
a) Becomes $$\dot{Ω} = – 0.00338 \frac{1}{n}$$
b) Becomes 0
c) Becomes $$\dot{Ω} = – 0.00338$$
d) Becomes $$\dot{Ω} = – 0.00338 (cos i)$$

Explanation: For a polar orbit, the inclination ‘i’ is 90 degrees, meaning that cos(i) = 0. So $$\dot{Ω}$$ becomes 0.

10. If we consider a circular orbit, the secular rate of change of the right ascension of the ascending node caused by the Moon is given by
$$\dot{Ω}_{moon} = – 0.00338 \frac{cos i}{n}$$
while that caused by the sun is given by
$$\dot{Ω}_{sun} = – 0.00154 \frac{cos i}{n}$$
The Value of $$(\frac{\dot{Ω}_{moon}}{\dot{Ω}_{sun}})$$ is a constant
a) True
b) False

Explanation: We have, for a given inclination and orbital altitude,
$$(\frac{\dot{Ω}_{moon}}{\dot{Ω}_{sun}}) = \frac{(-0.00338 \frac{cos i}{n})}{(- 0.00154 \frac{cos i}{n})}$$
$$(\frac{\dot{Ω}_{moon}}{\dot{Ω}_{sun}}) = \frac{-0.0033}{- 0.00154} = 2.1948$$
This is clearly a constant value.

Sanfoundry Global Education & Learning Series – Astronautics.

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