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Spaceflight Mechanics [ Question No: 1 To 8 ] :
An earth satellite has the instantaneous position vector \(\vec{r}\) and velocity vector \(\vec{v}\) as given below. Here \(\hat{p}\) and \(\hat{q}\) denote the unit vectors along the \(x\) and \(y\) axes of the perifocal frame, respectively. Assume that the value of gravitational parameter is 398600 km3/s2. Which one of the following trajectories does the satellite follow?
\(\vec{r} = (8000\hat{p} + 9000\hat{q})\) km and \(\vec{v} = (-6\hat{p} + 6\hat{q})\) km/s
At a point in the trajectory of an unpowered space vehicle moving about the Earth, the altitude above the mean sea level is 600 km, and the speed with reference to a coordinate system fixed to the center of mass of the Earth is 9 km/s. Assume that the Earth is a sphere with a radius 6400 km and \(G M_{Earth} = 3.98 \times 10^{14}\) m3/s2, where, G is the universal gravitational constant and \(M_{Earth}\) is mass of the Earth. The trajectory is:
Which one of the following figures represents the qualitative variation of absolute deceleration \(\left|\frac{dV}{dt}\right|\) with altitude \(h\) (measured from the mean sea level) for a space vehicle undergoing a ballistic entry into the Earth’s atmosphere?
In an elliptic orbit around any planet, the location at which a spacecraft has the maximum angular velocity is
Equation of the trajectory of a typical space object around any planet, in polar coordinates (r, θ) (i.e. a general conic section geometry), is given as follows. (h is angular momentum, μ is gravitational parameter, e is eccentricity, r is radial distance from the planet center, θ, is angle between vectors \(\vec{e}\) and \(\vec{r}\).
The tangential velocity component 'V' of a spacecraft, which is in a circular orbit of radius 'R' around a spherical Earth (μ = GM → gravitational parameter of Earth) is given by the following expression.
A satellite is injected at an altitude of 350 km above the Earth's surface, with a velocity of 8.0 km/s parallel to the local horizon. (Earth radius=6378 km, \(\mu_{E}\) (GM=Gravitational constant × Earth mass) = 3.986×1014 m3s-2). The satellite
The Hohmann ellipse used as earth-Mars transfer orbit has
Total number of questions : 8
Number of answered questions : 0
Number of unanswered questions : 8
Spaceflight Mechanics [ Question No: 1 To 8 ] :
An earth satellite has the instantaneous position vector \(\vec{r}\) and velocity vector \(\vec{v}\) as given below. Here \(\hat{p}\) and \(\hat{q}\) denote the unit vectors along the \(x\) and \(y\) axes of the perifocal frame, respectively. Assume that the value of gravitational parameter is 398600 km3/s2. Which one of the following trajectories does the satellite follow?
\(\vec{r} = (8000\hat{p} + 9000\hat{q})\) km and \(\vec{v} = (-6\hat{p} + 6\hat{q})\) km/s
Your Answer : (Not Answered)
Correct Answer : B
The satellite follows a hyperbola.
The type of orbit follows from the sign of the specific orbital energy \(\varepsilon = v^{2}/2 - \mu/r\). With the given vectors:
\[ \begin{aligned} r &= \sqrt{8000^{2} + 9000^{2}} \approx 12041.6\ \text{km} \\ v^{2} &= 6^{2} + 6^{2} = 72\ \text{km}^{2}/\text{s}^{2} \\ \varepsilon &= 36 - \frac{398600}{12041.6} \\ &\approx 36 - 33.10 = 2.90\ \text{km}^{2}/\text{s}^{2} \end{aligned} \]
The energy is positive, so the orbit is open and hyperbolic. Equivalently, the speed \(\sqrt{72} \approx 8.49\) km/s exceeds the local escape speed \(\sqrt{2\mu/r} \approx 8.14\) km/s.
A parabola needs exactly zero energy, and a circle needs negative energy with the speed equal to \(\sqrt{\mu/r} \approx 5.75\) km/s. A straight line would need zero angular momentum, but \(h = r_p v_q - r_q v_p = 8000(6) - 9000(-6) = 102000\ \text{km}^{2}/\text{s}\), which is not zero.
At a point in the trajectory of an unpowered space vehicle moving about the Earth, the altitude above the mean sea level is 600 km, and the speed with reference to a coordinate system fixed to the center of mass of the Earth is 9 km/s. Assume that the Earth is a sphere with a radius 6400 km and \(G M_{Earth} = 3.98 \times 10^{14}\) m3/s2, where, G is the universal gravitational constant and \(M_{Earth}\) is mass of the Earth. The trajectory is:
Your Answer : (Not Answered)
Correct Answer : B
The trajectory is elliptic.
The orbit type depends on how the speed compares with the circular and escape speeds at that radius. Here \(r = 6400 + 600 = 7000\) km.
\[ \begin{aligned} v_{c} &= \sqrt{\frac{GM}{r}} = \sqrt{\frac{3.98 \times 10^{14}}{7 \times 10^{6}}} \approx 7.54\ \text{km/s} \\ v_{esc} &= \sqrt{2}\,v_{c} \approx 10.66\ \text{km/s} \end{aligned} \]
The speed of 9 km/s is above the circular speed and below the escape speed, so the specific orbital energy is negative and the vehicle is on a closed, elliptic path.
A circular orbit would need exactly 7.54 km/s, a parabolic one exactly 10.66 km/s, and a hyperbolic one more than 10.66 km/s.
Which one of the following figures represents the qualitative variation of absolute deceleration \(\left|\frac{dV}{dt}\right|\) with altitude \(h\) (measured from the mean sea level) for a space vehicle undergoing a ballistic entry into the Earth’s atmosphere?
Your Answer : (Not Answered)
Correct Answer : D
The correct variation is graph (D): deceleration is near zero high up, rises to a single peak at an intermediate altitude and falls again lower down.
In a ballistic entry, deceleration is the drag force per unit mass, proportional to \(\rho V^{2}\). High in the atmosphere the density is tiny, so the deceleration is negligible even though the vehicle is very fast. As it descends, density grows exponentially and the deceleration rises sharply. But the drag also bleeds off speed, and lower down \(V^{2}\) falls faster than \(\rho\) grows, so the deceleration passes a maximum and decreases. Allen and Eggers’ analysis shows this peak occurs at a definite altitude, with a maximum value independent of the vehicle’s ballistic coefficient.
Graphs (B) and (C) are monotonic, missing the peak. Graph (A) shows a large deceleration at the very highest altitudes, where the air is too thin to slow the vehicle, and its largest value at sea level, by which time most of the speed has gone.
In an elliptic orbit around any planet, the location at which a spacecraft has the maximum angular velocity is
Your Answer : (Not Answered)
Correct Answer : B
The angular velocity is greatest at periapsis.
Angular momentum per unit mass is conserved in orbit: \(h = r^{2}\dot{\theta}\) is constant (Kepler's second law, equal areas in equal times). So the angular velocity is \(\dot{\theta} = h/r^{2}\), largest where r is smallest, at periapsis, the point of closest approach.
Apoapsis, the farthest point, is where the angular velocity is smallest. A point 45° from periapsis or 90° from apoapsis lies farther out than periapsis, so the angular velocity there is lower.
Equation of the trajectory of a typical space object around any planet, in polar coordinates (r, θ) (i.e. a general conic section geometry), is given as follows. (h is angular momentum, μ is gravitational parameter, e is eccentricity, r is radial distance from the planet center, θ, is angle between vectors \(\vec{e}\) and \(\vec{r}\).
Your Answer : (Not Answered)
Correct Answer : C
The trajectory equation is \(r = \dfrac{(h^{2}/\mu)}{1 + e\cos\theta}\).
Solving the two-body equation of motion gives the conic \(r = p/(1 + e\cos\theta)\), with the semi-latus rectum \(p = h^{2}/\mu\) and \(\theta\) (the true anomaly) measured from the eccentricity vector, which points to periapsis. At \(\theta = 0\) the denominator is largest, giving the closest point, \(r = p/(1 + e)\), as it must; \(e = 0\) gives a circle, \(0 \lt e \lt 1\) an ellipse, \(e = 1\) a parabola and \(e \gt 1\) a hyperbola.
The form with \(1 - e\cos\theta\) would put periapsis at \(\theta = 180^{\circ}\), which is what you get by measuring the angle from the opposite direction, not from \(\vec{e}\). The forms with \(e \pm \cos\theta\) are not conics in this form: with \(e = 0\) they do not even give a circle.
The tangential velocity component 'V' of a spacecraft, which is in a circular orbit of radius 'R' around a spherical Earth (μ = GM → gravitational parameter of Earth) is given by the following expression.
Your Answer : (Not Answered)
Correct Answer : B
The velocity is \(V = \sqrt{\mu/R}\).
In a circular orbit, gravity provides exactly the centripetal force:
\[ \begin{aligned} \frac{mV^{2}}{R} &= \frac{GMm}{R^{2}} = \frac{\mu m}{R^{2}} \\ V &= \sqrt{\frac{\mu}{R}} \end{aligned} \]
\(\sqrt{\mu/2R}\) is smaller than the circular speed by a factor \(\sqrt{2}\); it equals half the escape speed \(\sqrt{2\mu/R}\). \(\dfrac{2\pi}{\sqrt{\mu}}R^{3/2}\) is the orbital period, a time, not a velocity, and the version with \(R^{2/3}\) is that period with the exponent inverted.
A satellite is injected at an altitude of 350 km above the Earth's surface, with a velocity of 8.0 km/s parallel to the local horizon. (Earth radius=6378 km, \(\mu_{E}\) (GM=Gravitational constant × Earth mass) = 3.986×1014 m3s-2). The satellite
Your Answer : (Not Answered)
Correct Answer : B
The satellite forms an elliptic orbit.
At injection the radius is \(r = 6378 + 350 = 6728\) km. Compare the injection speed with the circular and escape speeds at that radius:
\[ \begin{aligned} v_{c} &= \sqrt{\mu_{E}/r} = \sqrt{\frac{3.986 \times 10^{14}}{6.728 \times 10^{6}}} \\ &\approx 7.70\ \text{km/s} \\ v_{esc} &= \sqrt{2}\,v_{c} \approx 10.89\ \text{km/s} \end{aligned} \]
The injection speed of 8.0 km/s, horizontal, lies between the two, so the orbit is an ellipse with the injection point as perigee. Exactly 7.70 km/s would give a circle, 10.89 km/s or more would escape, and a speed below circular at this height would make injection the apogee of a lower orbit, which is how a satellite might fall back.
The Hohmann ellipse used as earth-Mars transfer orbit has
Your Answer : (Not Answered)
Correct Answer : C
The transfer orbit has apogee at Mars and perigee at earth.
A Hohmann transfer is the ellipse tangent to both circular orbits. For an outward transfer from earth to Mars, the spacecraft leaves at the point of the ellipse closest to the Sun, which lies on earth's orbit, and arrives at the farthest point, which lies on Mars's orbit. (Strictly these are perihelion and aphelion for a Sun-centred orbit; the paper uses perigee and apogee.)
Reversing them describes a Mars-to-earth transfer, and an ellipse with both apsides at the same planet would never reach the other one.
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Total number of questions : 8
Number of answered questions : 0
Number of unanswered questions : 8