Questions & explanations
1. Calculate the total delta-v required for a Hohmann transfer from a 200 km altitude circular orbit around Earth to a 35,786 km altitude geostationary orbit. Assume Earth radius 6371 km, gravitational parameter μ = 3.986×10^5 km³/s².
First, radii: r1 = 6371+200 = 6571 km, r2 = 6371+35786 = 42157 km. Transfer orbit semi-major axis a = (r1+r2)/2 = (6571+42157)/2 = 24364 km. Velocity in initial circular orbit: v1 = sqrt(μ/r1) = sqrt(3.986e5/6571) ≈ 7.784 km/s. Velocity at perigee of transfer: vp = sqrt(μ*(2/r1 - 1/a)) = sqrt(3.986e5*(2/6571 - 1/24364)) = sqrt(3.986e5*(0.0003044 - 0.00004104)) = sqrt(3.986e5*0.0002634) = sqrt(105.0) ≈ 10.247 km/s. Delta-v1 = vp - v1 = 10.247 - 7.784 = 2.463 km/s. At apogee: va = sqrt(μ*(2/r2 - 1/a)) = sqrt(3.986e5*(2/42157 - 1/24364)) = sqrt(3.986e5*(0.00004745 - 0.00004104)) = sqrt(3.986e5*0.00000641) = sqrt(2.555) ≈ 1.598 km/s. Circular velocity at r2: v2 = sqrt(μ/r2) = sqrt(3.986e5/42157) ≈ 3.075 km/s. Delta-v2 = v2 - va = 3.075 - 1.598 = 1.477 km/s. Total delta-v = 2.463+1.477 = 3.940 km/s.
2. Explain how you would use patched conic approximation to plan a transfer from Earth to Jupiter. List the main segments.
First, consider the Earth departure phase: the spacecraft is in a parking orbit around Earth, then performs a burn to enter a hyperbolic escape trajectory relative to Earth. The excess velocity at Earth's sphere of influence becomes the heliocentric velocity. Second, the heliocentric transfer: the spacecraft follows an elliptical orbit around the Sun from Earth's orbit to Jupiter's orbit. This is typically a Hohmann-like transfer if timing allows. Third, Jupiter arrival: as the spacecraft approaches Jupiter, it enters Jupiter's sphere of influence, where a hyperbolic approach trajectory is used. A burn at perijove (closest approach to Jupiter) captures it into orbit. The patched conic approximation treats each segment independently, matching velocities at the boundaries.
3. What is the patched conic approximation used for in interplanetary missions?
The patched conic approximation simplifies interplanetary trajectory design by dividing the mission into segments, each dominated by one celestial body's gravity. For example, a Mars mission starts with an Earth-centered hyperbolic escape, then a heliocentric (Sun-centered) transfer ellipse, and finally a Mars-centered hyperbolic capture. At each segment, the spacecraft's path is approximated as a conic section (ellipse, parabola, or hyperbola) around the dominant body. The segments are 'patched' together at the sphere of influence boundaries, where the gravitational influence of one body becomes negligible compared to another. This method avoids solving the complex three-body problem and gives good preliminary results for delta-v and timing.
4. Compare a Hohmann transfer to a bi-elliptic transfer for moving from a low orbit to a very high orbit. Which one requires less delta-v and why?
For a very large increase in orbital radius (say, from LEO to beyond GEO), a bi-elliptic transfer can sometimes require less total delta-v than a Hohmann transfer. A bi-elliptic transfer uses three burns: first to raise apogee to a very high intermediate point, then a second burn at that apogee to raise perigee to the final orbit, and a third burn to circularize. The trade-off is that the bi-elliptic transfer takes much longer time. The delta-v saving occurs because the second burn at high altitude is more efficient due to the lower orbital speed. However, for typical transfers (radius ratio less than about 11.94), Hohmann is more efficient. For example, Earth to Mars (ratio ~1.5) Hohmann is best.
5. Compare the accuracy of patched conic approximation with a full three-body simulation. When is it acceptable to use the approximation?
Patched conic approximation is less accurate than a full three-body simulation because it ignores the continuous gravitational influence of multiple bodies. For example, during the heliocentric transfer, the Moon's gravity might slightly perturb the trajectory, but the approximation ignores that. However, it is acceptable for preliminary mission design and delta-v estimates because errors are typically small (a few percent) for most interplanetary transfers. It is also much faster and simpler to compute. Full three-body simulations are used for final trajectory design and navigation, where high precision is needed. So patched conic is a good starting point for feasibility studies and education.
6. Why does the Moon lie inside Earth's sphere of influence, and how does that affect lunar mission planning?
The Moon orbits at about 384,400 km from Earth, which is less than Earth's SOI radius of about 923,000 km. So the Moon is inside Earth's SOI, meaning Earth's gravity dominates over the Sun's in that region. For lunar missions, the patched conic approximation treats the spacecraft as moving under Earth's gravity until it enters the Moon's own SOI (about 66,000 km from the Moon). This simplifies the trajectory: first, an Earth-centered transfer orbit to the Moon's vicinity, then a Moon-centered approach. Without the SOI concept, we would need to solve the three-body problem of Earth, Moon, and spacecraft continuously. The SOI allows us to break the problem into two separate two-body problems.
7. What is the sphere of influence (SOI) of a planet, and why is it important in orbital mechanics?
The sphere of influence (SOI) is a region around a celestial body where its gravity dominates over the Sun's gravity. For a planet, the SOI radius is roughly the distance from the planet where the planet's gravitational pull equals the Sun's tidal force. It is important because it defines the boundary where we switch from a heliocentric (Sun-centered) to a planet-centered reference frame in the patched conic approximation. Inside the SOI, the spacecraft's motion is primarily governed by the planet's gravity; outside, the Sun's gravity dominates. This concept simplifies trajectory calculations by allowing us to treat the spacecraft as moving under the influence of only one body at a time.
8. How do you model the aerodynamic forces on a vibrating turbine blade for flutter analysis?
We use computational fluid dynamics (CFD) to solve the unsteady flow around the blade as it vibrates. The blade motion is prescribed as a sinusoidal oscillation at a given frequency and mode shape. The CFD code calculates the unsteady pressure on the blade surface, which is then integrated to get aerodynamic forces. These forces are used to compute the aerodynamic damping or work per cycle. If the work is positive, the blade gains energy from the flow and flutter may occur. The model must capture the phase between blade motion and pressure response accurately. Simplified methods like the influence coefficient method or traveling wave mode approach are also used for efficiency.
9. Compare the thrust of an NTR to a chemical rocket. Which one typically produces more thrust?
Chemical rockets typically produce much higher thrust than NTRs for the same engine size. For example, the Space Shuttle main engine produced about 2.2 million Newtons of thrust, while an NTR of similar size might produce only 100,000-300,000 Newtons. The reason is that chemical rockets burn fuel and oxidizer, releasing energy rapidly and creating high-pressure exhaust. NTRs rely on heating hydrogen, which has lower density and pressure. However, NTRs have higher specific impulse (about 900 seconds vs 450 seconds for chemical rockets), meaning they use propellant more efficiently. So NTRs are better for long-duration missions where fuel efficiency matters more than raw thrust.
10. What is a Hohmann transfer orbit and what makes it the minimum-energy transfer between two circular coplanar orbits?
A Hohmann transfer orbit is an elliptical orbit that connects two circular orbits around the same central body. It is the minimum-energy transfer because it requires the least total delta-v (change in velocity) for a given change in orbital radius. The transfer orbit is tangent to both the inner and outer circular orbits at its periapsis and apoapsis, respectively. For example, to go from a low Earth orbit (LEO) to geostationary orbit (GEO), you first burn at perigee to raise the apogee to GEO altitude, then burn again at apogee to circularize. The two burns are tangential, meaning no angle change, which minimizes fuel use. Any other transfer path would require more energy.
11. What is the role of the reduced frequency in turbomachinery flutter?
Reduced frequency is a dimensionless number that compares the blade vibration frequency to the flow convection time. It is defined as k = ω c / V, where ω is vibration frequency, c is blade chord, and V is flow velocity. High reduced frequency means the blade oscillates many times while the flow travels across the chord, leading to quasi-steady aerodynamics. Low reduced frequency means the flow adjusts quickly relative to blade motion, and unsteady effects are important. Flutter often occurs at low reduced frequencies where aerodynamic damping is sensitive. Engineers use reduced frequency to categorize flutter types and to scale model test results to full-scale conditions.
12. A structural element has σ_x = 100 MPa, σ_y = -40 MPa, τ_xy = 30 MPa. Use Mohr's circle to find the normal and shear stresses on a plane rotated 30° counterclockwise from the x-axis.
Center C = (100-40)/2 = 30 MPa. Radius R = sqrt(((100-(-40))/2)² + 30²) = sqrt(70²+30²)= sqrt(4900+900)= sqrt(5800)=76.16 MPa. The angle from the x-axis to the plane is 30°, so on Mohr's circle, we rotate by 2θ = 60°. The point on the circle at angle 2θ from the x-axis (which is at σ_x, τ_xy) gives the stresses. The x-axis point is at (100, 30). The angle of the x-point from the center is: tan(α) = τ_xy/(σ_x-C) = 30/(100-30)=30/70=0.4286, α=23.2°. So the new point is at angle α+60°=83.2° from the horizontal. Then σ_n = C + R cos(83.2°) = 30 + 76.16*0.118 = 30+8.99=38.99 MPa. τ_n = R sin(83.2°) = 76.16*0.993 = 75.63 MPa. So normal stress ≈ 39 MPa, shear stress ≈ 75.6 MPa.