Orbital Insertion Satellite Delta-V Thruster Burn (Aerospace Engineering) Solution

An aerospace mission analyst at the UNLV Howard R. Hughes College of Engineering is calculating the terminal orbit-raising propulsion burn for a communications satellite being inserted into geostationary transfer orbit (GTO). The satellite utilizes two bi-propellant reaction control thrusters aligned along orthogonal vehicle axes.

To minimize mechanical structural fatigue and tank pressure drop, mission controllers seek to minimize a quadratic propellant energy cost metric:

To achieve orbital capture and rendezvous with the required orbital plane, the velocity components must strictly satisfy the orbital mechanics flight constraint:

1. Formulate the gradient vectors $\nabla f(x, y)$ and $\nabla g(x, y)$ in component bracket form.
2. State the vector equation of Lagrange multipliers $\nabla f = \lambda \nabla g$ and decompose it into individual scalar component equations.
3. Eliminate $\lambda$ to establish a linear relationship between the required velocity burns $x$ and $y$.
4. Substitute this relationship into the flight constraint to solve for the optimal velocity increments $(x_{\text{opt}}, y_{\text{opt}})$ and the multiplier $\lambda$.
5. Compute the minimum propellant burn cost $f_{\min}$.
6. Interpret the physical and mission engineering meaning of the Lagrange multiplier $\lambda = \frac{d f_{\text{opt}}}{d k}$.

Step-by-Step Solution:

1. Compute Gradient Vectors: State the Two-Variable Gradient Formula (defined in The Gradient Vector):

   $$\nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle, \quad \nabla g = \left\langle \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right\rangle$$
   Differentiate the objective function $f(x, y) = 3x^2 + y^2$:
   $$\nabla f(x, y) = \langle 6x, 2y \rangle = 6x\hat{i} + 2y\hat{j}$$
   Differentiate the constraint function $g(x, y) = 2x + y$:
   $$\nabla g(x, y) = \langle 2, 1 \rangle = 2\hat{i} + \hat{j}$$

2. Formulate the Lagrange Multiplier System: State the Lagrange Multiplier Vector System (defined in Lagrange Multipliers):

   $$\nabla f = \lambda \nabla g \quad \text{and} \quad g(x, y) = 14$$
   Equate scalar components:
   $$\hat{i}: 6x = 2\lambda$$
   $$\hat{j}: 2y = \lambda$$
   $$\text{Constraint}: 2x + y = 14$$

3. Eliminate $\lambda$ and Relate Variables: From the $\hat{i}$ equation:

   $$\lambda = \frac{6x}{2} = 3x$$
   From the $\hat{j}$ equation:
   $$\lambda = 2y$$
   Equating the two expressions for $\lambda$:
   $$3x = 2y \implies y = 1.5x \quad \text{or} \quad x = \frac{2}{3}y$$

4. Substitute into the Constraint: Substitute $y = 1.5x$ into $2x + y = 14$:

   $$2x + 1.5x = 14 \implies 3.5x = 14 \implies \frac{7}{2}x = 14 \implies x = 14 \cdot \frac{2}{7} = 4\text{ km/s}$$
   Now calculate $y$:
   $$y = 1.5(4) = 6\text{ km/s}$$
   Compute the multiplier $\lambda$:
   $$\lambda = 3x = 3(4) = 12 \quad (\text{or } \lambda = 2(6) = 12)$$
   The optimal thruster burn allocation is $(x_{\text{opt}}, y_{\text{opt}}) = (4\text{ km/s}, 6\text{ km/s})$ with multiplier $\lambda = 12$.

5. Compute Minimum Propellant Burn Cost: Substitute the optimal coordinates into $f(x, y)$:

   $$f(4, 6) = 3(4)^2 + (6)^2 = 3(16) + 36 = 48 + 36 = 84$$
   The minimum propellant cost metric is $f_{\min} = 84$.

6. Physical Interpretation of $\lambda$: State the Lagrange Multiplier Sensitivity Theorem (defined in Lagrange Multipliers):

   $$\lambda = \frac{d f_{\text{opt}}}{d k}$$
   • Here, $k = 14\text{ km/s}$ represents the target orbital insertion parameter.
   • The multiplier $\lambda = 12$ means that if mission control increases the orbital insertion threshold from $14\text{ km/s}$ to $15\text{ km/s}$ ($\Delta k = +1\text{ km/s}$), the minimum fuel expenditure metric will increase by approximately $\Delta f \approx \lambda \Delta k = 12(1) = 12\text{ units}$.
   • Conversely, if orbit analysts can relax the required flight velocity constraint by $0.1\text{ km/s}$ using an aerodynamic gravity assist, the required thruster burn cost will decrease by approximately $12(0.1) = 1.2\text{ units}$.