# US Space Force Tracking Problem: Step-by-Step Solution Let's break down each part of the problem and solve it step by step. --- ## Given Data - **Miss distance, \( A = 24,000\,\text{km} \)** (distance of closest approach if no gravitational interaction) - **Closest point of approach (periapsis), \( r_p = 12,000\,\text{km} \)** - **Velocity vector on entry is colinear and in same direction as Earth's velocity around the Sun** - **Entry position vector is 135° from Earth's position vector (relative to Sun)** - **Assume Earth's sphere of influence (SOI) is much larger than miss distance** Let’s denote: - \( \mu_E \): Earth's gravitational parameter \( = 398,600\,\text{km}^3/\text{s}^2 \) - \( \mu_S \): Sun's gravitational parameter \( = 1.327 \times 10^{11}\,\text{km}^3/\text{s}^2 \) - \( r_{SOI} \): Earth's SOI radius \( \approx 924,000\,\text{km} \) - \( v_{\infty} \): Hyperbolic excess velocity (relative to Earth) --- ## a. **Semi-major Axis of the Object's Orbit w.r.t. Earth** The object's orbit about Earth is **hyperbolic** since it comes from interplanetary space and leaves Earth's SOI. For a hyperbolic trajectory: - Closest approach (periapsis) \( r_p \) - Miss distance (impact parameter or asymptote, \( b = 24,000\,\text{km} \)) - Periapsis \( r_p = 12,000\,\text{km} \) The semi-major axis for a hyperbolic orbit is negative: \[ a = -\frac{\mu_E}{v_{\infty}^2} \] But first, relate \( r_p \), \( v_{\infty} \), and \( a \): For a hyperbola: \[ r_p = a (e - 1) \] where \( e \) is the eccentricity \( (>1) \). The relationship between miss distance (impact parameter, \( b \)) and periapsis: \[ b = a \sqrt{e^2 - 1} \] But more commonly, the impact parameter with respect to a hyperbolic approach is: \[ A = a \sqrt{e^2 - 1} \] Let’s relate \( r_p \) and \( A \): \[ \frac{r_p}{A} = \frac{a(e-1)}{a\sqrt{e^2-1}} = \frac{e-1}{\sqrt{e^2-1}} \] Let \( x = e \): \[ \frac{12,000}{24,000} = \frac{x-1}{\sqrt{x^2-1}} \] \[ .5 = \frac{x-1}{\sqrt{x^2-1}} \] \[ .5 \sqrt{x^2-1} = x-1 \] \[ .25(x^2 - 1) = (x-1)^2 \] \[ .25x^2 - .25 = x^2 - 2x + 1 \] \[ .25x^2 - .25 - x^2 + 2x - 1 = \] \[ -.75x^2 + 2x - 1.25 = \] \[ .75x^2 - 2x + 1.25 = \] Quadratic formula: \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \times .75 \times 1.25}}{2 \times .75} \] \[ x = \frac{2 \pm \sqrt{4 - 3.75}}{1.5} \] \[ x = \frac{2 \pm \sqrt{.25}}{1.5} \] \[ x = \frac{2 \pm .5}{1.5} \] \[ x_1 = \frac{2.5}{1.5} = 1.\overline{6} \] \[ x_2 = \frac{1.5}{1.5} = 1 \] \( e = 1 \) is a degenerate case (parabola). So, \( e = 1.666\ldots \). **Now, solve for \( a \):** \[ r_p = a(e - 1) \implies a = \frac{r_p}{e-1} \] \[ a = \frac{12,000}{1.66667 - 1} = \frac{12,000}{.66667} \approx 18,000\,\text{km} \] ### **Final Answer (a):** \[ \boxed{a = -18,000\,\text{km}} \] (The negative sign denotes a hyperbolic orbit.) --- ## b. **Specific Mechanical Energy and Specific Angular Momentum** ### **Specific Mechanical Energy (per unit mass):** \[ \epsilon = \frac{v^2}{2} - \frac{\mu}{r} \] For a hyperbolic orbit: \[ \epsilon = \frac{\mu_E}{2|a|} \] But since \( a < \), \[ \epsilon = \frac{\mu_E}{2a} \] Plug in \( \mu_E = 398,600\,\text{km}^3/\text{s}^2 \), \( a = -18,000\,\text{km} \): \[ \epsilon = \frac{398,600}{2 \times (-18,000)} = \frac{398,600}{-36,000} \approx -11.07\,\text{km}^2/\text{s}^2 \] ### **Specific Angular Momentum:** \[ h = r_p v_p \] At periapsis, for a hyperbolic trajectory: \[ v_p = \sqrt{v_\infty^2 + \frac{2\mu_E}{r_p}} \] But \( a = -\frac{\mu_E}{v_\infty^2} \implies v_\infty = \sqrt{-\frac{\mu_E}{a}} \) \[ v_\infty = \sqrt{-\frac{398,600}{-18,000}} = \sqrt{22.144} \approx 4.71\,\text{km/s} \] Now, calculate \( v_p \): \[ v_p = \sqrt{v_\infty^2 + \frac{2\mu_E}{r_p}} \] \[ = \sqrt{(4.71)^2 + \frac{2 \times 398,600}{12,000}} \] \[ = \sqrt{22.144 + \frac{797,200}{12,000}} \] \[ = \sqrt{22.144 + 66.433} \] \[ = \sqrt{88.577} \approx 9.42\,\text{km/s} \] Now \( h = r_p v_p = 12,000 \times 9.42 = 113,040\,\text{km}^2/\text{s} \) ### **Final Answers (b):** - **Specific Mechanical Energy:** \( \boxed{-11.1\,\text{km}^2/\text{s}^2} \) - **Specific Angular Momentum:** \( \boxed{113,000\,\text{km}^2/\text{s}} \) (rounded to 3 sig figs) --- ## c. **Hyperbolic Excess Velocity (\( v_\infty \))** Already found in step (b): \[ \boxed{v_\infty = 4.71\,\text{km/s}} \] --- ## d. **Specific Mechanical Energy Relative to the Sun, and Orbit Type** When the object approaches Earth's SOI, its velocity vector is colinear and in the **same direction as Earth's velocity** (relative to Sun). So, in the Sun's frame, its velocity magnitude is the same as Earth's. - **Earth’s orbital speed:** \( v_E = \sqrt{\frac{\mu_S}{r_E}} \) - \( r_E = 1\,\text{AU} = 1.496 \times 10^8\,\text{km} \) \[ v_E = \sqrt{\frac{1.327 \times 10^{11}}{1.496 \times 10^8}} \] \[ v_E = \sqrt{887.5} \approx 29.83\,\text{km/s} \] Thus, the object’s heliocentric speed is also \( 29.83\,\text{km/s} \). Specific mechanical energy relative to the Sun: \[ \epsilon_{sun} = \frac{v^2}{2} - \frac{\mu_S}{r} \] \[ = \frac{(29.83)^2}{2} - \frac{1.327 \times 10^{11}}{1.496 \times 10^8} \] \[ = \frac{889.8}{2} - 887.5 = 444.9 - 887.5 = -442.6\,\text{km}^2/\text{s}^2 \] ### **Type of Orbit:** - Since \( \epsilon_{sun} < \), the object is on an **elliptical (bound) orbit around the Sun**. ### **Final Answers (d):** - **Specific Mechanical Energy relative to the Sun:** \( \boxed{-443\,\text{km}^2/\text{s}^2} \) - **Type of Orbit:** \( \boxed{\text{Elliptical (bound) orbit}} \) --- ## e. **Object’s Orbital Period Around the Sun** For an elliptical orbit: \[ T = 2\pi \sqrt{\frac{a_{sun}^3}{\mu_S}} \] But we need the semi-major axis (\( a_{sun} \)). For two-body orbits: \[ \epsilon_{sun} = -\frac{\mu_S}{2a_{sun}} \implies a_{sun} = -\frac{\mu_S}{2\epsilon_{sun}} \] Plug in values: \[ a_{sun} = -\frac{1.327 \times 10^{11}}{2 \times (-443)} \] \[ = \frac{1.327 \times 10^{11}}{886} \] \[ = 1.497 \times 10^8\,\text{km} \] This is essentially \( 1\,\text{AU} \). Now, calculate period: \[ T = 2\pi \sqrt{\frac{(1.497 \times 10^8)^3}{1.327 \times 10^{11}}} \] \[ (1.497 \times 10^8)^3 = 3.355 \times 10^{24} \] \[ \frac{3.355 \times 10^{24}}{1.327 \times 10^{11}} = 2.53 \times 10^{13} \] \[ \sqrt{2.53 \times 10^{13}} = 5.03 \times 10^6\,\text{s} \] \[ T = 2\pi \times 5.03 \times 10^6 = 3.16 \times 10^7\,\text{s} \] Convert to years: \[ \frac{3.16 \times 10^7}{3.154 \times 10^7} \approx 1.00\,\text{years} \] ### **Final Answer (e):** \[ \boxed{T = 1.00\,\text{year}} \] --- ## **Summary Table** | Part | Answer | |------|--------| | (a) | \( a = -18,000\,\text{km} \) (semi-major axis w.r.t. Earth) | | (b) | Specific mechanical energy: \( -11.1\,\text{km}^2/\text{s}^2 \) <br> Specific angular momentum: \( 113,000\,\text{km}^2/\text{s} \) | | (c) | Hyperbolic excess velocity: \( 4.71\,\text{km/s} \) | | (d) | Specific energy relative to Sun: \( -443\,\text{km}^2/\text{s}^2 \); **elliptical orbit** | | (e) | Orbital period around Sun: \( 1.00\,\text{year} \) | --- ###  *Hyperbolic approach (blue) and elliptical orbit (red) relative to attracting body*