Let's break the process into **3 steps** to find the **orbital elements** from the given position (\(\vec{r}\)) and velocity (\(\vec{v}\)) vectors. ### **Given:** - \(\vec{r} = -9000\hat{i} + 10000\hat{j} + 6000\hat{k}\) km - \(\vec{v} = -4\hat{i} + 3\hat{j} - 3\hat{k}\) km/s - Standard gravitational parameter for Earth, \(\mu = 398600 \ \text{km}^3/\text{s}^2\) --- ## **Step 1: Compute Key Vectors (h, e, n) and Magnitudes** ### **a) Angular Momentum Vector (\(\vec{h}\))** \[ \vec{h} = \vec{r} \times \vec{v} \] \[ \vec{h} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -9000 & 10000 & 6000 \\ -4 & 3 & -3 \\ \end{vmatrix} \] \[ h_x = (10000)(-3) - (6000)(3) = -30000 - 18000 = -48000 \] \[ h_y = -[( -9000)(-3) - (6000)(-4)] = -[27000 + 24000] = -51000 \] \[ h_z = ( -9000)(3) - (10000)(-4) = -27000 + 40000 = 13000 \] So, \[ \vec{h} = -48000\hat{i} - 51000\hat{j} + 13000\hat{k} \] \[ |\vec{h}| = \sqrt{(-48000)^2 + (-51000)^2 + (13000)^2} \approx 71,393 \ \text{km}^2/\text{s} \] **Explanation:** We computed the angular momentum vector (\(\vec{h}\)), which is perpendicular to the orbital plane and is necessary for calculating inclination and other elements. --- ## **Step 2: Compute Eccentricity Vector (\(\vec{e}\)), Node Vector (\(\vec{n}\)), and Their Magnitudes** ### **a) Node Vector (\(\vec{n}\))** \[ \vec{n} = \hat{k} \times \vec{h} = (0, 0, 1) \times (h_x, h_y, h_z) = (-h_y, h_x, 0) \] \[ \vec{n} = (51000, -48000, 0) \] \[ |\vec{n}| = \sqrt{(51000)^2 + (-48000)^2} \approx 70,923 \] ### **b) Eccentricity Vector (\(\vec{e}\))** \[ \vec{e} = \frac{1}{\mu}\left[ (\vec{v} \times \vec{h}) - \mu \frac{\vec{r}}{|\vec{r}|} \right] \] Calculate \(|\vec{r}|\): \[ |\vec{r}| = \sqrt{(-9000)^2 + 10000^2 + 6000^2} \approx 14,177 \ \text{km} \] Compute \(\vec{v} \times \vec{h}\): \[ \vec{v} \times \vec{h} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 3 & -3 \\ -48000 & -51000 & 13000 \\ \end{vmatrix} \] \[ i: (3)(13000) - (-3)(-51000) = 39000 - 153000 = -114000 \] \[ j: -[(-4)(13000) - (-3)(-48000)] = -(-52000 - 144000) = 196000 \] \[ k: (-4)(-51000) - (3)(-48000) = 204000 + 144000 = 348000 \] So, \[ \vec{v} \times \vec{h} = -114000\hat{i} + 196000\hat{j} + 348000\hat{k} \] Now, \[ \vec{e} = \frac{1}{398600}\left[ (-114000, 196000, 348000) - 398600 \frac{(-9000, 10000, 6000)}{14177} \right] \] Calculate \(398600 \frac{(-9000, 10000, 6000)}{14177} \approx ( -252,859, 280,957, 168,574 )\) So, \[ \vec{e} \approx \frac{1}{398600}\left[ (-114000 + 252859, 196000 - 280957, 348000 - 168574) \right] \] \[ \vec{e} \approx \frac{1}{398600}\left[ 138,859, -84,957, 179,426 \right] \] \[ \vec{e} \approx (0.3485, -0.2133, 0.4503) \] \[ |\vec{e}| = \sqrt{0.3485^2 + (-0.2133)^2 + 0.4503^2} \approx 0.604 \] **Explanation:** We found the node vector (\(\vec{n}\)) and the eccentricity vector (\(\vec{e}\)). The magnitude of \(\vec{e}\) gives the orbit's eccentricity. --- ## **Step 3: Calculate Orbital Elements** - **Semi-major axis (\(a\)):** \[ \text{Specific energy: } \epsilon = \frac{v^2}{2} - \frac{\mu}{r} \] \[ |\vec{v}| = \sqrt{(-4)^2 + 3^2 + (-3)^2} = \sqrt{16 + 9 + 9} = \sqrt{34} \approx 5.831 \ \text{km/s} \] \[ \epsilon = \frac{(5.831)^2}{2} - \frac{398600}{14177} = 17.023/2 - 28.113 = 8.511 - 28.113 = -19.60 \ \text{km}^2/\text{s}^2 \] \[ a = -\frac{\mu}{2\epsilon} = -\frac{398600}{2 \times -19.60} = \frac{398600}{39.20} \approx 10,170 \ \text{km} \] - **Inclination (\(i\)):** \[ i = \cos^{-1}\left(\frac{h_z}{|\vec{h}|}\right) = \cos^{-1}\left(\frac{13000}{71393}\right) \approx \cos^{-1}(0.1821) \approx 79.5^\circ \] - **Longitude of ascending node (\(\Omega\)):** \[ \Omega = \cos^{-1}\left(\frac{n_x}{|\vec{n}|}\right) = \cos^{-1}\left(\frac{51000}{70923}\right) \approx \cos^{-1}(0.719) \approx 44.1^\circ \] As \(n_y < 0\), \(\Omega = 360^\circ - 44.1^\circ = 315.9^\circ\) - **Argument of periapsis (\(\omega\)):** \[ \omega = \cos^{-1}\left(\frac{\vec{n} \cdot \vec{e}}{|\vec{n}| |\vec{e}|}\right) \] \[ \vec{n} \cdot \vec{e} = 51000 \cdot 0.3485 + (-48000) \cdot (-0.2133) = 17773.5 + 10238.4 = 28011.9 \] \[ \omega = \cos^{-1}\left(\frac{28011.9}{70923 \times 0.604}\right) = \cos^{-1}(0.658) \approx 49.0^\circ \] As \(e_z > 0\), \(\omega = 49.0^\circ\) - **True anomaly (\(\nu\)):** \[ \nu = \cos^{-1}\left(\frac{\vec{e} \cdot \vec{r}}{|\vec{e}| |\vec{r}|}\right) \] \[ \vec{e} \cdot \vec{r} = 0.3485 \cdot (-9000) + (-0.2133) \cdot 10000 + 0.4503 \cdot 6000 = -3136.5 - 2133 + 2701.8 = -2567.7 \] \[ \nu = \cos^{-1}\left(\frac{-2567.7}{0.604 \times 14177}\right) = \cos^{-1}(-0.302) \approx 107.6^\circ \] As \(\vec{r} \cdot \vec{v} < 0\), \(\nu = 360^\circ - 107.6^\circ = 252.4^\circ\) --- ## **Final Answer (Orbital Elements):** 1. **Semi-major axis (\(a\))**: **10,170 km** 2. **Eccentricity (\(e\))**: **0.604** 3. **Inclination (\(i\))**: **79.5°** 4. **Longitude of ascending node (\(\Omega\))**: **315.9°** 5. **Argument of periapsis (\(\omega\))**: **49.0°** 6. **True anomaly (\(\nu\))**: **252.4°** --- ### **Summary:** 1. **Step 1**: Calculated angular momentum vector and its magnitude. 2. **Step 2**: Found node and eccentricity vectors, and their magnitudes. 3. **Step 3**: Used these vectors to compute all 6 classical orbital elements. **Final Answer:** See above for all 6 orbital elements.