Let's break this into **2 steps** as requested, with explanations and all parts addressed at each step. --- ## **Step 1: Derivations and Ratios** ### **a) Conservation of Momentum (Snowplow Model)** - **Swept-up mass:** \( M = \rho \frac{4}{3}\pi r^3 \) - **Original ejected mass:** \( M_ \) at velocity \( v_ \) - **Current shell mass:** \( M + M_ \) at velocity \( v \) **Conservation of momentum:** \[ (M + M_)v = M_ v_ \] --- ### **b) Kinetic Energies** - **Original kinetic energy \( E_ \):** \[ E_ = \frac{1}{2} M_ v_^2 \] - **Current kinetic energy \( E \):** \[ E = \frac{1}{2} (M + M_) v^2 \] --- ### **c) Ratio of Energies and Velocities** Using momentum conservation: \[ (M + M_)v = M_ v_ \implies v = \frac{M_ v_}{M + M_} \] \[ \frac{E}{E_} = \frac{\frac{1}{2} (M + M_) v^2}{\frac{1}{2} M_ v_^2} = \frac{(M + M_) v^2}{M_ v_^2} \] Substitute \( v \) from above: \[ E = \frac{1}{2}(M + M_)\left(\frac{M_ v_}{M + M_}\right)^2 = \frac{1}{2} \frac{M_^2 v_^2}{M + M_} \] So, \[ \frac{E}{E_} = \frac{M_^2 v_^2}{M_ (M + M_) v_^2} = \frac{M_}{M + M_} \] Also, \[ \frac{v}{v_} = \frac{M_}{M + M_} \] **Explanation:** - The shell slows as it sweeps up mass, conserving momentum but losing kinetic energy (lost as heat, radiation, or work on the ISM). --- #### **End of Step 1** **We have shown:** - **Momentum:** \( (M + M_)v = M_ v_ \) - **Kinetic energy ratio:** \( \frac{E}{E_} = \frac{v}{v_} = \frac{M_}{M + M_} \) --- ## **Step 2: Numerical Answers and Explanations** ### **Given Values** - \( M_ = 4M_\odot \) - \( v_ = 500\, \text{km/s} = 5\times10^8\, \text{cm/s} \) - \( v = 10\, \text{km/s} = 1\times10^6\, \text{cm/s} \) - \( \rho = 2 \times 10^{-24}\, \text{g/cm}^3 \) - \( M_\odot = 2 \times 10^{33}\, \text{g} \) --- ### **a) Original Kinetic Energy \( E_ \)** \[ E_ = \frac{1}{2} M_ v_^2 = \frac{1}{2} \times 8 \times 10^{33}\, \text{g} \times (5 \times 10^8)^2\, \text{cm}^2/\text{s}^2 \] \[ = 4 \times 10^{33} \times 2.5 \times 10^{17} = 1 \times 10^{51}\, \text{erg} \] --- ### **b) Mass \( M \) swept up for \( v = 10\, \text{km/s} \)** \[ \frac{v}{v_} = \frac{M_}{M + M_} \implies M + M_ = M_ \frac{v_}{v} \] \[ M = M_ \left(\frac{v_}{v} - 1\right) = 8 \times 10^{33} \left(\frac{5 \times 10^8}{1 \times 10^6} - 1\right) = 8 \times 10^{33} \times (500 - 1) = 8 \times 10^{33} \times 499 = 3.992 \times 10^{36}\, \text{g} \] --- ### **c) Energy Reduction Factor** \[ \frac{E}{E_} = \frac{v}{v_} = \frac{1 \times 10^6}{5 \times 10^8} = 2 \times 10^{-3} \] **Kinetic energy is reduced by a factor of 500.** **Lost energy** is radiated away or heats the swept-up ISM. --- ### **d) Radius \( r \) when \( v = 10\, \text{km/s} \)** \[ M = \rho \frac{4}{3}\pi r^3 \implies r^3 = \frac{3M}{4\pi\rho} \] \[ r^3 = \frac{3 \times 3.992 \times 10^{36}}{4\pi \times 2 \times 10^{-24}} \] \[ = \frac{1.1976 \times 10^{37}}{2.513 \times 10^{-23}} = 4.76 \times 10^{59} \] \[ r = (4.76 \times 10^{59})^{1/3} \approx 1.68 \times 10^{20}\, \text{cm} \] **Convert to parsecs:** (1 pc = \( 3.086 \times 10^{18} \) cm) \[ r = \frac{1.68 \times 10^{20}}{3.086 \times 10^{18}} \approx 54.4\, \text{pc} \] --- ### **e) Time dependence (\( v \propto r^{-3} \)), Age \( t \)** For \( v \propto r^{-3} \): \[ \frac{dr}{dt} \propto r^{-3} \implies dr \propto r^{-3} dt \implies r^3 dr \propto dt \] \[ \int_^r r'^3 dr' \propto \int_^t dt' \implies \frac{1}{4} r^4 \propto t \implies t \propto r^4 \] **If expansion were at constant speed:** \[ t = \frac{r}{v} \] **But with deceleration (\( v \propto r^{-3} \)),** \[ t = \frac{1}{4} \frac{r}{v} \] So, **age is 4 times shorter** than for constant speed. --- ## **Final Answers** ### **Summary Table** | Quantity | Value/Expression | |-----------------------|---------------------------------------------------------------------------------| | Original kinetic energy (\( E_ \)) | \( 1 \times 10^{51} \) erg | | Swept-up mass (\( M \)) | \( 3.99 \times 10^{36} \) g | | Energy reduction factor (\( E/E_ \)) | \( 2 \times 10^{-3} \) (reduced by 500x) | | Radius at \( v = 10\, \text{km/s} \) | \( 1.68 \times 10^{20} \) cm \( \approx 54.4 \) pc | | Age reduction (\( t \)) | Age is 4 times shorter than \( t = r/v \) due to deceleration | --- **Explanation (Step 2):** - The shell slows as it sweeps up ISM, losing most kinetic energy. - The lost energy goes into heating the ISM or is radiated away. - The remnant's radius when slowed to 10 km/s is about 54 pc. - The remnant is much younger than a shell expanding at constant speed. --- ## **Final Answer:** **All steps have been shown: conservation of momentum and kinetic energy loss, numerical calculations for swept-up mass, energy, and radius, and explanation of the remnant's age.**

# Solution to the Supernova Remnant Problem ## Step 1: Derivations and Ratios ### a) Conservation of Momentum (Snowplow Model) - **Swept-up mass:** \[ M = \rho \frac{4}{3}\pi r^3 \] - **Original ejected mass:** \( M_0 \) at velocity \( v_0 \) - **Current shell mass:** \( M + M_0 \) at velocity \( v \) **Conservation of momentum equation:** \[ (M + M_0)v = M_0 v_0 \] ### b) Kinetic Energies - **Original kinetic energy \( E_0 \):** \[ E_0 = \frac{1}{2} M_0 v_0^2 \] - **Current kinetic energy \( E \):** \[ E = \frac{1}{2} (M + M_0) v^2 \] ### c) Ratio of Energies and Velocities By using the conservation of momentum: \[ (M + M_0)v = M_0 v_0 \implies v = \frac{M_0 v_0}{M + M_0} \] Substituting \( v \) into the kinetic energy ratio: \[ \frac{E}{E_0} = \frac{\frac{1}{2} (M + M_0) v^2}{\frac{1}{2} M_0 v_0^2} = \frac{(M + M_0) v^2}{M_0 v_0^2} \] Substituting the expression for \( v \): \[ E = \frac{1}{2}(M + M_0)\left(\frac{M_0 v_0}{M + M_0}\right)^2 = \frac{1}{2} \frac{M_0^2 v_0^2}{M + M_0} \] Thus, we have: \[ \frac{E}{E_0} = \frac{M_0}{M + M_0} \] Also, we find: \[ \frac{v}{v_0} = \frac{M_0}{M + M_0} \] **Explanation:** - The shell slows down as it sweeps up ambient mass, conserving momentum while losing kinetic energy. --- ## Step 2: Numerical Answers and Explanations ### Given Values - \( M_0 = 4M_\odot \) where \( M_\odot = 2 \times 10^{33} \, \text{g} \) - \( v_0 = 5000 \, \text{km/s} = 5 \times 10^8 \, \text{cm/s} \) - \( v = 10 \, \text{km/s} = 1 \times 10^6 \, \text{cm/s} \) - Ambient mass density \( \rho = 2 \times 10^{-24} \, \text{g/cm}^3 \) ### a) Original Kinetic Energy \( E_0 \) \[ E_0 = \frac{1}{2} M_0 v_0^2 = \frac{1}{2} \times (8 \times 10^{33} \, \text{g}) \times (5 \times 10^8 \, \text{cm/s})^2 \] \[ = 4 \times 10^{33} \times 2.5 \times 10^{17} = 1 \times 10^{51} \, \text{erg} \] ### b) Mass \( M \) swept up for \( v = 10 \, \text{km/s} \) Using the momentum ratio: \[ \frac{v}{v_0} = \frac{M_0}{M + M_0} \implies M + M_0 = M_0 \frac{v_0}{v} \] Thus, \[ M = M_0 \left(\frac{v_0}{v} - 1\right) = 8 \times 10^{33} \, \text{g} \left(\frac{5 \times 10^8}{1 \times 10^6} - 1\right) = 8 \times 10^{33} \, \text{g} \times 499 = 3.992 \times 10^{36} \, \text{g} \] ### c) Energy Reduction Factor The energy ratio is: \[ \frac{E}{E_0} = \frac{v}{v_0} = \frac{1 \times 10^6}{5 \times 10^8} = 0.002 \] This implies the kinetic energy is reduced by a factor of \( 500 \). **Lost energy** is transformed into heat or radiation. ### d) Radius \( r \) when \( v = 10 \, \text{km/s} \) From the mass: \[ M = \rho \frac{4}{3}\pi r^3 \implies r^3 = \frac{3M}{4\pi\rho} \] Substituting for \( M \): \[ r^3 = \frac{3 \times 3.992 \times 10^{36}}{4\pi \times 2 \times 10^{-24}} = \frac{1.1976 \times 10^{37}}{2.513 \times 10^{-23}} \approx 4.76 \times 10^{59} \] Taking the cube root: \[ r \approx (4.76 \times 10^{59})^{1/3} \approx 1.68 \times 10^{20} \, \text{cm} \] **Convert to parsecs:** \[ r \approx \frac{1.68 \times 10^{20}}{3.086 \times 10^{18}} \approx 54.4 \, \text{pc} \] ### e) Time dependence and Age \( t \) From \( v \propto r^{-3} \): \[ \frac{dr}{dt} \propto r^{-3} \implies dr \propto r^{-3} dt \implies \int dr \propto \int r^{-3} dt \] This implies: \[ t \propto r^4 \] If expanded at constant speed: \[ t = \frac{r}{v} \] But with deceleration: \[ t = \frac{1}{4} \frac{r}{v} \] Thus, **the age of the remnant is 4 times shorter than if it expanded at a constant speed.** --- ## Final Answers ### Summary Table | Quantity | Value/Expression | |--------------------------------------|-------------------------------------------------| | Original kinetic energy \( E_0 \) | \( 1 \times 10^{51} \, \text{erg} \) | | Swept-up mass \( M \) | \( 3.99 \times 10^{36} \, \text{g} \) | | Energy reduction factor \( E/E_0 \) | \( 0.002 \) (reduced by factor of 500) | | Radius at \( v = 10 \, \text{km/s} \) | \( 1.68 \times 10^{20} \, \text{cm} \approx 54.4 \, \text{pc} \) | | Age reduction \( t \) | Age is 4 times shorter than \( t = \frac{r}{v} \) | ### Conclusion - All steps have been resolved accurately, showing the conservation of momentum, kinetic energy relations, numerical calculations, and implications for the supernova remnant's properties.