| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The escape velocity from the Earth can be calculated using the formula: $$v_e = \sqrt{\frac{2GM}{r}}$$ where $v_e$ is the escape velocity, $G$ is the gravitational constant ($6.67 \times 10^{-11} \text{ m}^3\text{kg}^{-1}\text{s}^{-2}$), $M$ is the mass of the Earth ($5.97 \times 10^{24} \text{ kg}$), and $r$ is the radius of the Earth ($6.38 \times 10^6 \text{ m}$). Plugging in the values given in the problem, we get: $$v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.38 \times 10^6}}$$ Simplifying this expression, we get: $$v_e = 1.1186 \times 10^4 \text{ m/s}$$ Therefore, the minimum muzzle speed that would allow a shell to escape from the Earth completely is approximately $1.1186 \times 10^4 \text{ m/s}$, or $1.1 \times 10^4 \text{ m/s}$ when rounded to the nearest tenth, which is the required unit. Thus, the answer is $\boxed{1.1}$. |
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| deepseekr1 | To find the minimum muzzle speed (escape speed) required for a shell to escape Earth's gravity, we use the formula derived from equating kinetic energy to gravitational potential energy: \[ v = \sqrt{\frac{2 G M_E}{R_E}} \] where: - \( G \) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2\)), - \( M_E \) is the mass of the Earth (\(5.97 \times 10^{24} \, \text{kg}\)), - \( R_E \) is the radius of the Earth (\(6.38 \times 10^6 \, \text{m}\)). First, calculate the numerator: \[ 2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24} = 7.974756 \times 10^{14} \] Next, divide by the Earth's radius: \[ \frac{7.974756 \times 10^{14}}{6.38 \times 10^6} = 1.24902 \times 10^8 \] Then, take the square root: \[ \sqrt{1.24902 \times 10^8} \approx 1.1176 \times 10^4 \, \text{m/s} \] Rounding to three significant figures (since the given values for \( R_E \) and \( M_E \) have three significant figures), we get: \[ 1.12 \times 10^4 \, \text{m/s} \] Thus, the minimum muzzle speed required for the shell to escape Earth's gravity is \(\boxed{1.12} \times 10^4 \, \text{m/s}\). |
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| gpto1 | To find the minimum muzzle speed required for a shell to escape from Earth completely (the escape speed), we use the formula for escape velocity: \[ v_{\text{esc}} = \sqrt{\dfrac{2GM_E}{R_E}} \] Where: - \( G \) is the gravitational constant (\(6.67430 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2\)), - \( M_E \) is the mass of the Earth (\(5.97 \times 10^{24} \, \text{kg}\)), - \( R_E \) is the radius of the Earth (\(6.38 \times 10^6 \, \text{m}\)). **Step 1: Calculate the numerator \( 2GM_E \)** \[ 2GM_E = 2 \times (6.67430 \times 10^{-11}) \times (5.97 \times 10^{24}) \] \[ 2GM_E = 2 \times 6.67430 \times 5.97 \times 10^{-11 + 24} \] \[ 2GM_E = 2 \times 6.67430 \times 5.97 \times 10^{13} \] First, multiply \(6.67430 \times 5.97\): \[ 6.67430 \times 5.97 = 39.845571 \] Then multiply by 2: \[ 2 \times 39.845571 = 79.691142 \] So, \[ 2GM_E = 79.691142 \times 10^{13} \] **Step 2: Calculate the escape velocity squared \( v_{\text{esc}}^2 \)** \[ v_{\text{esc}}^2 = \dfrac{2GM_E}{R_E} = \dfrac{79.691142 \times 10^{13}}{6.38 \times 10^6} \] Divide the exponents: \[ 10^{13} / 10^6 = 10^{7} \] Now, divide the numbers: \[ \dfrac{79.691142}{6.38} \approx 12.5 \] So, \[ v_{\text{esc}}^2 \approx 12.5 \times 10^{7} \] **Step 3: Calculate the escape velocity \( v_{\text{esc}} \)** \[ v_{\text{esc}} = \sqrt{12.5 \times 10^{7}} = \sqrt{1.25 \times 10^{8}} \] \[ v_{\text{esc}} = \sqrt{1.25} \times \sqrt{10^{8}} = \sqrt{1.25} \times 10^4 \] Calculate \( \sqrt{1.25} \): \[ \sqrt{1.25} = \sqrt{\dfrac{5}{4}} = \dfrac{\sqrt{5}}{2} \approx \dfrac{2.23607}{2} = 1.11803 \] Therefore, \[ v_{\text{esc}} = 1.11803 \times 10^4 \, \text{m/s} \] **Answer in units of \(10^4 \, \text{m/s}\):** \[ v_{\text{esc}} = 1.12 \times 10^4 \, \text{m/s} \] **Final Answer:** \(\boxed{1.12}\) |
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| GPT-4o | To determine the minimum muzzle speed required for a shell to escape Earth's gravitational field, we need to calculate the escape velocity. The escape velocity (\(v_{esc}\)) is the speed at which an object must travel to break free from a planet's gravitational influence without further propulsion. The formula for escape velocity is derived from the conservation of energy principles and is given by: \[ v_{esc} = \sqrt{\frac{2GM}{R}} \] where: - \(G\) is the gravitational constant, \(6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\), - \(M\) is the mass of the Earth, \(5.97 \times 10^{24} \, \text{kg}\), - \(R\) is the radius of the Earth, \(6.38 \times 10^6 \, \text{m}\). Let's plug in these values to find the escape velocity: \[ v_{esc} = \sqrt{\frac{2 \cdot 6.674 \times 10^{-11} \cdot 5.97 \times 10^{24}}{6.38 \times 10^6}} \] First, calculate the numerator inside the square root: \[ 2 \cdot 6.674 \times 10^{-11} \cdot 5.97 \times 10^{24} = 7.973 \times 10^{14} \] Now, divide by the radius of the Earth: \[ \frac{7.973 \times 10^{14}}{6.38 \times 10^6} \approx 1.250 \times 10^8 \] Finally, take the square root of the result: \[ v_{esc} = \sqrt{1.250 \times 10^8} \approx 1.118 \times 10^4 \, \text{m/s} \] Therefore, the minimum muzzle speed required to escape Earth's gravitational field is approximately \( 1.118 \times 10^4 \, \text{m/s} \). Converting this to the specified unit (10^4 m/s): \[ v_{esc} \approx 1.12 \times 10^4 \, \text{m/s} \] Thus, the minimum muzzle speed that would allow the shell to escape from the Earth completely is \(1.12 \times 10^4 \, \text{m/s}\). |
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| llama3.1-70B-Instruct | ## Step 1: Understand the problem and the concept of escape speed The problem asks for the minimum muzzle speed required for a shell to escape the Earth's gravitational pull completely. The escape speed is the speed at which an object must be traveling to break free from the gravitational pull of a celestial body, in this case, the Earth. ## Step 2: Recall the formula for escape speed The formula for escape speed is given by $v_{esc} = \sqrt{\frac{2Gm}{r}}$, where $G$ is the gravitational constant, $m$ is the mass of the celestial body (in this case, the Earth), and $r$ is the radius of the celestial body. ## Step 3: Plug in the values for the Earth's mass and radius Given that the Earth's mass $m_E = 5.97 \times 10^{24} kg$ and the Earth's radius $R_E = 6.38 \times 10^6 m$, we can substitute these values into the formula. ## Step 4: Use the gravitational constant The gravitational constant $G = 6.674 \times 10^{-11} N \cdot m^2 / kg^2$. We will use this value in the calculation. ## Step 5: Calculate the escape speed Substituting the given values into the formula, we get $v_{esc} = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6.38 \times 10^6}}$. ## Step 6: Perform the calculation $v_{esc} = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6.38 \times 10^6}} = \sqrt{\frac{79.89 \times 10^{13}}{6.38 \times 10^6}} = \sqrt{12.53 \times 10^7} = 3.54 \times 10^4 m/s$. The final answer is: $\boxed{3.54}$ |
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| llama3.1-405B-Instruct | To find the minimum muzzle speed that would allow a shell to escape from the Earth completely, we need to calculate the escape velocity from the Earth's surface. The escape velocity is the speed at which an object must be traveling to break free from the gravitational pull of a celestial body, in this case, the Earth. The formula for escape velocity is given by: v = sqrt(2 \* G \* M / r) where: * v is the escape velocity * G is the gravitational constant (6.67408e-11 N\*m^2/kg^2) * M is the mass of the Earth (5.97 x 10^24 kg) * r is the radius of the Earth (6.38 x 10^6 m) Plugging in the values, we get: v = sqrt(2 \* 6.67408e-11 \* 5.97 x 10^24 / 6.38 x 10^6) = sqrt(1.18 x 10^8) = 1.08 x 10^4 m/s Therefore, the minimum muzzle speed required for a shell to escape from the Earth completely is approximately 1.08 x 10^4 m/s. |
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| DeepSeek-Coder-V2 | To find the minimum muzzle speed (escape speed) that would allow a shell to escape from the Earth completely, we can use the formula for escape speed: \[ v_e = \sqrt{\frac{2GM}{R}} \] where: - \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \) - \( M \) is the mass of the Earth, \( 5.97 \times 10^{24} \, \text{kg} \) - \( R \) is the radius of the Earth, \( 6.38 \times 10^6 \, \text{m} \) Substituting the given values into the formula: \[ v_e = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \times 5.97 \times 10^{24} \, \text{kg}}{6.38 \times 10^6 \, \text{m}}} \] First, calculate the numerator: \[ 2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24} = 7.98 \times 10^{14} \, \text{m}^3 \text{s}^{-2} \] Next, divide by the radius of the Earth: \[ \frac{7.98 \times 10^{14}}{6.38 \times 10^6} = 1.25 \times 10^8 \, \text{m}^2 \text{s}^{-2} \] Finally, take the square root to find the escape speed: \[ v_e = \sqrt{1.25 \times 10^8} = 1.12 \times 10^4 \, \text{m/s} \] Thus, the minimum muzzle speed required to escape the Earth completely is: \[ \boxed{1.12} \] |
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| Qwen2-72B-Instruct | The escape velocity from the Earth can be calculated using the formula: $$v_e = \sqrt{\frac{2GM}{r}}$$ where $v_e$ is the escape velocity, $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the radius of the Earth. Plugging in the values given in the problem, we get: $$v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.38 \times 10^6}}$$ Simplifying this expression, we get: $$v_e = 1.1186 \times 10^4 \text{ m/s}$$ Therefore, the minimum muzzle speed that would allow a shell to escape from the Earth completely is approximately $1.1186 \times 10^4 \text{ m/s}$, or $1.1186 \times 10^2 \text{ km/s}$, which is equivalent to $1.1186 \times 10^1 \text{ km/s}$ when expressed in units of $10^4 \text{ m/s}$. However, since the answer is required in units of $10^4 \text{ m/s}$, we can round this value to the nearest tenth and report it as: $$v_e = 1.1 \times 10^4 \text{ m/s}$$ Therefore, the answer is $\boxed{1.1}$. |
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| baichuan4 | To find the escape speed from Earth, we can use the formula for the escape velocity (ve) from the surface of a celestial body: \[ v_e = \sqrt{\frac{2GM}{R}} \] where: - G is the gravitational constant, - M is the mass of the Earth, - R is the radius of the Earth. Given: - \( R_E = 6.38 \times 10^6 \, \text{m} \), - \( M_E = 5.97 \times 10^{24} \, \text{kg} \). The gravitational constant (G) is approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \). Plugging in the values: \[ v_e = \sqrt{\frac{2 \times (6.674 \times 10^{-11}) \times (5.97 \times 10^{24})}{6.38 \times 10^6}} \] \[ v_e = \sqrt{1.118 \times 10^{14}} \] \[ v_e = 3.34 \times 10^6 \, \text{m/s} \] Converting to the units required (\( 10^4 \, \text{m/s} \)): \[ v_e = \frac{3.34 \times 10^6}{10^4} \] \[ v_e = 334 \times 10^4 \, \text{m/s} \] So, the minimum muzzle speed (escape speed) to allow a shell to completely escape from Earth's gravity is approximately \( 334 \times 10^4 \, \text{m/s} \). |
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| doubao | The escape speed $v_e$ is given by the formula $v_e = \sqrt{\frac{2GM}{R}}$, where $G$ is the gravitational constant ($6.67 \times 10^{-11} \text{ N m}^2 \text{/kg}^2$), $M$ is the mass of the earth, and $R$ is the radius of the earth. Substituting the given values, we have: $v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.38 \times 10^6}} \approx 11.2 \times 10^3 \text{ m/s} = 11.2 \text{ 10}^4 \text{ m/s}$. So the answer is 11.2. |
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| minimax | To find the escape velocity from Earth, we can use the formula for escape speed, which is derived from equating the kinetic energy needed to overcome the gravitational potential energy of the Earth. The escape speed \( v_{\text{esc}} \) is given by: \[ v_{\text{esc}} = \sqrt{\frac{2GM_E}{R_E}} \] where: - \( G \) is the gravitational constant, \( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), - \( M_E \) is the mass of the Earth, \( M_E = 5.97 \times 10^{24} \, \text{kg} \), - \( R_E \) is the radius of the Earth, \( R_E = 6.38 \times 10^6 \, \text{m} \). Plugging in the values: \[ v_{\text{esc}} = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6.38 \times 10^6}} \] \[ v_{\text{esc}} = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6.38 \times 10^6}} \] \[ v_{\text{esc}} = \sqrt{\frac{7.988 \times 10^{14}}{6.38 \times 10^6}} \] \[ v_{\text{esc}} = \sqrt{1.252 \times 10^8} \] \[ v_{\text{esc}} = 1.119 \times 10^4 \, \text{m/s} \] So the minimum muzzle speed that would allow a shell to escape from the Earth completely is approximately \( 1.119 \times 10^4 \, \text{m/s} \), or in the units requested (10^4 m/s), it is 1.119 units. |
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| kimi | To find the minimum muzzle speed required for a shell to escape Earth's gravitational pull, we need to calculate the escape velocity. The escape velocity is the minimum speed needed to break free from the gravitational attraction of a celestial body without further propulsion. The formula for escape velocity ($v_e$) from a celestial body is given by: \[ v_e = \sqrt{\frac{2GM}{R}} \] where: - \( G \) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\)) - \( M \) is the mass of the Earth - \( R \) is the radius of the Earth Given the values: - \( R_E = 6.38 \times 10^6 \, \text{m} \) - \( m_E = 5.97 \times 10^{24} \, \text{kg} \) We can plug these values into the formula: \[ v_e = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \times 5.97 \times 10^{24} \, \text{kg}}{6.38 \times 10^6 \, \text{m}}} \] \[ v_e = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6.38 \times 10^6}} \] \[ v_e = \sqrt{\frac{2 \times 6.674 \times 5.97 \times 10^{13}}{6.38}} \] \[ v_e = \sqrt{\frac{79.86 \times 10^{13}}{6.38}} \] \[ v_e = \sqrt{12.52 \times 10^{13}} \] \[ v_e = \sqrt{12.52} \times 10^{6.5} \] \[ v_e \approx 3.54 \times 10^{6.5} \, \text{m/s} \] Since \( 10^{0.5} \approx 3.16 \), we can approximate \( 10^{6.5} \) as \( 10^6 \times 10^{0.5} \approx 3.16 \times 10^6 \, \text{m/s} \). So, the escape velocity is approximately: \[ v_e \approx 3.54 \times 3.16 \, \text{m/s} \] \[ v_e \approx 11.13 \times 10^3 \, \text{m/s} \] Thus, the minimum muzzle speed required for the shell to escape Earth's gravity is approximately \( 11.13 \times 10^3 \, \text{m/s} \) or \( 11.13 \, \text{km/s} \). |
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| wenxinyiyan | To calculate the escape velocity from the Earth, we use the formula for escape velocity: $$ v_{\text{escape}} = \sqrt{\frac{2GM}{R}} $$ where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $R$ is the radius of the Earth. Given the values for the Earth's mass and radius: $$ m_E = 5.97 \times 10^{24} \, \text{kg}, \quad R_E = 6.38 \times 10^6 \, \text{m} $$ and the gravitational constant: $$ G = 6.674 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2} $$ we can substitute these into the escape velocity formula: $$ v_{\text{escape}} = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2} \times 5.97 \times 10^{24} \, \text{kg}}{6.38 \times 10^6 \, \text{m}}} $$ $$ v_{\text{escape}} = \sqrt{1.23 \times 10^8 \, \text{m}^2\text{s}^{-2}} $$ $$ v_{\text{escape}} = 1.11 \times 10^4 \, \text{m/s} $$ Therefore, the minimum muzzle speed that would allow a shell to escape from the Earth completely is $1.11 \times 10^4 \, \text{m/s}$. |
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| zhipu | To find the escape speed from the Earth, we can use the concept of gravitational potential energy and kinetic energy. The escape speed is the minimum speed needed for an object to escape the gravitational pull of the Earth without any further propulsion. The total energy (kinetic + potential) of the object at the surface of the Earth should be zero for it to just escape the gravitational field. This means that the kinetic energy given to the object should be equal to the gravitational potential energy it needs to overcome. The gravitational potential energy (U) at the surface of the Earth is given by: \[ U = -\frac{G m_E m}{R_E} \] where: - \( G \) is the gravitational constant, \( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \), - \( m_E \) is the mass of the Earth, \( 5.97 \times 10^{24} \, \text{kg} \), - \( m \) is the mass of the object (shell), - \( R_E \) is the radius of the Earth, \( 6.38 \times 10^6 \, \text{m} \). The kinetic energy (K) of the object is given by: \[ K = \frac{1}{2} m v^2 \] where \( v \) is the escape speed. For the object to escape, the kinetic energy must equal the absolute value of the gravitational potential energy: \[ \frac{1}{2} m v^2 = \frac{G m_E m}{R_E} \] We can cancel the mass \( m \) of the object from both sides: \[ \frac{1}{2} v^2 = \frac{G m_E}{R_E} \] Solving for \( v \): \[ v^2 = \frac{2 G m_E}{R_E} \] \[ v = \sqrt{\frac{2 G m_E}{R_E}} \] Now we can plug in the given values: \[ v = \sqrt{\frac{2 \times 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \times 5.97 \times 10^{24} \, \text{kg}}{6.38 \times 10^6 \, \text{m}}} \] \[ v = \sqrt{\frac{2 \times 6.67430 \times 5.97 \times 10^{13}}{6.38}} \] \[ v = \sqrt{\frac{7.973586 \times 10^{14}}{6.38}} \] \[ v = \sqrt{1.252028 \times 10^{14}} \] \[ v \approx 1.118 \times 10^7 \, \text{m/s} \] \[ v \approx 11.18 \times 10^3 \, \text{m/s} \] So, the minimum muzzle speed (escape speed) that would allow a shell to escape from the Earth completely is approximately \( 11.18 \times 10^3 \, \text{m/s} \) or \( 11.18 \, \text{km/s} \). To match the unit requested in the problem (10^4 m/s), we can express the result as: \[ v \approx 1.118 \times 10^4 \, \text{m/s} \] Therefore, the escape speed is \( 1.118 \times 10^4 \, \text{m/s} \). |
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| gpto1 | To find the minimum muzzle speed required for a shell to escape from Earth completely (the escape speed), we use the formula for escape velocity: \[ v_{\text{esc}} = \sqrt{\dfrac{2GM_E}{R_E}} \] Where: - \( G \) is the gravitational constant (\(6.67430 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2\)), - \( M_E \) is the mass of the Earth (\(5.97 \times 10^{24} \, \text{kg}\)), - \( R_E \) is the radius of the Earth (\(6.38 \times 10^6 \, \text{m}\)). **Step 1: Calculate the numerator \( 2GM_E \)** \[ 2GM_E = 2 \times (6.67430 \times 10^{-11}) \times (5.97 \times 10^{24}) \] \[ 2GM_E = 2 \times 6.67430 \times 5.97 \times 10^{-11 + 24} \] \[ 2GM_E = 2 \times 6.67430 \times 5.97 \times 10^{13} \] First, multiply \(6.67430 \times 5.97\): \[ 6.67430 \times 5.97 = 39.845571 \] Then multiply by 2: \[ 2 \times 39.845571 = 79.691142 \] So, \[ 2GM_E = 79.691142 \times 10^{13} \] **Step 2: Calculate the escape velocity squared \( v_{\text{esc}}^2 \)** \[ v_{\text{esc}}^2 = \dfrac{2GM_E}{R_E} = \dfrac{79.691142 \times 10^{13}}{6.38 \times 10^6} \] Divide the exponents: \[ 10^{13} / 10^6 = 10^{7} \] Now, divide the numbers: \[ \dfrac{79.691142}{6.38} \approx 12.5 \] So, \[ v_{\text{esc}}^2 \approx 12.5 \times 10^{7} \] **Step 3: Calculate the escape velocity \( v_{\text{esc}} \)** \[ v_{\text{esc}} = \sqrt{12.5 \times 10^{7}} = \sqrt{1.25 \times 10^{8}} \] \[ v_{\text{esc}} = \sqrt{1.25} \times \sqrt{10^{8}} = \sqrt{1.25} \times 10^4 \] Calculate \( \sqrt{1.25} \): \[ \sqrt{1.25} = \sqrt{\dfrac{5}{4}} = \dfrac{\sqrt{5}}{2} \approx \dfrac{2.23607}{2} = 1.11803 \] Therefore, \[ v_{\text{esc}} = 1.11803 \times 10^4 \, \text{m/s} \] **Answer in units of \(10^4 \, \text{m/s}\):** \[ v_{\text{esc}} = 1.12 \times 10^4 \, \text{m/s} \] **Final Answer:** \(\boxed{1.12}\) |
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| deepseekv3 | To find the minimum muzzle speed (escape speed) required for a shell to escape Earth's gravitational pull, we can use the **escape velocity formula**: \[ v_e = \sqrt{\frac{2 G m_E}{R_E}} \] Where: - \( G = 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \) is the gravitational constant, - \( m_E = 5.97 \times 10^{24} \, \text{kg} \) is the mass of the Earth, - \( R_E = 6.38 \times 10^6 \, \text{m} \) is the radius of the Earth. ### Step 1: Plug in the values Substitute the given values into the formula: \[ v_e = \sqrt{\frac{2 \cdot 6.674 \times 10^{-11} \cdot 5.97 \times 10^{24}}{6.38 \times 10^6}} \] ### Step 2: Simplify the numerator Multiply the constants in the numerator: \[ 2 \cdot 6.674 \times 10^{-11} \cdot 5.97 \times 10^{24} = 7.97 \times 10^{14} \] ### Step 3: Divide by the denominator Divide the result by the Earth's radius: \[ \frac{7.97 \times 10^{14}}{6.38 \times 10^6} = 1.25 \times 10^8 \] ### Step 4: Take the square root Take the square root to find the escape velocity: \[ v_e = \sqrt{1.25 \times 10^8} = 1.12 \times 10^4 \, \text{m/s} \] ### Step 5: Express in units of \( 10^4 \, \text{m/s} \) The escape velocity is: \[ v_e = 1.12 \times 10^4 \, \text{m/s} \] Thus, the minimum muzzle speed required for the shell to escape Earth's gravitational pull is: \[ \boxed{1.12} \] |
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