| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the gravitational force \(F\) between two masses, we use Newton's law of universal gravitation: \[F = G \frac{m_1 m_2}{r^2}\] where - \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}\)), - \(m_1\) and \(m_2\) are the masses of the two objects, - \(r\) is the distance between the centers of the two objects. Given: - \(m_1 = 5.97 \times 10^{24} \, \text{kg}\) (mass of Earth), - \(m_2 = 7.35 \times 10^{22} \, \text{kg}\) (mass of the Moon), - \(r = 3.84 \times 10^5 \, \text{km} = 3.84 \times 10^8 \, \text{m}\) (mean distance between Earth and Moon). Substitute the values into the formula: \[F = 6.674 \times 10^{-11} \times \frac{(5.97 \times 10^{24}) \times (7.35 \times 10^{22})}{(3.84 \times 10^8)^2}\] \[F = 6.674 \times 10^{-11} \times \frac{4.37795 \times 10^{47}}{1.47456 \times 10^{17}}\] \[F = 6.674 \times 10^{-11} \times 2.9653 \times 10^{29}\] \[F = 2.0007 \times 10^{19}\] \[F = 2.0007 \times 10^{20} \, \text{N} / 100\] \[F \approx 2.0007 \times 10^{20} \, \text{N}\] Therefore, \(X = 2.0007\). Rounded to two significant figures (as per the precision of the given constants), \(X = 2.0\). |
1.99 |
| deepseekr1 | The gravitational force exerted by Earth on the Moon can be calculated using Newton's law of universal gravitation: \[ F = G \frac{m_1 m_2}{r^2} \] **Given values:** - \( G = 6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 \) - Earth's mass \( m_1 = 5.97 \times 10^{24} \, \text{kg} \) - Moon's mass \( m_2 = 7.35 \times 10^{22} \, \text{kg} \) - Distance \( r = 3.84 \times 10^8 \, \text{m} \) (converted from kilometers to meters) **Steps:** 1. **Calculate the numerator \( m_1 m_2 \):** \[ 5.97 \times 10^{24} \times 7.35 \times 10^{22} = 43.8795 \times 10^{46} \, \text{kg}^2 \] 2. **Multiply by \( G \):** \[ 6.674 \times 10^{-11} \times 43.8795 \times 10^{46} = 292.851783 \times 10^{35} \, \text{N·m}^2 \] 3. **Calculate the denominator \( r^2 \):** \[ (3.84 \times 10^8)^2 = 14.7456 \times 10^{16} \, \text{m}^2 \] 4. **Divide numerator by denominator:** \[ \frac{292.851783 \times 10^{35}}{14.7456 \times 10^{16}} = 19.86 \times 10^{19} \, \text{N} = 1.986 \times 10^{20} \, \text{N} \] **Significant Figures:** All input values have three significant figures. Rounding \( 1.986 \times 10^{20} \, \text{N} \) to three significant figures gives: \[ X = 1.99 \] **Answer:** \( \boxed{1.98} \) **Note:** After a more precise calculation and considering standard references, the value rounds to \( 1.98 \times 10^{20} \, \text{N} \), hence \( X = 1.98 \). |
1.99 |
| gpto1 | To solve this problem, we'll use Newton's law of universal gravitation, which states that the gravitational force \( F \) between two masses is given by: \[ F = G \frac{M_e M_m}{r^2} \] 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 Earth (\( 5.97 \times 10^{24} \, \text{kg} \)) - \( M_m \) is the mass of the Moon (\( 7.35 \times 10^{22} \, \text{kg} \)) - \( r \) is the distance between Earth's center and the Moon (\( 3.84 \times 10^5 \, \text{km} \)) **Step 1: Convert the distance to meters.** \[ r = 3.84 \times 10^5 \, \text{km} = 3.84 \times 10^5 \times 10^3 \, \text{m} = 3.84 \times 10^8 \, \text{m} \] **Step 2: Calculate \( r^2 \).** \[ r^2 = (3.84 \times 10^8 \, \text{m})^2 = 1.47456 \times 10^{17} \, \text{m}^2 \] **Step 3: Calculate the numerator \( G M_e M_m \).** Compute the numeric part: \[ (G M_e M_m) = (6.67430 \times 10^{-11}) \times (5.97 \times 10^{24}) \times (7.35 \times 10^{22}) \] First, multiply the masses: \[ 5.97 \times 10^{24} \times 7.35 \times 10^{22} = 43.8795 \times 10^{46} \] Now multiply by \( G \): \[ (6.67430 \times 10^{-11}) \times 43.8795 \times 10^{46} = 292.677 \times 10^{35} \, \text{N}\cdot\text{m}^2 \] **Step 4: Calculate the force \( F \).** \[ F = \frac{292.677 \times 10^{35}}{1.47456 \times 10^{17}} \] Compute the division: \[ \frac{292.677}{1.47456} \approx 198.36 \] Compute the exponent part: \[ 10^{35} / 10^{17} = 10^{18} \] So: \[ F \approx 198.36 \times 10^{18} \, \text{N} \] **Step 5: Express \( F \) in terms of \( X \times 10^{20} \, \text{N} \).** \[ F \approx 1.9836 \times 10^{20} \, \text{N} \] So, \( X \approx 1.98 \). **Answer:** 1.98 |
1.99 |
| GPT-4o | To solve this problem, we can use Newton's law of universal gravitation. The formula for the gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by: \[ F = G \frac{m_1 m_2}{r^2} \] where \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{N} \, (\text{m}^2 / \text{kg}^2) \). Given: - Mass of Earth (\( m_1 \)) = \( 5.97 \times 10^{24} \) kg - Mass of the Moon (\( m_2 \)) = \( 7.35 \times 10^{22} \) kg - Mean distance between Earth and Moon (\( r \)) = \( 3.84 \times 10^5 \) km = \( 3.84 \times 10^8 \) meters We can plug these values into the formula: \[ F = 6.674 \times 10^{-11} \frac{ (5.97 \times 10^{24}) (7.35 \times 10^{22}) }{ (3.84 \times 10^8)^2 } \] First, calculate the denominator: \[ (3.84 \times 10^8)^2 = 1.47456 \times 10^{17} \, \text{m}^2 \] Next, calculate the numerator: \[ (5.97 \times 10^{24}) (7.35 \times 10^{22}) = 4.38795 \times 10^{47} \, \text{kg}^2 \] Now, divide the numerator by the denominator: \[ \frac{4.38795 \times 10^{47}}{1.47456 \times 10^{17}} = 2.97727 \times 10^{30} \] Finally, multiply by \( G \): \[ F = 6.674 \times 10^{-11} \times 2.97727 \times 10^{30} = 1.988 \times 10^{20} \, \text{N} \] Given that the magnitude of the gravitational force is \( X \times 10^{20} \) N, we find: \[ X = 1.988 \] Therefore, \( X \) is approximately \( 1.988 \). |
1.99 |
| llama3.1-70B-Instruct | To find the magnitude of the gravitational force exerted by Earth on the Moon, we can use the formula for gravitational force, which is given by Newton's law of universal gravitation: F = G * (m1 * m2) / r^2 where: - F is the gravitational force between the masses, - G is the gravitational constant (6.674 * 10^-11 N m^2 kg^-2), - m1 is the mass of Earth (5.97 * 10^24 kg), - m2 is the mass of the Moon (7.35 * 10^22 kg), - r is the mean distance from the center of Earth to the Moon (3.84 * 10^8 m, converted from km to m). Plugging the given values into the formula: F = (6.674 * 10^-11 N m^2 kg^-2) * ((5.97 * 10^24 kg) * (7.35 * 10^22 kg)) / (3.84 * 10^8 m)^2 First, calculate the numerator: (6.674 * 10^-11) * (5.97 * 10^24) * (7.35 * 10^22) = 6.674 * 5.97 * 7.35 * 10^(-11 + 24 + 22) = 6.674 * 5.97 * 7.35 * 10^(35) ≈ 291.32 * 10^35 Now, calculate the denominator: (3.84 * 10^8)^2 = 14.7456 * 10^16 Now, divide the numerator by the denominator: F ≈ (291.32 * 10^35) / (14.7456 * 10^16) ≈ 19.774 * 10^19 ≈ 1.9774 * 10^20 N Therefore, X is approximately 1.9774, but since the problem asks for a numeric value without specifying the precision, we can round it to a more reasonable form. |
1.99 |
| llama3.1-405B-Instruct | To find the magnitude of the gravitational force exerted by Earth on the Moon, we can use the formula for gravitational force, which is given by: F = G \* (m1 \* m2) / r^2 where: - F is the gravitational force, - G is the gravitational constant (6.674 * 10^-11 N m^2 kg^-2), - m1 and m2 are the masses of the two objects (in this case, Earth and the Moon), - r is the distance between the centers of the two objects. Given: - m1 = 5.97 * 10^24 kg (mass of Earth), - m2 = 7.35 * 10^22 kg (mass of the Moon), - r = 3.84 * 10^5 km = 3.84 * 10^8 m (mean distance of the Moon from the center of Earth). Plugging in the values: F = (6.674 * 10^-11 N m^2 kg^-2) * ((5.97 * 10^24 kg) * (7.35 * 10^22 kg)) / (3.84 * 10^8 m)^2 F = (6.674 * 10^-11) * (5.97 * 7.35 * 10^(24+22)) / (3.84^2 * 10^(8*2)) F = (6.674 * 10^-11) * (43.8635 * 10^46) / (14.7456 * 10^16) F = (6.674 * 43.8635 * 10^(46-11)) / (14.7456 * 10^16) F = (293.096 * 10^35) / (14.7456 * 10^16) F = 293.096 * 10^(35-16) / 14.7456 F = 293.096 * 10^19 / 14.7456 F = 19.872 * 10^20 N Therefore, X = 19.872. |
1.99 |
| DeepSeek-Coder-V2 | To solve the problem, we need to calculate the gravitational force exerted by Earth on the Moon using the formula for gravitational force: \[ F = G \frac{m_1 m_2}{r^2} \] where: - \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \) - \( m_1 \) is the mass of Earth, \( 5.97 \times 10^{24} \, \text{kg} \) - \( m_2 \) is the mass of the Moon, \( 7.35 \times 10^{22} \, \text{kg} \) - \( r \) is the distance between the centers of Earth and the Moon, \( 3.84 \times 10^5 \, \text{km} \) First, we need to convert the distance from kilometers to meters: \[ r = 3.84 \times 10^5 \, \text{km} = 3.84 \times 10^8 \, \text{m} \] Now, we can substitute the values into the gravitational force formula: \[ F = 6.674 \times 10^{-11} \frac{(5.97 \times 10^{24})(7.35 \times 10^{22})}{(3.84 \times 10^8)^2} \] Calculate the numerator: \[ (5.97 \times 10^{24})(7.35 \times 10^{22}) = 4.38795 \times 10^{47} \] Calculate the denominator: \[ (3.84 \times 10^8)^2 = 1.47456 \times 10^{17} \] Now, divide the numerator by the denominator: \[ F = 6.674 \times 10^{-11} \frac{4.38795 \times 10^{47}}{1.47456 \times 10^{17}} \] \[ F = 6.674 \times 10^{-11} \times 2.9758 \times 10^{30} \] \[ F = 1.985 \times 10^{20} \, \text{N} \] The magnitude of the gravitational force is \( 1.985 \times 10^{20} \, \text{N} \). Therefore, the value of \( X \) is: \[ X = 1.985 \] So, the final answer is: \[ \boxed{1.985} \] |
1.99 |
| Qwen2-72B-Instruct | To find the gravitational force \(F\) between two masses, we use Newton's law of universal gravitation: \[F = G \frac{m_1 m_2}{r^2}\] where - \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}\)), - \(m_1\) and \(m_2\) are the masses of the two objects, - \(r\) is the distance between the centers of the two objects. Given: - \(m_1 = 5.97 \times 10^{24} \, \text{kg}\) (mass of Earth), - \(m_2 = 7.35 \times 10^{22} \, \text{kg}\) (mass of the Moon), - \(r = 3.84 \times 10^5 \, \text{km} = 3.84 \times 10^8 \, \text{m}\) (mean distance between Earth and Moon). Substitute the values into the formula: \[F = 6.674 \times 10^{-11} \times \frac{(5.97 \times 10^{24}) \times (7.35 \times 10^{22})}{(3.84 \times 10^8)^2}\] \[F = 6.674 \times 10^{-11} \times \frac{4.37795 \times 10^{47}}{1.47456 \times 10^{17}}\] \[F = 6.674 \times 10^{-11} \times 2.9653 \times 10^{29}\] \[F = 2.0007 \times 10^{19}\] \[F = 2.0007 \times 10^{20} \, \text{N} / 100\] \[F \approx 2.0007 \times 10^{20} \, \text{N}\] Therefore, \(X = 2.0007\). Rounded to two significant figures (as per the precision of the given constants), \(X = 2.0\). |
1.99 |
| baichuan4 | To calculate the gravitational force exerted by Earth on the Moon, we can use Newton's law of gravitation: F = G * (m1 * m2) / r^2 where: - F is the gravitational force - G is the gravitational constant, approximately 6.674 x 10^-11 N(m/kg)^2 - m1 and m2 are the masses of the two objects (in this case, Earth and the Moon) - r is the distance between the centers of the two objects Given: - Mass of Earth (m1): 5.97 x 10^24 kg - Mass of the Moon (m2): 7.35 x 10^22 kg - Mean distance of the Moon from the center of Earth (r): 3.84 x 10^5 km = 3.84 x 10^8 m (since 1 km = 1000 m) Plugging in the values: F = 6.674 x 10^-11 * (5.97 x 10^24 * 7.35 x 10^22) / (3.84 x 10^8)^2 F = 6.674 x 10^-11 * (4.40895 x 10^46) / (1.47456 x 10^17) F = 6.674 x 10^-11 * 2.9903 x 10^29 F = 1.9914 x 10^20 N So, the magnitude of the gravitational force exerted by Earth on the Moon is approximately 1.9914 x 10^20 N. Therefore, X is approximately 1.9914. |
1.99 |
| doubao | The gravitational force between two objects is given by the formula F = G * (m1 * m2) / r^2, where G is the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2), m1 and m2 are the masses of the two objects, and r is the distance between them. Converting the distance from km to m: 3.84 × 10^5 km = 3.84 × 10^8 m. The masses are 5.97 × 10^24 kg for Earth and 7.35 × 10^22 kg for the Moon. Calculating the force: F = 6.67430 × 10^-11 * (5.97 × 10^24 * 7.35 × 10^22) / (3.84 × 10^8)^2 = 1.98375 × 10^20 N. So X = 1.98375. |
1.99 |
| minimax | To solve this problem, we can use Newton's law of universal gravitation, which states that the force (F) between two masses (m1 and m2) separated by a distance (r) is given by: \[ F = G \frac{m_1 m_2}{r^2} \] where G is the gravitational constant, approximately equal to \(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\). Given: - Mass of Earth (\(m_1\)) = \(5.97 \times 10^{24}\) kg - Mass of the Moon (\(m_2\)) = \(7.35 \times 10^{22}\) kg - Mean distance of the Moon from the center of Earth (\(r\)) = \(3.84 \times 10^5\) km = \(3.84 \times 10^8\) m (since 1 km = 1000 m) We can plug these values into the formula to find the gravitational force (F): \[ F = G \frac{m_1 m_2}{r^2} \] \[ F = (6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) \frac{(5.97 \times 10^{24} \, \text{kg})(7.35 \times 10^{22} \, \text{kg})}{(3.84 \times 10^8 \, \text{m})^2} \] \[ F = (6.674 \times 10^{-11}) \frac{(5.97 \times 10^{24})(7.35 \times 10^{22})}{(3.84 \times 10^8)^2} \] \[ F = (6.674 \times 10^{-11}) \frac{(4.39 \times 10^{47})}{(1.47456 \times 10^{17})} \] \[ F = (6.674 \times 10^{-11}) \times (2.976 \times 10^{30}) \] \[ F = 1.986 \times 10^{20} \, \text{N} \] So, the magnitude of the gravitational force exerted by Earth on the Moon is \(1.986 \times 10^{20}\) N. Therefore, X is approximately 1.986. |
1.99 |
| kimi | To find the magnitude of the gravitational force exerted by Earth on the Moon, we can use Newton's law of universal gravitation: \[ F = G \frac{m_1 m_2}{r^2} \] where: - \( F \) is the gravitational force between the two masses, - \( G \) is the gravitational constant (\(6.674 \times 10^{-11} \text{N}(m/kg)^2\)), - \( m_1 \) is the mass of the Earth, - \( m_2 \) is the mass of the Moon, - \( r \) is the distance between the centers of the two masses. Given: - \( m_1 = 5.97 \times 10^{24} \) kg (mass of Earth), - \( m_2 = 7.35 \times 10^{22} \) kg (mass of the Moon), - \( r = 3.84 \times 10^8 \) m (mean distance from the center of Earth to the Moon, converted from km to m). Now, let's plug in the values: \[ F = 6.674 \times 10^{-11} \frac{(5.97 \times 10^{24}) (7.35 \times 10^{22})}{(3.84 \times 10^8)^2} \] \[ F = 6.674 \times 10^{-11} \frac{4.37495 \times 10^{47}}{14.7456 \times 10^{16}} \] \[ F = 6.674 \times 10^{-11} \times 2.97 \times 10^{30} \] \[ F = 1.99 \times 10^{20} \] So, \( X = \boxed{1.99} \). |
1.99 |
| wenxinyiyan | To solve this problem, we need to use Newton's law of universal gravitation, which states that the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. The formula for the gravitational force F is: F = G * (m1 * m2) / r^2 where: - F is the gravitational force - G is the gravitational constant, approximately 6.674 * 10^(-11) N * (m^2 / kg^2) - m1 and m2 are the masses of the two objects - r is the distance between the centers of the two objects In this case, m1 is the mass of Earth (5.97 * 10^24 kg), m2 is the mass of the Moon (7.35 * 10^22 kg), and r is the mean distance of the Moon from the center of Earth (3.84 * 10^5 km, which needs to be converted to meters by multiplying by 1000). Let's perform the calculation: F = 6.674 * 10^(-11) * (5.97 * 10^24 * 7.35 * 10^22) / (3.84 * 10^8)^2 = 6.674 * 10^(-11) * 4.38915 * 10^47 / 1.47456 * 10^17 = 1.98892 * 10^31 / 1.47456 * 10^17 = 1.34841 * 10^14 N However, the problem asks for the answer in the form of X * 10^20 N, so we need to convert our answer to this form. Our answer, 1.34841 * 10^14 N, can be written as 1.34841 * 10^(-6) * 10^20 N. Therefore, X = 1.34841 * 10^(-6). So, the numeric value of X is approximately 1.34841 * 10^(-6). |
1.99 |
| zhipu | To find the magnitude of the gravitational force exerted by Earth on the Moon, we can use Newton's law of universal gravitation, which is given by the formula: \[ F = G \frac{m_1 m_2}{r^2} \] where: - \( F \) is the gravitational force between two masses, - \( G \) is the gravitational constant, approximately \( 6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \), - \( m_1 \) and \( m_2 \) are the masses of the two objects (in this case, Earth and the Moon), - \( r \) is the distance between the centers of the two masses. Given: - Mass of Earth, \( m_1 = 5.97 \times 10^{24} \, \text{kg} \), - Mass of the Moon, \( m_2 = 7.35 \times 10^{22} \, \text{kg} \), - Mean distance from the center of Earth to the Moon, \( r = 3.84 \times 10^5 \, \text{km} \). First, we need to convert the distance from kilometers to meters, since the gravitational constant \( G \) uses meters: \[ r = 3.84 \times 10^5 \, \text{km} \times 10^3 \, \text{m/km} = 3.84 \times 10^8 \, \text{m} \] Now we can plug the values into the formula: \[ F = 6.67430 \times 10^{-11} \frac{5.97 \times 10^{24} \times 7.35 \times 10^{22}}{(3.84 \times 10^8)^2} \] First, calculate the numerator: \[ 5.97 \times 10^{24} \times 7.35 \times 10^{22} = 4.39 \times 10^{47} \] Then, calculate the denominator: \[ (3.84 \times 10^8)^2 = 1.47456 \times 10^{17} \] Now, divide the numerator by the denominator: \[ \frac{4.39 \times 10^{47}}{1.47456 \times 10^{17}} \approx 2.98 \times 10^{30} \] Finally, multiply by \( G \): \[ F \approx 6.67430 \times 10^{-11} \times 2.98 \times 10^{30} \] \[ F \approx 1.99 \times 10^{20} \, \text{N} \] Therefore, the value of \( X \) is approximately: \[ X \approx 1.99 \] However, since we need to return a numeric value without approximation, let's use a more precise calculation: \[ F \approx 6.67430 \times 10^{-11} \times \frac{4.39 \times 10^{47}}{1.47456 \times 10^{17}} \] \[ F \approx 1.982 \times 10^{20} \, \text{N} \] So, the numeric value of \( X \) is: \[ X = 1.982 \] Thus, the final answer is: \[ 1.982 \] |
1.99 |
| gpto1 | To solve this problem, we'll use Newton's law of universal gravitation, which states that the gravitational force \( F \) between two masses is given by: \[ F = G \frac{M_e M_m}{r^2} \] 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 Earth (\( 5.97 \times 10^{24} \, \text{kg} \)) - \( M_m \) is the mass of the Moon (\( 7.35 \times 10^{22} \, \text{kg} \)) - \( r \) is the distance between Earth's center and the Moon (\( 3.84 \times 10^5 \, \text{km} \)) **Step 1: Convert the distance to meters.** \[ r = 3.84 \times 10^5 \, \text{km} = 3.84 \times 10^5 \times 10^3 \, \text{m} = 3.84 \times 10^8 \, \text{m} \] **Step 2: Calculate \( r^2 \).** \[ r^2 = (3.84 \times 10^8 \, \text{m})^2 = 1.47456 \times 10^{17} \, \text{m}^2 \] **Step 3: Calculate the numerator \( G M_e M_m \).** Compute the numeric part: \[ (G M_e M_m) = (6.67430 \times 10^{-11}) \times (5.97 \times 10^{24}) \times (7.35 \times 10^{22}) \] First, multiply the masses: \[ 5.97 \times 10^{24} \times 7.35 \times 10^{22} = 43.8795 \times 10^{46} \] Now multiply by \( G \): \[ (6.67430 \times 10^{-11}) \times 43.8795 \times 10^{46} = 292.677 \times 10^{35} \, \text{N}\cdot\text{m}^2 \] **Step 4: Calculate the force \( F \).** \[ F = \frac{292.677 \times 10^{35}}{1.47456 \times 10^{17}} \] Compute the division: \[ \frac{292.677}{1.47456} \approx 198.36 \] Compute the exponent part: \[ 10^{35} / 10^{17} = 10^{18} \] So: \[ F \approx 198.36 \times 10^{18} \, \text{N} \] **Step 5: Express \( F \) in terms of \( X \times 10^{20} \, \text{N} \).** \[ F \approx 1.9836 \times 10^{20} \, \text{N} \] So, \( X \approx 1.98 \). **Answer:** 1.98 |
1.99 |
| deepseekv3 | To calculate the gravitational force exerted by Earth on the Moon, we can use **Newton's Law of Gravitation**: \[ F = G \cdot \frac{m_1 \cdot m_2}{r^2} \] Where: - \( F \) is the gravitational force, - \( G \) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2\)), - \( m_1 \) is the mass of Earth (\(5.97 \times 10^{24} \, \text{kg}\)), - \( m_2 \) is the mass of the Moon (\(7.35 \times 10^{22} \, \text{kg}\)), - \( r \) is the distance between the centers of Earth and the Moon (\(3.84 \times 10^5 \, \text{km} = 3.84 \times 10^8 \, \text{m}\)). ### Step 1: Substitute the values into the formula \[ F = (6.674 \times 10^{-11}) \cdot \frac{(5.97 \times 10^{24}) \cdot (7.35 \times 10^{22})}{(3.84 \times 10^8)^2} \] ### Step 2: Simplify the numerator and denominator \[ F = (6.674 \times 10^{-11}) \cdot \frac{4.39 \times 10^{47}}{1.47 \times 10^{17}} \] ### Step 3: Perform the division \[ F = (6.674 \times 10^{-11}) \cdot (2.98 \times 10^{30}) \] ### Step 4: Multiply the terms \[ F = 1.99 \times 10^{20} \, \text{N} \] Thus, the gravitational force exerted by Earth on the Moon is \(1.99 \times 10^{20} \, \text{N}\). Therefore, \(X = 1.99\). |
1.99 |