QUESTION 5 (Start on a new page.) The acceleration due to gravity on planet \( X \) is \( 2,7 \mathrm{~m} \cdot \mathrm{~s}^{-2} \). The radius of this planet is a third \( (1 / 3) \) of the radius of Earth. \( \begin{array}{ll}5.1 & \text { Explain the difference between weight and mass. } \\ 5.2 & \text { Calculate the mass of planet } X \text {. } \\ \begin{array}{ll}\text { (2) } \\ 5.3 & \text { Determine the factor by which the weight of an object on planet } X \text { will differ } \\ \text { from the weight of the same object on Earth. }\end{array}\end{array} \begin{array}{l}\text { (2) }\end{array} \)

Let's break down the problem step by step. ### 5.1 Explain the difference between weight and mass. **Mass** is a measure of the amount of matter in an object, typically measured in kilograms (kg). It is a scalar quantity and does not change regardless of the object's location in the universe. **Weight**, on the other hand, is the force exerted on an object due to gravity. It is calculated as the product of mass and the acceleration due to gravity. Weight is a vector quantity and is measured in newtons (N). The formula for weight \( W \) is given by: \[ W = m \cdot g \] where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity. ### 5.2 Calculate the mass of planet \( X \). To calculate the mass of planet \( X \), we can use the formula for gravitational acceleration: \[ g = \frac{G \cdot M}{R^2} \] where: - \( g \) is the acceleration due to gravity on planet \( X \) (given as \( 2.7 \, \text{m/s}^2 \)), - \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \, \text{m}^3 \cdot \text{kg}^{-1} \cdot \text{s}^{-2} \)), - \( M \) is the mass of planet \( X \), - \( R \) is the radius of planet \( X \). Given that the radius of planet \( X \) is \( \frac{1}{3} \) of the radius of Earth, we can denote the radius of Earth as \( R_E \) and the radius of planet \( X \) as: \[ R = \frac{1}{3} R_E \] The average radius of Earth \( R_E \) is approximately \( 6.371 \times 10^6 \, \text{m} \). Therefore, the radius of planet \( X \) is: \[ R = \frac{1}{3} \times 6.371 \times 10^6 \, \text{m} \] Now, substituting \( g \) and \( R \) into the gravitational formula, we can solve for \( M \): \[ 2.7 = \frac{6.674 \times 10^{-11} \cdot M}{\left(\frac{1}{3} \times 6.371 \times 10^6\right)^2} \] Let's calculate \( R \) and then solve for \( M \). ### 5.3 Determine the factor by which the weight of an object on planet \( X \) will differ from the weight of the same object on Earth. The weight of an object on Earth \( W_E \) is given by: \[ W_E = m \cdot g_E \] where \( g_E \) (acceleration due to gravity on Earth) is approximately \( 9.81 \, \text{m/s}^2 \). The weight of the same object on planet \( X \) \( W_X \) is given by: \[ W_X = m \cdot g_X \] To find the factor by which the weight differs, we can calculate: \[ \text{Factor} = \frac{W_X}{W_E} = \frac{g_X}{g_E} \] Now, let's perform the calculations for \( R \), \( M \), and the weight factor. Calculate the value by following steps: - step0: Calculate: \(\frac{2.7}{9.81}\) - step1: Convert the expressions: \(\frac{\frac{27}{10}}{\frac{981}{100}}\) - step2: Multiply by the reciprocal: \(\frac{27}{10}\times \frac{100}{981}\) - step3: Reduce the numbers: \(3\times \frac{10}{109}\) - step4: Multiply: \(\frac{3\times 10}{109}\) - step5: Multiply: \(\frac{30}{109}\) Calculate or simplify the expression \( (1/3) * 6.371 * 10^6 \). Calculate the value by following steps: - step0: Calculate: \(\frac{1}{3}\times 6.371\times 10^{6}\) - step1: Rewrite the expression: \(\frac{1}{3}\times 6371\times 10^{3}\) - step2: Multiply: \(\frac{6371000}{3}\) Solve the equation \( 2.7 = (6.674 * 10^(-11) * M) / ((1/3 * 6.371 * 10^6)^2) \). Solve the equation by following steps: - step0: Solve for \(M\): \(2.7=\frac{\left(6.674\times 10^{-11}M\right)}{\left(\left(\frac{1}{3}\times 6.371\times 10^{6}\right)^{2}\right)}\) - step1: Evaluate: \(2.7=\frac{\left(6.674\times 10^{-11}M\right)}{\left(\frac{1}{3}\times 6.371\times 10^{6}\right)^{2}}\) - step2: Remove the parentheses: \(2.7=\frac{6.674\times 10^{-11}M}{\left(\frac{1}{3}\times 6.371\times 10^{6}\right)^{2}}\) - step3: Simplify: \(2.7=\frac{30033M}{32\times 10^{14}\times 796375^{2}}\) - step4: Swap the sides: \(\frac{30033M}{32\times 10^{14}\times 796375^{2}}=2.7\) - step5: Convert the expressions: \(\frac{30033M}{32\times 10^{14}\times 796375^{2}}=\frac{27}{10}\) - step6: Multiply both sides of the equation by \(32\times 10^{14}\times 796375^{2}:\) \(\frac{30033M}{32\times 10^{14}\times 796375^{2}}\times 32\times 10^{14}\times 796375^{2}=\frac{27}{10}\times 32\times 10^{14}\times 796375^{2}\) - step7: Multiply the terms: \(30033M=\frac{27\times 32\times 10^{14}\times 796375^{2}}{10}\) - step8: Divide the terms: \(30033M=7077888\times 5^{13}\times 796375^{2}\) - step9: Divide both sides: \(\frac{30033M}{30033}=\frac{7077888\times 5^{13}\times 796375^{2}}{30033}\) - step10: Divide the numbers: \(M=\frac{786432\times 5^{13}\times 796375^{2}}{3337}\) ### 5.2 Calculate the mass of planet \( X \). First, we calculated the radius of planet \( X \): \[ R = \frac{1}{3} \times 6.371 \times 10^6 \approx 2.123666 \times 10^6 \, \text{m} \] Next, we used the gravitational formula to find the mass \( M \) of planet \( X \): \[ 2.7 = \frac{6.674 \times 10^{-11} \cdot M}{\left(2.123666 \times 10^6\right)^2} \] Solving for \( M \): \[ M \approx 1.824527 \times 10^{23} \, \text{kg} \] ### 5.3 Determine the factor by which the weight of an object on planet \( X \) will differ from the weight of the same object on Earth. We calculated the factor by which the weight differs: \[ \text{Factor} = \frac{g_X}{g_E} = \frac{2.7}{9.81} \approx 0.275229 \] ### Summary of Results - **Mass of planet \( X \)**: \( M \approx 1.824527 \times 10^{23} \, \text{kg} \) - **Weight factor**: The weight of an object on planet \( X \) is approximately \( 0.275 \) times the weight of the same object on Earth. This means that an object will weigh significantly less on planet \( X \) compared to its weight on Earth.