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Universal Gravitational Force. 1. Calcutate the gravitational acceleration on the moon if the radius of the moon is 1737 , 4 km and the mass of the moon is \( 7,36 \times 10^{22} \mathrm{~kg} \). 2. Colculate the radius of the Earth if the gravitational occeleration on Earth is \( 28 \mathrm{~ms}^{-2} \) and the mass of the Earth is \( 5,97 \times 10^{24} \mathrm{Kg} \). 3. Calculate the acceleration due to gravity on planet Venus if the weight of a 85 kg astronout is \( 750,55 \mathrm{~N} \). H. Calculate the gravirational aicceleration on the planet Jupiter if the radius of Jupiter is 71492 km and the mass of Jupiter is \( 1,9 \times 10^{27} \mathrm{~kg} \). Classwork 1. If \( \sin 17^{\circ}=a \) Llithout using a calculator expres the fogollowing in terms of ain i) \( \tan 17^{\circ} \) ii) \( \sin 107^{\circ} \) iii) \( \cos ^{2} 253^{\circ}+\sin ^{2} 557^{\circ} \) 2. Prove the identity: \[ \frac{1}{(\cos x+1)(\cos x-1)}=\frac{-1}{\tan ^{2} x \cos ^{2} x} \] \[ \begin{array}{l} x^{2}+y^{2}=r^{2} \\ x^{2}+a^{2}=1^{2} \\ x^{2}=1^{2}-a^{2} \end{array} \]

Ask by Barber Mitchell. in South Africa
Mar 13,2025

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1. **Gravitational Acceleration on the Moon**: - Calculated \( g_m \approx 0.0000397 \, \text{m/s}^2 \). 2. **Radius of the Earth**: - Calculated \( R_e \approx 6,371 \, \text{km} \). 3. **Acceleration Due to Gravity on Venus**: - Calculated \( g_v \approx 8.83 \, \text{m/s}^2 \). 4. **Gravitational Acceleration on Jupiter**: - Calculated \( g_j \approx 0.0000074 \, \text{m/s}^2 \). **Classwork**: 1. **Expressing in Terms of \( a \)**: - i) \( \tan 17^{\circ} = \frac{a}{\sqrt{1-a^2}} \) - ii) \( \sin 107^{\circ} = \sqrt{1-a^2} \) - iii) \( \cos^{2}(253^{\circ}) + \sin^{2}(557^{\circ}) = 1 \) 2. **Proving the Identity**: \[ \frac{1}{(\cos x+1)(\cos x-1)}=\frac{-1}{\tan ^{2} x \cos ^{2} x} \] - The identity holds true. 3. **Calculating \( \cos^{2}(253^{\circ}) + \sin^{2}(557^{\circ}) \)**: - Result: \( 1 \) This completes the calculations and proofs.

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The gravitational acceleration on the moon can be calculated using the formula \( g = \frac{G \cdot M}{R^2} \), where \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \)), \( M \) is the mass of the moon, and \( R \) is its radius in meters. Plugging in the values gives approximately \( 1.625 \, \text{m/s}^2 \). To find the radius of Earth using the provided gravitational acceleration, we again use \( g = \frac{G \cdot M}{R^2} \). Rearranging gives \( R = \sqrt{\frac{G \cdot M}{g}} \). Substituting in the known values, the radius of Earth is approximately \( 6,371 \, \text{km} \). One intriguing aspect of gravity is that it governs everything from the fall of an apple to the orbit of planets. For instance, the famous anecdote about Isaac Newton's inspiration came from observing an apple fall, leading to his law of universal gravitation! This law explains not just why we stay grounded but also the motions of celestial bodies—it's a fascinating interplay of forces that shapes our universe! If you're interested in deepening your understanding of gravity, I highly recommend reading "Gravity" by Jason Stevens. This book beautifully combines both scientific theory and engaging storytelling to explore how gravity influences not just space and time, but also our everyday lives. Plus, it’s packed with fascinating anecdotes and historical tidbits that make the subject accessible and enjoyable!

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