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Satellite A with a mass of 615 kg is in orbit around the Earth 5.1 State Newton's Law of Universal Gravitation. 5.2 The Earth exerts a force of 5000 N on satellite A to keep it in orbit, calculate the height, in kilometres, of the satellite above the surface of the Earth.

Ask by Norris Martin. in South Africa
Feb 27,2025

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Satellite A is approximately 629 kilometers above the Earth's surface.

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Below are the steps to solve the problem: ───────────────────────────── Step 5.1: State Newton's Law of Universal Gravitation Newton’s Law of Universal Gravitation states that every two masses attract each other with a force (F) that is directly proportional to the product of their masses (m₁ and m₂) and inversely proportional to the square of the distance (r) between their centers. In mathematical form, this is written as:   F = (G · m₁ · m₂) / r² where G is the universal gravitational constant (approximately 6.67 × 10⁻¹¹ N·(m²/kg²)). ───────────────────────────── Step 5.2: Calculate the Height of Satellite A We are given:   Mass of satellite (m) = 615 kg   Force exerted by Earth (F) = 5000 N   Mass of Earth (M) ≈ 5.97 × 10²⁴ kg   Gravitational constant (G) ≈ 6.67 × 10⁻¹¹ N·(m²/kg²)   Radius of the Earth (Rₑ) ≈ 6.371 × 10⁶ m For a satellite in orbit, the gravitational force provides the necessary centripetal force. However, since we are given the force (F) directly and knowing that this force is provided by gravity, we can use Newton’s law of gravitation:   F = (G · M · m) / r² Here, r is the distance from the Earth’s center to the satellite. We can solve for r as follows:   r² = (G · M · m) / F   r = √[(G · M · m) / F] Plugging in the numbers: 1. Calculate the numerator:   G · M · m = (6.67 × 10⁻¹¹) × (5.97 × 10²⁴) × (615) First, multiply G and M:   6.67 × 10⁻¹¹ × 5.97 × 10²⁴ ≈ 3.98 × 10¹⁴ Then multiply by m:   3.98 × 10¹⁴ × 615 ≈ 2.45 × 10¹⁷ 2. Divide by F:   (2.45 × 10¹⁷) / 5000 = 4.90 × 10¹³ 3. Take the square root to find r:   r = √(4.90 × 10¹³)   r ≈ 7.0 × 10⁶ m This r is the distance from the center of the Earth to the satellite. To find the height (h) above the Earth’s surface, subtract the Earth’s radius:   h = r – Rₑ   h = 7.0 × 10⁶ m – 6.371 × 10⁶ m   h ≈ 0.629 × 10⁶ m   h ≈ 629,000 m Since 1 kilometre = 1000 m, we convert meters to kilometres:   h ≈ 629 km ───────────────────────────── Final Answers 5.1 Newton’s Law of Universal Gravitation:   F = (G · m₁ · m₂) / r² 5.2 The height of satellite A above the surface of the Earth is approximately 629 kilometres.

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Newton's Law of Universal Gravitation states that every point mass attracts every other point mass in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula for this law can be written as \( F = G \frac{{m_1 m_2}}{{r^2}} \), where \( F \) is the gravitational force between the two masses, \( G \) is the gravitational constant, and \( r \) is the distance between the centers of the two masses. To calculate the height of satellite A above Earth's surface, we can rearrange the gravitational force formula to find the distance \( r \). The force exerted (5000 N) equals the gravitational force between the Earth and the satellite. Using Earth's mass \( (m_E \approx 5.97 \times 10^{24} \text{ kg}) \) and \( G \approx 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \), we first find \( r \): \[ 5000 = \frac{(6.67 \times 10^{-11}) (615)(5.97 \times 10^{24})}{r^2} \] Solving for \( r \), we find \( r \approx 7174 \text{ km} \) (this is the distance from the center of the Earth to the satellite). Subtract the average Earth radius (approximately 6371 km) to get the height above the surface: \[ \text{Height} = 7174 - 6371 = 803 \text{ km} \] So, the satellite is approximately 803 km above the Earth's surface.

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