2) At what distance from Earth's surface is the acceleration due to gravity 7.33m/s^2 (9.97times 10^5m)

Here's how to solve this problem:**1. Understand the concept:**The acceleration due to gravity (g) decreases as you move further away from the Earth's center. The relationship is given by Newton's Law of Universal Gravitation:g = (G * M) / r²Where:* g is the acceleration due to gravity* G is the gravitational constant (6.674 x 10⁻¹¹ N⋅m²/kg²)* M is the mass of the Earth (5.972 x 10²⁴ kg)* r is the distance from the center of the Earth**2. Set up the equation:**We are given g = 7.33 m/s² and we want to find r. We know G and M.7.33 m/s² = (6.674 x 10⁻¹¹ N⋅m²/kg² * 5.972 x 10²⁴ kg) / r²**3. Solve for r:*** Multiply both sides by r²:7.33 m/s² * r² = (6.674 x 10⁻¹¹ N⋅m²/kg² * 5.972 x 10²⁴ kg)* Divide both sides by 7.33 m/s²:r² = (6.674 x 10⁻¹¹ N⋅m²/kg² * 5.972 x 10²⁴ kg) / 7.33 m/s²r² ≈ 5.456 x 10¹³ m²* Take the square root of both sides:r ≈ 7.386 x 10⁶ m**4. Calculate the distance from Earth's surface:**The value of 'r' we calculated is the distance from the *center* of the Earth. To find the distance from the *surface*, we need to subtract the Earth's radius (R), which is approximately 6.371 x 10⁶ m.Distance from surface = r - RDistance from surface ≈ 7.386 x 10⁶ m - 6.371 x 10⁶ mDistance from surface ≈ 1.015 x 10⁶ m or 1015 km**Therefore, the distance from the Earth's surface where the acceleration due to gravity is 7.33 m/s² is approximately 1.015 x 10⁶ m or 1015 km.** The provided answer of 9.97 x 10⁵ m is incorrect.