2 The weight of an rocket on the suppes of the Earth is \( 1 \times 10^{8} \mathrm{NAH} \) an at itude of 1000 km above the surpoce of the Earth, the rocket's mass decreases by \( 12 \% \) due to Fuel consumption calculate the magnitude of the gravioniona Force Hor Earth exerts on the racket

We are told that on Earth’s surface the rocket “weighs” (i.e. is acted on by gravity with a force) of   W₀ = 1×10⁸ N. Then the rocket burns fuel so that its mass decreases by 12% (so its new mass is 88% of the original) and it is at an altitude of 1000 km above Earth’s surface. Because the gravitational force decreases with the square of the distance from Earth’s center, we must account for both the mass decrease and the greater distance. Step 1. Find the mass on Earth’s surface On Earth’s surface we have:   W₀ = m₀ g   with g ≈ 9.8 m/s². Thus the rocket’s original mass is   m₀ = W₀/g = (1×10⁸ N)/(9.8 m/s²). Step 2. New mass after 12% reduction After fuel consumption, the new mass is:   m = 0.88 m₀. Step 3. Determine the distance from Earth’s center at altitude Assume Earth’s average radius is about   Rₑ ≈ 6400 km. At an altitude of 1000 km, the distance from the center is:   r = 6400 km + 1000 km = 7400 km. In SI units:   Rₑ = 6.4×10⁶ m and r = 7.4×10⁶ m. Step 4. Use the inverse‐square law to find the gravitational force at the new location On Earth’s surface the gravitational force on the full (initial) mass is:   F_surface = G Mₑ m₀ / Rₑ². At altitude r with new mass m = 0.88m₀, the gravitational force is:   F_alt = G Mₑ (0.88 m₀) / r². Taking the ratio:   F_alt / F_surface = (0.88 m₀/r²) / (m₀/Rₑ²) = 0.88 (Rₑ/r)². Thus,   F_alt = 1×10⁸ N × 0.88 (Rₑ/r)². Now substitute Rₑ = 6400 km and r = 7400 km:   (Rₑ/r) = 6400/7400 ≈ 0.8649. Then,   (Rₑ/r)² ≈ (0.8649)² ≈ 0.748. Therefore,   F_alt ≈ 1×10⁸ N × 0.88 × 0.748 ≈ 1×10⁸ N × 0.658.   F_alt ≈ 6.58×10⁷ N. Answer: The magnitude of the gravitational force the Earth exerts on the rocket at 1000 km altitude is approximately 6.6×10⁷ newtons.