The escape speed for an object at the surface of Earth is 11.2kms. – [Free] B15
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Question:
The escape speed for an object at the surface of Earth is 11.2 km/s. How is this value derived using energy considerations?
Answer:
🌍 Understanding the Concept of Escape Speed:
To escape Earth’s gravity, an object must possess enough kinetic energy to overcome Earth’s gravitational potential energy. This leads to the equation:
½mv2esc = GMm / R
- G = Gravitational Constant =
6.67 × 10−11 m³/kg·s² - M = Mass of Earth =
5.98 × 1024 kg - R = Radius of Earth =
6.37 × 106 m
🧮 Deriving the Escape Speed Formula:
Cancel mass m from both sides and solve for vesc:
vesc = √(2GM / R)
📐 Step-by-Step Calculation:
Step 1: Multiply constants in the numerator:
2 × 6.67 × 10−11 × 5.98 × 1024 = 7.973 × 1014
Step 2: Divide by radius:
7.973 × 1014 / 6.37 × 106 ≈ 1.251 × 108
Step 3: Take the square root:
vesc = √(1.251 × 108) ≈ 1.118 × 104 m/s
Therefore, vesc ≈ 11,180 m/s ≈ 11.2 km/s
✅ Final Answer:
The escape speed from Earth’s surface is approximately: 11.2 km/s