Answer:
1.12 × 10⁴ m/s
Explanation:
The escape velocity of the object v = √(2GM/R) where G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of the Earth = 6 × 10²⁴ kg and R = radius of the Earth = 6.4 × 10⁶ m
Since v = √(2GM/R)
Substituting the values of the variables into the equation, we have
v = √(2GM/R)
v = √(2 × 6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg/6.4 × 10⁶ m)
v = √(13.34 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg/6.4 × 10⁶ m)
v = √(80.04 × 10⁻¹¹ × 10²⁴Nm²/kg/6.4 × 10⁶ m)
v = √(80.04 × 10¹³Nm²/kg ÷ 6.4 × 10⁶ m)
v = √(80.04 ÷ 6.4 × 10¹³ ÷ 10⁶Nm/kg)
v = √(12.50625 × 10⁷ Nm/kg)
v = √(125.0625 × 10⁶ Nm/kg)
v = 11.18 × 10³ m/s
v = 1.118 × 10 × 10³ m/s
v = 1.118 × 10⁴ m/s
v ≅ 1.12 × 10⁴ m/s
