If an object has a weight of 4050 N near the surface of the Earth, what is its mass?

417.5 kg

Explanation:

Since the gravitational for of attraction F, equals the weight of the object W when its close to the earth, F = GMm/R² = W = mg where m = mass of object and g = acceleration due to gravity close to the earth = 9.8 m/s².

g = GM/R² where  g = acceleration due to gravity close to the earth, G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of earth = 5.927 × 10²⁴ kg and R = radius of earth = 6.4 × 10⁶ m

So, g = GM/R²

= 6.67 × 10⁻¹¹ Nm²/kg² × 5.927 × 10²⁴ kg/(6.4 × 10⁶ m)²

= 39.53 × 10¹³ Nm²/kg ÷ 40.96 × 10¹² m²

= 9.65 N/kg

≅ 9.7 N/kg

= 9.7 m/s²

Since W = 4050 N and W = mg

m = W/g = 4050 N/9.7 m/s²

= 417.53 kg

≅ 417.5 kg

So, the mass of the object is 417.5 kg

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