Given the Moon's radius 1,730 km and mass 7.34 x 10^22 kg, the acceleration due to gravity on the Moon is approximately which value?

Given the Moon's radius 1,730 km and mass 7.34 x 10^22 kg, the acceleration due to gravity on the Moon is approximately which value?

• A. 1.6 m/s^2
• B. 2.6 m/s^2
• C. 3.6 m/s^2
• D. 0.6 m/s^2

Answer: (A)

Gravity on a body's surface comes from Newton's law of gravitation and is given by g = G M / R^2, where G is the gravitational constant, M is the body's mass, and R is its radius. For the Moon, convert the radius to meters and plug in the values: R = 1.730×10^6 m, M = 7.34×10^22 kg, and G = 6.674×10^-11 N m^2/kg^2. Compute the product G M ≈ 6.674×10^-11 × 7.34×10^22 ≈ 4.90×10^12. Compute R^2 ≈ (1.730×10^6)^2 ≈ 2.99×10^12. Divide: g ≈ (4.90×10^12) / (2.99×10^12) ≈ 1.64 m/s^2. Thus the Moon’s surface gravity is about 1.6 m/s^2, which is much weaker than Earth's due to its much smaller mass relative to its size.

Gravity on a body's surface comes from Newton's law of gravitation and is given by g = G M / R^2, where G is the gravitational constant, M is the body's mass, and R is its radius.

For the Moon, convert the radius to meters and plug in the values: R = 1.730×10^6 m, M = 7.34×10^22 kg, and G = 6.674×10^-11 N m^2/kg^2.

Compute the product G M ≈ 6.674×10^-11 × 7.34×10^22 ≈ 4.90×10^12. Compute R^2 ≈ (1.730×10^6)^2 ≈ 2.99×10^12. Divide: g ≈ (4.90×10^12) / (2.99×10^12) ≈ 1.64 m/s^2.

Thus the Moon’s surface gravity is about 1.6 m/s^2, which is much weaker than Earth's due to its much smaller mass relative to its size.