Question

Two identical coins of mass 8 g are 50 cm apart on a tabletop. How many times larger is the weight of one coin than the gravitational attraction of the other coin for it? (G = \( 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), g = 9.81 m/s\(^2\)):

• A. \( 4.6 \times 10^{12} \)
• B. \( 4.6 \times 10^{10} \)
• C. \( 4.6 \times 10^{14} \)
• D. None

Hint:

Gravitational attraction is calculated using Newton's law of gravitation, and the weight is simply the product of mass and gravitational acceleration.

Answer 1

The Correct Option is A

The weight of one coin is: \[ W = m \cdot g = 0.008 \cdot 9.81 = 0.07848 \, \text{N} \] The gravitational attraction between the coins is: \[ F = \frac{G m^2}{r^2} = \frac{6.67 \times 10^{-11} \cdot (0.008)^2}{(0.5)^2} = 8.53 \times 10^{-13} \, \text{N} \] The ratio is: \[ \frac{W}{F} = \frac{0.07848}{8.53 \times 10^{-13}} \approx 4.6 \times 10^{12} \]