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

To find how many times larger the weight of one coin is than the gravitational attraction between the two coins, we start by calculating both the weight of one coin and the gravitational force between the two coins.

Step 1: Calculate the weight of one coin.

The weight \( W \) of one coin is given by:

\( W = m \cdot g \)

where \( m = 8 \, \text{g} = 0.008 \, \text{kg} \) (since 1 g = 0.001 kg) and \( g = 9.81 \, \text{m/s}^2 \).

Substituting the values:

\( W = 0.008 \times 9.81 = 0.07848 \, \text{N} \)

Step 2: Calculate the gravitational attraction between the two coins.

The gravitational force \( F \) between two objects is calculated using Newton's law of universal gravitation:

\( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \)

where \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), \( m_1 = m_2 = 0.008 \, \text{kg} \), and \( r = 50 \, \text{cm} = 0.5 \, \text{m} \).

Substituting the values:

\( F = \frac{6.67 \times 10^{-11} \times 0.008 \times 0.008}{0.5^2} = \frac{4.2688 \times 10^{-15}}{0.25} = 1.70752 \times 10^{-14} \, \text{N} \)

Step 3: Calculate the ratio of the weight to the gravitational attraction.

The ratio \( R \) is given by:

\( R = \frac{W}{F} \)

Substituting the values we calculated:

\( R = \frac{0.07848}{1.70752 \times 10^{-14}} \approx 4.6 \times 10^{12} \)

Thus, the weight of one coin is approximately \( 4.6 \times 10^{12} \) times larger than the gravitational attraction between the two coins.

Answer 2

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: \[ \frac{W}{F} = \frac{0.07848}{8.53 \times 10^{-13}} \approx 4.6 \times 10^{12} \]