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

Answer 3

This question compares the weight of a coin (its usual gravitational pull toward Earth) with the tiny gravitational attraction it feels from a second identical coin sitting half a metre away. Instead of computing the weight and the inter-coin force as two independent numbers and then dividing them, it is quicker to write the ratio symbolically first and cancel out one factor of the coin's mass before touching a calculator.

Weight: \( W = mg \). Gravitational pull of coin 2 on coin 1: \( F = \dfrac{Gm^2}{r^2} \). Their ratio is

\[ \frac{W}{F} = \frac{mg}{\dfrac{Gm^2}{r^2}} = \frac{g r^2}{G m} \]

Only one mass survives in the final expression, which cuts down the arithmetic. Substituting \( g = 9.81 \, \text{m/s}^2 \), \( r = 0.5 \, \text{m} \), \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), and \( m = 0.008 \, \text{kg} \):

\[ \frac{W}{F} = \frac{9.81 \times (0.5)^2}{6.67 \times 10^{-11} \times 0.008} = \frac{9.81 \times 0.25}{5.336 \times 10^{-13}} = \frac{2.4525}{5.336 \times 10^{-13}} \approx 4.6 \times 10^{12} \]

1. Option 1 (\(4.6 \times 10^{12}\)): matches the ratio obtained above exactly, so this is the value the calculation supports.
2. Option 2 (\(4.6 \times 10^{10}\)): two powers of ten too small; this would only arise if the separation were mistakenly used as 50 m instead of 0.5 m, inflating the denominator by \(10^4\) and driving the ratio down by the same factor.
3. Option 3 (\(4.6 \times 10^{14}\)): two powers of ten too large; this happens if the mass were left in grams instead of kilograms in the calculation, shifting the answer the wrong way.
4. Option 4 (None): ruled out because Option 1 reproduces the calculated ratio exactly.

The weight of a coin is therefore about \(4.6 \times 10^{12}\) times larger than the gravitational pull the neighbouring coin exerts on it, a good illustration of just how feeble gravity is between everyday-sized objects.

So the correct answer is \(4.6 \times 10^{12}\).