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

Step 1: Write both quantities in scientific notation. Weight \( W = mg = (8 \times 10^{-3})(9.81) = 7.848 \times 10^{-2} \, \text{N} \).

Step 2: Gravitational force \( F = \dfrac{Gm^2}{r^2} = \dfrac{(6.67 \times 10^{-11})(8 \times 10^{-3})^2}{(5 \times 10^{-1})^2} = \dfrac{(6.67 \times 10^{-11})(6.4 \times 10^{-5})}{2.5 \times 10^{-1}} \).

Step 3: Multiply the mantissas and add the exponents separately: numerator \( = 42.688 \times 10^{-16} = 4.2688 \times 10^{-15} \); dividing by \(2.5 \times 10^{-1}\) gives \( F = 1.708 \times 10^{-14} \, \text{N} \).

Step 4: Divide the two results, again handling mantissa and exponent separately: \( \dfrac{7.848 \times 10^{-2}}{1.708 \times 10^{-14}} = 4.596 \times 10^{12} \).\[ \boxed{\dfrac{W}{F} \approx 4.6 \times 10^{12}} \]

Answer 2

Rather than comparing forces directly, compare the two accelerations a coin would feel: Earth pulls it down with acceleration \( g \), while the other coin pulls it sideways with acceleration \( a' = \dfrac{Gm}{r^2} \) (only one mass appears here, since acceleration doesn't depend on the mass being accelerated). Since both forces act on the same coin, the ratio of forces equals the ratio of accelerations, so \( \dfrac{W}{F} = \dfrac{g}{a'} \).

Computing \( a' = \dfrac{(6.67 \times 10^{-11})(0.008)}{(0.5)^2} = \dfrac{5.336 \times 10^{-13}}{0.25} = 2.134 \times 10^{-12} \, \text{m/s}^2 \), an acceleration many trillion times smaller than \( g = 9.81 \, \text{m/s}^2 \).

Taking the ratio: \( \dfrac{g}{a'} = \dfrac{9.81}{2.134 \times 10^{-12}} \approx 4.6 \times 10^{12} \), confirming the same result reached from the force comparison.\[ \boxed{4.6 \times 10^{12}} \]