Question

there is a gravitational force between a lamp and a dog. if the distance between them were doubled, what would happen to the gravitational force?
$f = -g \frac{m_1m_2}{d^2}$, $f \propto \frac{1}{d^2}$
it would go to zero
it would decrease
it would be the same
it would increase

Explanation:

Step1: Recall the gravitational force relationship

The gravitational force \( F \) is proportional to \( \frac{1}{d^2} \), where \( d \) is the distance between the two objects. Mathematically, this is given by \( F \propto \frac{1}{d^2} \).

Step2: Analyze the effect of doubling the distance

Let the original distance be \( d_1 \) and the new distance be \( d_2 = 2d_1 \). The original force \( F_1 \propto \frac{1}{d_1^2} \), and the new force \( F_2 \propto \frac{1}{d_2^2} = \frac{1}{(2d_1)^2} = \frac{1}{4d_1^2} \).

Step3: Compare the new force with the original force

We can find the ratio of the new force to the original force: \( \frac{F_2}{F_1} = \frac{\frac{1}{4d_1^2}}{\frac{1}{d_1^2}} = \frac{1}{4} \). This shows that the new force \( F_2 \) is \( \frac{1}{4} \) of the original force \( F_1 \), meaning the force decreases.

Answer:

it would decrease