Question

there is a gravitational force between a lamp and a dog. if the distance between them were doubled, how much would the gravitational force change?
$f = -g \frac{m_1m_2}{d^2}$, $f \propto \frac{1}{d^2}$
it would be the same
twice as much
a fourth as much
half as much

Explanation:

Step1: Recall the gravitational force relationship

The gravitational force \( F \) is proportional to \( \frac{1}{d^2} \), i.e., \( F \propto \frac{1}{d^2} \). Let the initial distance be \( d_1 = d \) and the initial force be \( F_1 \), so \( F_1 \propto \frac{1}{d_1^2}=\frac{1}{d^2} \).

Step2: Determine the new distance

The new distance \( d_2 = 2d \) (since the distance is doubled). Let the new force be \( F_2 \), then \( F_2 \propto \frac{1}{d_2^2}=\frac{1}{(2d)^2} \).

Step3: Find the ratio of new force to initial force

Calculate the ratio \( \frac{F_2}{F_1} \). Since \( F_1 \propto \frac{1}{d^2} \) and \( F_2 \propto \frac{1}{(2d)^2} \), we have:
\[
\frac{F_2}{F_1}=\frac{\frac{1}{(2d)^2}}{\frac{1}{d^2}}=\frac{d^2}{(2d)^2}=\frac{d^2}{4d^2}=\frac{1}{4}
\]
So \( F_2=\frac{1}{4}F_1 \), meaning the gravitational force becomes a fourth as much.

Answer:

a fourth as much