Question

compared to the bodies in figure 1, the gravitational attraction between the bodies in figure 2 is
a one - half as strong.
b one - fourth as strong.
c twice as strong.
d four times as strong.
question #10
stephen learned that any two objects exert a gravitational force on each other. if the distance between
two objects triples, the gravitational force between them will change by a factor of what?
$f = g\frac{m_1m_2}{r^2}$
a $\frac{1}{9}$
b $\frac{1}{8}$
c 6
d 9

Explanation:

Step1: Analyze the formula for gravitational force

The formula for gravitational force is \(F = G\frac{m_1m_2}{r^2}\), where \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the two objects, and \(r\) is the distance between them.

Step2: Consider the change in distance

Let the initial distance be \(r_1\) and the final distance be \(r_2 = 3r_1\). The initial force \(F_1=G\frac{m_1m_2}{r_1^2}\), and the final force \(F_2 = G\frac{m_1m_2}{r_2^2}\). Substitute \(r_2 = 3r_1\) into the formula for \(F_2\): \(F_2=G\frac{m_1m_2}{(3r_1)^2}=G\frac{m_1m_2}{9r_1^2}\).

Step3: Find the ratio of the forces

Divide \(F_2\) by \(F_1\): \(\frac{F_2}{F_1}=\frac{G\frac{m_1m_2}{9r_1^2}}{G\frac{m_1m_2}{r_1^2}}\). The \(G\), \(m_1\), and \(m_2\) terms cancel out, leaving \(\frac{F_2}{F_1}=\frac{1}{9}\).

Answer:

A. \(\frac{1}{9}\)