Question

the force of gravity between two objects is given by $f_g = \frac{-gm_1m_2}{r^2}$, where $g$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the objects, and $r$ is the distance between the objects centers. there is a vertical asymptote at $r = 0$. what does this mean in context of the problem?

as the distance between
the objects approaches 0,
the gravitational force
between them approaches
infinity.

as the masses of the
objects approach 0, the
gravitational force between
them approaches infinity.

as the masses of the
objects approach 0, the
gravitational force between
them approaches 0.

as the distance between
the objects approaches 0,
the gravitational force
between them approaches
0.

Explanation:

To determine the meaning of the vertical asymptote at \( r = 0 \) for the gravitational force formula \( F_g=\frac{-Gm_1m_2}{r^2} \), we analyze the behavior of the function as \( r \) approaches 0. A vertical asymptote at \( r = 0 \) means the function's value (here, \( F_g \)) will approach either positive or negative infinity as \( r \) gets very close to 0 (from the right, since distance can't be negative).

• The first option says "As the distance between the objects approaches 0, the gravitational force between them approaches infinity." This matches the behavior of the function \( \frac{1}{r^2} \), which tends to infinity as \( r \to 0^+ \) (since \( r \) is in the denominator, making the whole term grow without bound).
• The second option incorrectly focuses on masses approaching 0, but the asymptote is about \( r \), not \( m_1 \) or \( m_2 \).
• The third option also focuses on masses approaching 0 and claims the force approaches 0, which is incorrect.
• The fourth option claims the force approaches 0 as \( r \to 0 \), but the function \( \frac{1}{r^2} \) goes to infinity, not 0, as \( r \to 0 \).

Answer:

A. As the distance between the objects approaches 0, the gravitational force between them approaches infinity.