Question

there is a gravitational force between a 5 kg ball and a nearby 63,000 kg building. if the ball were 2.5 kg instead, how would the gravitational force change?
$f = -g \frac{m_1m_2}{d^2}$
four times as much twice as much
it would be the same half as much

Explanation:

Step1: Recall Gravitational Force Formula

The gravitational force formula is \( F = -G\frac{m_1m_2}{d^2} \), where \( m_1 \) and \( m_2 \) are the masses of the two objects, \( d \) is the distance between them, and \( G \) is the gravitational constant. Here, the mass of the building (\( m_2 \)) and the distance (\( d \)) remain constant, only the mass of the ball (\( m_1 \)) changes. Let the initial mass of the ball be \( m_{1i} = 5\space kg \) and the final mass be \( m_{1f}=2.5\space kg \). The initial force \( F_i=-G\frac{m_{1i}m_2}{d^2} \), and the final force \( F_f=-G\frac{m_{1f}m_2}{d^2} \).

Step2: Find the Ratio of Final Force to Initial Force

Calculate the ratio \( \frac{F_f}{F_i} \):
\[
\frac{F_f}{F_i}=\frac{-G\frac{m_{1f}m_2}{d^2}}{-G\frac{m_{1i}m_2}{d^2}}
\]
The \( -G \), \( m_2 \), and \( d^2 \) terms cancel out, leaving \( \frac{F_f}{F_i}=\frac{m_{1f}}{m_{1i}} \).

Step3: Substitute the Mass Values

Substitute \( m_{1i} = 5\space kg \) and \( m_{1f}=2.5\space kg \) into the ratio:
\[
\frac{F_f}{F_i}=\frac{2.5}{5}=\frac{1}{2}
\]
This means \( F_f=\frac{1}{2}F_i \), so the gravitational force becomes half as much as the initial force.

Answer:

half as much