Question

putting it together: gravity, mass, and distance
there is a gravitational force between a 1,500 kg car and a nearby 230 kg lawn mower. if the car were 750 kg instead, how would the gravitational force change?
$f = -g\frac{m_1m_2}{d^2}$
a fourth as much\ta half as much
it would be the same\ttwice as much

Explanation:

Step1: Identify the formula and variables

The gravitational force formula is \( F = -G\frac{m_1m_2}{d^2} \). Here, \( m_1 \) is the mass of the car, \( m_2 \) is the mass of the lawn mower, \( G \) is the gravitational constant, and \( d \) is the distance between them. We assume \( G \), \( m_2 \), and \( d \) remain constant.

Step2: Calculate the ratio of the new force to the original force

Let the original mass of the car be \( m_{1_{original}} = 1500 \, \text{kg} \) and the new mass be \( m_{1_{new}} = 750 \, \text{kg} \). The original force \( F_{original} = -G\frac{m_{1_{original}}m_2}{d^2} \), and the new force \( F_{new} = -G\frac{m_{1_{new}}m_2}{d^2} \).
To find the ratio \( \frac{F_{new}}{F_{original}} \), we divide the two equations:
\[
\frac{F_{new}}{F_{original}}=\frac{-G\frac{m_{1_{new}}m_2}{d^2}}{-G\frac{m_{1_{original}}m_2}{d^2}}
\]
The \( -G \), \( m_2 \), and \( d^2 \) terms cancel out, leaving:
\[
\frac{F_{new}}{F_{original}}=\frac{m_{1_{new}}}{m_{1_{original}}}
\]
Substitute \( m_{1_{new}} = 750 \, \text{kg} \) and \( m_{1_{original}} = 1500 \, \text{kg} \):
\[
\frac{F_{new}}{F_{original}}=\frac{750}{1500}=\frac{1}{2}
\]
This means the new force \( F_{new} \) is \( \frac{1}{2} \) of the original force \( F_{original} \), so the gravitational force becomes a half as much.

Answer:

a half as much