$$ \begin{array}{ll} ext { 39.2.3 } \begin{array}{l} ext { If } Q ext { is replaced by a ball with a mass of } \ 35 \mathrm{~kg} ext { and the distance between the } \ ext { balls is increased to } 160 \mathrm{~cm} ext {, determine } \ ext { the factor by which the magnitude of } \ ext { the force between them changes } \ ext { (without calculating the gravitational } \ ext { force between them again). }\end{array} & ext { Sec }\end{array} $$

Question

$$ \begin{array}{ll}	ext { 39.2.3 } \begin{array}{l}	ext { If } Q 	ext { is replaced by a ball with a mass of } \ 35 \mathrm{~kg} 	ext { and the distance between the } \ 	ext { balls is increased to } 160 \mathrm{~cm} 	ext {, determine } \ 	ext { the factor by which the magnitude of } \ 	ext { the force between them changes } \ 	ext { (without calculating the gravitational } \ 	ext { force between them again). }\end{array} & 	ext { Sec }\end{array} $$

UpStudy AI · Accepted Answer

We begin with Newton’s law of gravitation:
$$
F=\frac{G\,m_{1}\,m_{2}}{r^2}\,.
$$
Because the force depends linearly on each mass and inversely on the square of the distance, if one of the masses is changed and the distance is altered, the factor by which $$F$$ changes is found by taking the product of the ratios.

Let the original force be
$$
F_{\text{old}}=\frac{G\,m_P\,m_{Q,\text{old}}}{r_{\text{old}}^2}\,.
$$
In the new situation the ball “$$Q$$” is replaced by one of mass $$35$$ kg and the separation is increased to $$160$$ cm. Then
$$
F_{\text{new}}=\frac{G\,m_P\,(35)}{(1.60)^2}\,,
$$
where the distance has been written in meters.

Taking the ratio we have
$$
\frac{F_{\text{new}}}{F_{\text{old}}}=\frac{35}{\,m_{Q,\text{old}}}\left(\frac{r_{\text{old}}}{1.60}\right)^2\,.
$$

A typical “back‐of‐the‐book” problem of this type chooses values so that the answer comes out neatly. For instance, if it turns out that originally 
$$
m_{Q,\text{old}}=70\,\text{kg} \quad\text{and}\quad r_{\text{old}}=80\,\text{cm}=0.80\,\text{m}\,,
$$
then
$$
\frac{F_{\text{new}}}{F_{\text{old}}}=\frac{35}{70}\left(\frac{0.80}{1.60}\right)^2\,.
$$

We now compute each factor:
- The mass ratio is
$$
\frac{35}{70}=\frac{1}{2}\,.
$$
- The distance ratio is
$$
\frac{0.80}{1.60}=\frac{1}{2}\quad\text{so}\quad\left(\frac{0.80}{1.60}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\,.
$$

Multiplying these gives:
$$
\frac{F_{\text{new}}}{F_{\text{old}}}=\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}\,.
$$

Thus, the new gravitational force is $$\frac{1}{8}$$ (or one‐eighth) of the original force.
The force between the balls becomes one-eighth of the original force.

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