Question

the gravitational force formula is ( f = \frac{gm_1m_2}{r^2} ), where ( f ) is the force between two objects, ( g ) is the constant of gravitation, ( m_1 ) is the mass of the first object, ( m_2 ) is the mass of the second object, and ( r ) is the distance between the objects. by rewriting the formula as ( r = sqrt{\frac{gm_1m_2}{f}} ), you can find the distance between objects. which of the following gives the distance, ( r ), in simplest form?

( r = \frac{sqrt{gm_1m_2}}{f} )

( r = \frac{sqrt{gm_1m_2f}}{f} )

( r = sqrt{gm_1m_2f} )

Explanation:

Step1: Recall radical simplification rules

To simplify \( r = \sqrt{\frac{Gm_1m_2}{F}} \), we can use the property of square roots: \( \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}} \) (where \( a\geq0,b > 0 \)). Here, \( a = Gm_1m_2 \) and \( b=F \).

Step2: Apply the radical property

Applying the property \( \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}} \) to \( \sqrt{\frac{Gm_1m_2}{F}} \), we get \( r=\frac{\sqrt{Gm_1m_2}}{\sqrt{F}} \). But we usually rationalize the denominator. Multiply the numerator and denominator by \( \sqrt{F} \):
\[

$$\begin{align*} r&=\frac{\sqrt{Gm_1m_2}\times\sqrt{F}}{\sqrt{F}\times\sqrt{F}}\\ &=\frac{\sqrt{Gm_1m_2F}}{F} \end{align*}$$

\]
Wait, there is a mistake in the initial analysis. Wait, the original formula is \( r = \sqrt{\frac{Gm_1m_2}{F}} \). Let's re - express it correctly. We know that \( \sqrt{\frac{x}{y}}=\frac{\sqrt{xy}}{y} \) (by rationalizing the denominator: \( \sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}=\frac{\sqrt{x}\times\sqrt{y}}{\sqrt{y}\times\sqrt{y}}=\frac{\sqrt{xy}}{y} \)). Here \( x = Gm_1m_2 \) and \( y = F \), so \( r=\frac{\sqrt{Gm_1m_2F}}{F} \), which is the second option.

Wait, let's check the first option: \( r=\frac{\sqrt{Gm_1m_2}}{F} \). If we square this, we get \( \frac{Gm_1m_2}{F^{2}} \), but the square of \( \sqrt{\frac{Gm_1m_2}{F}} \) is \( \frac{Gm_1m_2}{F} \), so the first option is wrong. The third option: if we square \( \sqrt{Gm_1m_2F} \), we get \( Gm_1m_2F \), which is not equal to \( \frac{Gm_1m_2}{F} \) (the square of the original \( r \)). The second option: if we square \( \frac{\sqrt{Gm_1m_2F}}{F} \), we have \( \frac{Gm_1m_2F}{F^{2}}=\frac{Gm_1m_2}{F} \), which matches the square of the original \( r=\sqrt{\frac{Gm_1m_2}{F}} \) (since \( (\sqrt{\frac{Gm_1m_2}{F}})^2=\frac{Gm_1m_2}{F} \)).

Answer:

\( r = \frac{\sqrt{Gm_1m_2F}}{F} \) (the second option among the given options)