Question

cphs : advanced algebra: concepts and connections - block (27.0831030)
dividing radicals
solving a real - world problem
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: Start with the formula \( F = \frac{Gm_1m_2}{r^2} \)

We need to solve for \( r \). First, multiply both sides by \( r^2 \) to get \( F r^2 = G m_1 m_2 \).

Step2: Divide both sides by \( F \)

This gives \( r^2 = \frac{G m_1 m_2}{F} \).

Step3: Take the square root of both sides

Taking the square root of both sides, we have \( r = \sqrt{\frac{G m_1 m_2}{F}} \). We can also rewrite the square root of a fraction as the fraction of the square roots: \( r=\frac{\sqrt{Gm_1m_2}}{\sqrt{F}} \). To rationalize the denominator, we multiply the numerator and denominator by \( \sqrt{F} \), so \( r = \frac{\sqrt{Gm_1m_2}\times\sqrt{F}}{\sqrt{F}\times\sqrt{F}}=\frac{\sqrt{Gm_1m_2F}}{F} \)? Wait, no, wait. Wait, original step: \( r^2=\frac{Gm_1m_2}{F} \), so \( r = \sqrt{\frac{Gm_1m_2}{F}} \). Let's check the options. Wait, maybe I made a mistake. Wait, let's re-express \( \sqrt{\frac{Gm_1m_2}{F}} \) as \( \frac{\sqrt{Gm_1m_2}}{\sqrt{F}} \), then multiply numerator and denominator by \( \sqrt{F} \): \( \frac{\sqrt{Gm_1m_2}\times\sqrt{F}}{\sqrt{F}\times\sqrt{F}}=\frac{\sqrt{Gm_1m_2F}}{F} \)? No, wait, \( \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}} \), and if we rationalize, \( \frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{ab}}{b} \). So \( \sqrt{\frac{Gm_1m_2}{F}}=\frac{\sqrt{Gm_1m_2F}}{F} \)? Wait, no, \( \sqrt{\frac{a}{b}}=\frac{\sqrt{a}\times\sqrt{b}}{\sqrt{b}\times\sqrt{b}}=\frac{\sqrt{ab}}{b} \). So here, \( a = Gm_1m_2 \), \( b = F \), so \( \sqrt{\frac{Gm_1m_2}{F}}=\frac{\sqrt{Gm_1m_2F}}{F} \)? Wait, no, \( \sqrt{\frac{Gm_1m_2}{F}}=\frac{\sqrt{Gm_1m_2}}{\sqrt{F}}=\frac{\sqrt{Gm_1m_2}\times\sqrt{F}}{F}=\frac{\sqrt{Gm_1m_2F}}{F} \). Wait, but let's check the options. Wait, the first option is \( r=\frac{\sqrt{Gm_1m_2}}{F} \)? No, wait the first option's numerator is \( \sqrt{Gm_1m_2} \), denominator \( F \)? No, the first option in the image (top box) is \( r=\frac{\sqrt{Gm_1m_2}}{F} \)? Wait, no, looking at the image:

First box: \( r = \frac{\sqrt{Gm_1m_2}}{F} \)? No, wait the first box's formula is \( r=\frac{\sqrt{Gm_1m_2}}{F} \)? Wait, no, maybe the options are:

First option: \( r=\frac{\sqrt{Gm_1m_2}}{F} \) – no, that can't be. Wait, second option: \( r=\frac{\sqrt{Gm_1m_2F}}{F} \)? Wait, third option: \( r=\sqrt{Gm_1m_2F} \)?

Wait, let's redo the algebra correctly.

Starting with \( F = \frac{Gm_1m_2}{r^2} \)

1. Multiply both sides by \( r^2 \): \( F r^2 = G m_1 m_2 \)
2. Divide both sides by \( F \): \( r^2 = \frac{G m_1 m_2}{F} \)
3. Take square root: \( r = \sqrt{\frac{G m_1 m_2}{F}} \)

We can write this as \( r=\frac{\sqrt{Gm_1m_2}}{\sqrt{F}} \). To rationalize the denominator, multiply numerator and denominator by \( \sqrt{F} \):

\( r = \frac{\sqrt{Gm_1m_2} \times \sqrt{F}}{\sqrt{F} \times \sqrt{F}} = \frac{\sqrt{Gm_1m_2F}}{F} \)

Wait, so that's \( r=\frac{\sqrt{Gm_1m_2F}}{F} \), which would be the second option? Wait, the second box in the image: \( r=\frac{\sqrt{Gm_1m_2F}}{F} \)? Wait, the third box is \( r=\sqrt{Gm_1m_2F} \), first is \( r=\frac{\sqrt{Gm_1m_2}}{F} \). Wait, no, maybe I messed up the options. Wait, let's check the original problem again.

Wait, the problem says "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?"

Wait, the first option: \( r=\frac{\sqrt{Gm_1m_2}}{F} \) – no, that's not. Wait, \( \sqrt{\frac{Gm_1m_2}{F}}=\frac{\sqrt{Gm_1m_2}}{\sqrt{F}} \), and if we rationalize, it's \( \frac{\sqrt{Gm_1m_2F}}{F} \). So the second option (middle box) is \( r=\frac{\sqrt{Gm_1m_2F}}{F} \)? Wait, or maybe the third option? Wait, no, let's compute again.

Wait, \( r^2=\frac{Gm_1m_2}{F} \), so \( r = \sqrt{\frac{Gm_1m_2}{F}} \). Let's square each option to see which one gives \( r^2=\frac{Gm_1m_2}{F} \).

First option: \( r=\frac{\sqrt{Gm_1m_2}}{F} \). Then \( r^2=\frac{Gm_1m_2}{F^2} \) – not equal to \( \frac{Gm_1m_2}{F} \).

Second option: \( r=\frac{\sqrt{Gm_1m_2F}}{F} \). Then \( r^2=\frac{Gm_1m_2F}{F^2}=\frac{Gm_1m_2}{F} \) – which matches.

Third option: \( r=\sqrt{Gm_1m_2F} \). Then \( r^2=Gm_1m_2F \) – not matching.

So the correct answer is the second option (middle box), which is \( r = \frac{\sqrt{Gm_1m_2F}}{F} \).

Answer:

The middle box (second option) with \( r = \frac{\sqrt{Gm_1m_2F}}{F} \)