Problem

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Problem 78 Hard Difficulty

According to Newton's Law of Universal Gravitation, the attractive force $F$ between two bodies is given by $F=G \frac{m_{1} m_{2}}{r^{2}}$ where $m_{1}, m_{2}=$ the masses of the two bodies $\begin{aligned} r=& \text { distance between the two bodies } \\ G=& \text { gravitational constant }=6.6742 \times 10^{-11} \\ & \text { newtons } \cdot \text { meter }^{2} \cdot \text { kilogram }^{-2} \end{aligned}$ Suppose an object is traveling directly from Earth to the moon. The mass of Earth is $5.9742 \times 10^{24}$ kilograms, the mass of the moon is $7.349 \times 10^{22}$ kilograms, and the mean distance from Earth to the moon is $384,400$ kilometers. For an object between Earth and the moon, how far from Earth is the force on the object due to the moon greater than the force on the object due to Earth?

Answer

$$
192,400 \text { and } 384,400
$$ $$
\text { kilometers. }
$$

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College Algebra 10e

Chapter 5

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Section 4

Polynomial and Rational Inequalities

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Video Transcript

According to Newton's Law of Gravitation, the force of attraction between two objects has given us G multiplied with M one m two divided by r squared very modern him two of the mass of the two objects Jeez, universal gravitation, constant and RSD distance between the centres off these two bodies. So let's speak X as the distance from the earth. Very the force off attraction Due to the moon, it will be greater than the force off attraction due to the earth. Now we need to find X such that force off attraction doodle The moon at this distance will be greater than force of attraction due to the earth at this distance. So we can say that we need to find the force of attraction due to moon divided by the force of attraction due to ought at this distance should be greater than one. So putting the values according to the formula for the force of attraction. As per Newton, we'll get G multiplied wit m where m is the moss off the body m m being the moss off moon divided by R minus X squared, divided by G. M m E, where m is the mass off the earth divided by X squares should be greater than one. So then g, m and G m will get cancelled because they're the same. So the inequality of allow change toe m m multiplied with X squared is greater than M E multiplied with ar minus. X holds Quit. Now let's all photo Now if we take away mm X squared from both sides so the inequality will become M M X squared minus M m X Trail is greater than M E R Square plus M E X squared minus twice off m E multiplied with ar minus X minus M m X square. So now cancelling out and rearranging, we are going to get the final inequality as zero is greater than M E multiplied with R squared minus to our X plus X squared minus m m exquisite, which will become Mars offered. Miners mass off moon multiplied with X squared minus two are must off Earth multiplied. Videx plus R squared multiplied with Marceau Effort is less than zero now, putting the values form us off earth mass off Moon on the radius. We're going to get 5.9742 multiplied with 10 to the power of 24 miners, 7.349 multiplied with 10 to the power off 20 to minus two multiplied with 384,400 multiplied with 5.9742 multiplied with tend to the part of 24 plus 384,000 and 400 squared multiplied with five point 9749 into 10. Raise to the part of 24 will be less than zero, which will give us on solving that X should lie between one point through seven multiplied with 10 days to the bar seven on day zero. So therefore the solution said will be zero on 1.27 into 10 race to the par off seven kilometers.

Ankit G.

Related Courses

Algebra

College Algebra 10e

Chapter 5

Polynomial and Rational Functions

Section 4

Polynomial and Rational Inequalities

Problem

Field Trip Mrs. West has decided to take her fift…

Problem 78 Hard Difficulty

Answer

$$
192,400 \text { and } 384,400
$$ $$
\text { kilometers. }
$$

Related Courses

College Algebra 10e

Related Topics

Discussion

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Heather Z.

Kayleah T.

Michael J.

Algebra Bootcamp

Absolute Value - Example 1

Absolute Value - Example 2

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Watch More Solved Questions in Chapter 5

Video Transcript

College Algebra 10e

Grace H.

Heather Z.

Kayleah T.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

$$192,400 \text { and } 384,400$$ $$\text { kilometers. }$$

College Algebra 10e

Grace H.

Heather Z.

Kayleah T.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

Grace H.

Heather Z.

Kayleah T.

Michael J.

$$
192,400 \text { and } 384,400
$$ $$
\text { kilometers. }
$$