Question

$7.349\times {10}^{22}$Gravitational Force 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, $r=$distance between the two bodies,$\text{G=gravitational}\text{constant}=6.6742\times {10}^{-11}\text{newtons meter}{}^2\text{kilogram}^{-2}$

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 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?

Short answer

The distance of the earth to the object is 346022.35 km.

Step-by-step solution

Given information

The attractive force $F$between two bodies is given by:

$F=G\frac{m_1{m}_2}{r^2}$

Determine the relation between masses and radius.

The gravitational pull of the earth to the object is equal to the gravitational pull of the moon of the object, thus:

$F_1=F_2$

Using the Law of Gravitation, let $m_e$the mass of earth, $m_m$the mass of moon, $r_1$the distance from object to earth and $r_2$the distance from the object to moon.

$\mathrm{G}\frac{m_e{m}_1}{r_1^2}=\mathrm{G}\frac{m_m{m}_1}{r_2^2}$

Let $m_1$be the mass of the object. Since the mass of the object is not stated, it is assumed to be negligible.

$\frac{m_e}{r_1^2}=\frac{m_m}{r_2^2}$

Let $x$be the distance of the object to Earth.

Step 3

Since the distance from from of earth to moon is 384400 km, let $384400-x$ be the distance of the object to moon.

Substitute $m_e=5.9742\times {10}^{24},m_m=7.349\times {10}^{22},r_1=x$and $r_2=384400-x$.

$\frac{5.9742\times {10}^{24}}{x^2}=\frac{7.349\times {10}^{22}}{(384400-x{)}^2}$

Solve for x

$\begin{array}{rcl}\frac{5.9742\times {10}^{24}}{x^2}& =& \frac{7.349\times {10}^{22}}{(384400-x{)}^2}\\ \frac{\sqrt{5.9742\times {10}^{24}}}{x}& =& \frac{\sqrt{7.349\times {10}^{22}}}{(384400-x)}\\ (384400-x)\sqrt{5.9742\times {10}^{24}}& =& x\sqrt{7.349\times {10}^{22}}\\ (384400-x)2444217666248.2417& =& 271090390829.33206x\\ x& =& 346022.34853\end{array}$

Write the conclusion

Since the distance of earth to moon is 384400 km, use the smaller value of $x$. The distance of the earth to the object is 346022.35 km.