Chapter 4.6, Problem 80AYU

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

$G=gravitationalconstant=6.6742\times {10}^{-11}newtons\cdot mete{r}^2\cdot kilogra{m}^{-2}$

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 $7.349\times {10}^{22}$ 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?

Source: www.solarviews.com;en.wikipedia.org

To determine

To find: 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 to Problem 80AYU

Solution:

$x=202142956.94, 3907477078.59$

Explanation of Solution

Given:

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

$F=\frac{G}{r^2}{m}_1{m}_2$

where $m_1, m_2=$the masses of the two bodies, $r=$ distance between the two bodies, $G=$ gravitational constant $=6.6742\times {10}^{-11}\text{newtons}{\text{meter}}^2{\text{kilogram}}^{-2}$.

Suppose an object is traveling directly from Earth to the moon. The mass of Earth is $5.9742\times {10}^{24}\text{kilograms}$, the mass of the moon is $7.349\times {10}^{22}\text{kilograms}$, and the mean distance from Earth to the moon is 384,400 kilometers.

Calculation:

Let the object at a distance $r$ from the planet. Then the object will be $384400-x$ km from the satellite.

Let the mass of the object be $m$ and its distance from moon be $x$.

Force on object due to planet moon $=F=\frac{G}{r^2}{m}_1{m}_2$

$=\frac{6.6742\times {10}^{-11}\times m\times 7.349\times {10}^{22}}{x^2}$

$=\frac{49.0487\times {10}^{11}\times m}{x^2}$

Force on object due to earth $=\frac{6.6742\times {10}^{-11}\times m\times 5.9742\times {10}^{24}}{{\left(384400\times {10}^3-x\right)}^2}$

$=\frac{39.873\times {10}^{13}\times m}{{\left(384400\times {10}^3-x\right)}^2}$

We want $\frac{49.0487\times {10}^{11}\times m}{x^2}>\frac{39.873\times {10}^{13}\times m}{{\left(384400\times {10}^3-x\right)}^2}$

$\frac{49.0487}{x^2}>\frac{39.873\times {10}^2}{{\left(384400\times {10}^3-x\right)}^2}$

$49.0487{\left(3844\times {10}^5-x\right)}^2>3987.3\times x^2$

Solve the inequality we get $x=202142956.94,3907477078.59$.