Question

Two spherical objects have a combined mass of $150kg.$The gravitational attraction between them is $8.00\times {10}^{-6}N$when their centers are $20cm$apart. What is the mass of each?

Short answer

The mass of each spherical object is$m_1=104kgor46kg;m_2=46kgor104kg$

Step-by-step solution

Given information

Combined mass $m_1+m_2=150kg$

The gravitational force $F_{12}=8.00\times {10}^{-6}N$

center distance $r=20cm=0.2m$

Gravitational constant$G=6.67\times {10}^{-11}{m}^3/kg{s}^2$

Explanation

The combined mass of the spherical object is given by

$$\begin{gathered} m_1+m_2=150kg \\ {m}_2=(150-m_1)kg\left(1\right) \end{gathered}$$

The magnitude of force is given by

$$\begin{gathered} F_{12}=G\frac{m_1{m}_2}{r^2}=\frac{6.67\times {10}^{-11}\times m_1(150-m_1)}{{(0.2)}^2} \\ 150m_1-{m_1}^2=\frac{8.00\times {10}^{-6}\times {(0.2)}^2}{6.67\times {10}^{-11}} \\ 150m_1-{m_1}^2=4784 \\ {m_1}^2-150m_1+4784=0 \end{gathered}$$

On simplifying the quadratic equations we get two real roots,

$$\begin{gathered} {m_1}^2-150m_1+4784=0 \\ {m_1}^2-104m_1-46m_1+4784=0 \\ {m}_1(m_1-104)-46(m_1-104)=0 \\ (m_1-104)(m_1-46)=0 \\ {m}_1=104kgor{m}_1=46kg \end{gathered}$$

Substituting the values of $m_1$in Eqn. $\left(1\right)$, we get

$m_2=46kgor{m}_1=104kg$