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Chapter 13: Q.39 - Excercises And Problems (page 354)

Two spherical objects have a combined mass of $150 k g .$ The gravitational attraction between them is $8.00 脳 10^{- 6} N$ when their centers are $20 c m$ apart. What is the mass of each?

Short Answer

Expert verified

The mass of each spherical object is $m_{1} = 104 k g o r 46 k g; m_{2} = 46 k g o r 104 k g$

Step by step solution

01

Given information

Combined mass $m_{1} + m_{2} = 150 k g$

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

center distance $r = 20 c m = 0.2 m$

Gravitational constant $G = 6.67 脳 10^{- 11} m^{3} / k g s^{2}$

02

Explanation

The combined mass of the spherical object is given by

$m_{1} + m_{2} = 150 k g m_{2} = (150 - m_{1}) k g (1)$

The magnitude of force is given by

$F_{12} = G \frac{m_{1} m_{2}}{r^{2}} = \frac{6.67 脳 10^{- 11} 脳 m_{1} (150 - m_{1})}{{(0.2)}^{2}} 150 m_{1} - {m_{1}}^{2} = \frac{8.00 脳 10^{- 6} 脳 {(0.2)}^{2}}{6.67 脳 10^{- 11}} 150 m_{1} - {m_{1}}^{2} = 4784 {m_{1}}^{2} - 150 m_{1} + 4784 = 0$

On simplifying the quadratic equations we get two real roots,

${m_{1}}^{2} - 150 m_{1} + 4784 = 0 {m_{1}}^{2} - 104 m_{1} - 46 m_{1} + 4784 = 0 m_{1} (m_{1} - 104) - 46 (m_{1} - 104) = 0 (m_{1} - 104) (m_{1} - 46) = 0 m_{1} = 104 k g o r m_{1} = 46 k g$

Substituting the values of $m_{1}$ in Eqn. $(1)$ , we get

$m_{2} = 46 k g o r m_{1} = 104 k g$

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