Mathematico | Mathematico

When placed $r$ meters apart, the gravitational force, $F,$ between two particles with masses $m_{1}$ and $m_{2}$ is given by

F = \frac{G m_{1} m_{2}}{r^{2}}

where $G = 6.674 \times 10^{- 11}$ is known as the gravitational constant, and $F$ is measured in Newtons.

Two particles are placed $1.75 m$ apart, and the force between them is measured as $1.9 \times 10^{- 9} N .$ Given that one particle is $18 kg$ heavier than the other, find the mass of the lighter particle.

Give your answer in $kg$ , correct to three significant figures.

Hint

Let the lighter particle have mass $m,$ so the heavier particle has mass $m + 18$ . Substitute all of the information into the formula and solve the resulting quadratic.

Solution

Let the lighter particle have mass $m,$ so the heavier particle has mass $m + 18$ . Then the formula gives

\begin{aligned} 1.9 \times 10^{- 9} & = \frac{6.674 \times 10^{- 11} \times m (m + 18)}{{1.75}^{2}} \\ m (m + 18) & = \frac{1.9 \times 10^{- 9} \times {1.75}^{2}}{6.674 \times 10^{- 11}} \\ = 87.185 \dots \\ m^{2} + 18 m - 87.185 \dots & = 0 \end{aligned}

Now by the quadratic formula

\begin{aligned} m & = \frac{- 18 \pm \sqrt{18^{2} - 4 (1) (- 87.185 \dots)}}{2} \\ = 3.97 (3 s.f.) \end{aligned}