Question

Two ocean liners, each with a mass of 40 000 metric tons, are moving on parallel courses 100 m apart. What is the magnitude of the acceleration of one of the liners toward the other due to their mutual gravitational attraction? Model the ships as particles.

Short answer

Acceleration at any ocean liners is$a=2.66\times {10}^{-7}{\text{m/s}}^{\text{2}}$

Step-by-step solution

Newton’s law

Newton’s law of universal gravitation states that the gravitational force of attraction between any two particles of masses $M_1$and $M_2$separated by a distance r has the magnitude:

$F_g=G\frac{m_1.m_2}{r^2}$

Where, $G=6.674\times {10}^{-11}\text{N}·{\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$is the universal gravitational constant. This equation enables us to calculate the force of attraction between masses under many circumstances.

Find the magnitude of the acceleration

Given mass of ocean liners: $m=40000metrictons$

$$\begin{gathered} 1metrictons={10}^{3}kg \\ m=40000\times {10}^{3}kg \end{gathered}$$

Distance between them $d=100\text{m}$

According to Newton’s law of universal gravitation, we have

$$\begin{gathered} F_g=G\frac{m_1.m_2}{r^2} \\ {F}_g=6.67\times {10}^{-11}\frac{40\times {10}^6.40\times {10}^6}{{100}^2} \\ {F}_g=\frac{106.72\times {10}^{-11}\times {10}^{14}}{{10}^4} \\ {F}_g=106.72\times {10}^{14-11-4} \\ {F}_g=106.72\times {10}^{-1} \\ {F}_g=10.672\text{N} \end{gathered}$$

This is the force by which they pull each other.

We know, acceleration at any ocean liners can be given by Newton’s second law.

localid="1663665513980" $$\begin{gathered} F_{net}=ma \\ 10.67=40\times {10}^6\times a \\ a=\frac{10.67}{40\times {10}^6} \\ a=2.66\times {10}^{-7}{\text{m/s}}^{\text{2}} \end{gathered}$$

$a=2.66\times {10}^{-7}{\text{m/s}}^{\text{2}}$

Hence, the Acceleration at any ocean liners is$a=2.66\times {10}^{-7}{\text{m/s}}^{\text{2}}$