Question

Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the other? Treat the ships as particles and assume each has a mass of 39,000 metric tons. (Give the magnitude of your answer in m/s2.)

Answer 1

Answer:

Acceleration of the ship, $a=2.14\times 10^{-7}\ m/s^2$

Explanation:

It is given that,

Mass of both ships, $m=39000\ metric\ tons=39\times 10^6\ kg$

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

$F=G\dfrac{m^2}{d^2}$

$F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}$

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

$a=\dfrac{F}{m}$

$a=\dfrac{8.38\ N}{39\times 10^6\ kg}$

$a=2.14\times 10^{-7}\ m/s^2$

So, the acceleration of either ship due to the gravitational attraction of the other is $2.14\times 10^{-7}\ m/s^2$. Hence, this is the required solution.

Answer 2

The Acceleration of the ship will be   $a=2.14\times10^{-7} \dfrac{m}{s}$

What will be the acceleration of the ship?

It is given that,

Mass of both ships,   $m=39000MT=39\times10^{6} KG$

Distance between two ships,  $d= 110 m$

The gravitational force between two ships is given by

$F=\dfrac{Gm_{1} m_{2} }{d^{2} }$

$F=\dfrac{6.67\times10^{-11} \times 39\times10^{} }{110^{2} }$

$F=8.38N$

Now newtons second law

$F=m\times a$

Here is the acceleration

$a=\dfrac{f}{m}$

$a=\dfrac{8.38}{39\times10^{6} }$

$a=2.14\times10^{-7} \dfrac{m}{s^{2} }$

Thus the Acceleration of the ship will be   $a=2.14\times10^{-7} \dfrac{m}{s}$

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