You see, Its moving finally!

Classical Mechanics Level 4

Congratulations! You just bought a brand new iron rod that has a length of 1 lightyear .

Now, suppose you pushed one end of the rod.

How much time (in seconds) will pass before you see that the other end moves?

Speed of Light, c 0 = 3 × 1 0 8 m / s c_0=3\times 10^8 m/s

Your eye can see anything at any distance, so you don't need to worry about that!

You must use this Resource for the standard data - Engineering Tools Box

Its suggested that you solve Dinesh Nath Goswami's Will it Move?


The answer is 1.844E+12.

Muhammad Arifur Rahman by Muhammad Arifur Rahman , 23, Bangladesh

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1 solution

You might've known that Force travels with speed of sound . Yes, in Newtonian Mechanics its an important fact. So if you push one end of the rod, you apply a force. This force travels 1 light year with speed of sound in the rod=5130m/s , which is included in the problem statement - "Engineering Tools Box". Then the light from the other end reaches your eyes with speed of light . It would take 1 year , as the rod itself is 1 light year!

Now, the calculations.

Given,

Speed of sound, v s = 5130 m / s v_s=5130m/s

Speed of light, c 0 = 3 × 1 0 8 m / s c_0=3\times 10^8 m/s

Distance, d = 1 l i g h t y e a r d=1 light year = ( 24 × 3600 × 365 × 3 × 1 0 8 ) m =(24\times 3600\times 365\times 3\times 10^8)m = 9.46 × 1 0 15 m =9.46\times 10^{15} m

So, Total Time , T = d v s + d c 0 T= \frac{d}{v_s} +\frac{d}{c_0} = 1.844 × 1 0 12 s =\boxed{1.844 \times 10^{12} s}

......And you see! Finally its MOVING!!!

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Perhaps you should mention that it is to 4 significant digits or to 3 decimal places in scientific notation? Because your speed of light has one significant digit and your speed of sound has 3 significant digits so someone could easily write 2 × 1 0 12 2\times 10^{12} or 1.84 × 1 0 12 1.84\times 10^{12} if we assume that the speed of light is exactly 3 × 1 0 8 m / s 3\times 10^8 m/s .

Manvith Narahari - 5 years, 5 months ago

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