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Setler79 [48]
3 years ago
12

How fast must an object move before its length appears to be contracted to one-fourth its proper length? (Give your answer in te

rms of c.)
Physics
1 answer:
Tresset [83]3 years ago
3 0

Answer:

<em>0.97c</em>

<em></em>

Explanation:

From the relativistic equation for length contraction, we have

l = l_{0}\sqrt{1 - \beta }

where

l is the final length of the object

l_{0} is the original length of the object before contraction

β = v^{2} /c^2

where v is the speed of the object

c is the speed of light in free space = 3 x 10^8 m/s

The equation can be re-written as

l/l_{0} = \sqrt{1 - \beta }

For the length to contract to one-fourth of the proper length, then

l/l_{0} = 1/4

substituting into the equation, we'll have

1/4 = \sqrt{1 - \beta }

substituting for β, we'll have

1/4 = \sqrt{1 - v^2/c^2 }

squaring both side of the equation, we'll have

1/16 = 1 - v^2/c^2

v^2/c^2 = 1 - 1/16

v^2/c^2 = 15/16

square root both sides of the equation, we have

v/c = 0.968

v = <em>0.97c</em>

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We need to be careful here.
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<span>
and now we're ready to go !

       Gravitational force = 

                   G  M₁ M₂ / R²

= (6.67 x 10⁻¹¹ N-m²/kg²)(</span><span>5.972 x 10²⁴ kg)(7.348 x 10²²  kg)/</span>(8.098 x 10⁶ m)²

= (6.67 · 5.972 · 7.348 / 8.098²) · (10²³)      Newtons

=    (I get ...)        4.463 x 10²³ Newtons

That's almost exactly   10²³ pounds 

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Those are big numbers. 
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8 0
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