Chapter 2: Q118CE (page 71)

Question: A rocket maintains a constant thrust F, giving it an acceleration of g
$(i . e ., 9.8 m / s^{2})$ .
(a) If classical physics were valid, how long would it take for the rocket’s speed to reach $0.99 c ?$ ?
(b) Using the result of exercise 117(c), how long would it really take to reach $0.99 c ?$ ?
$u = \frac{1}{\sqrt{1 + {(F t / m c)}^{2}}} \frac{F}{T} t$

Short Answer

Expert verified

If classical approximations were valid for high speeds,it would take almost a year for the rocket to reach $0.99 c,$
But in reality o.e., relativistic-ally,it would take around seven years.

Step by step solution

Determine the time required to reach 0.99 cclassically

(a) Using classical kinematic equations,

$v = u + a t$

Let’s consider the rocket was initially at rest, and then due to constant thrust, it traveled at an acceleration of $g .$

$v = g t t = \frac{v}{g} t = \frac{0.99 \times 3 \times 10^{8} \frac{m}{s}}{9.8 \frac{m}{s^{2}}} t = 3.03 \times 10^{7} s$

Therefore, classically,the time required to reach the velocity of is0.96 years or around350 days.

Determine the time required to reach 0.99 crelativistically

(b) Now,relativistic-ally,

$u = \frac{1}{\sqrt{1 + {(F t / m c)}^{2}}} \frac{F}{T} t = \frac{m u}{F} \sqrt{1 + {(\frac{F t}{m c})}^{2}} t^{2} = {(\frac{m u}{F})}^{2} (1 + {(\frac{F t}{m c})}^{2})$

Here, $F = m a = m g$

$t^{2} = {(\frac{u}{g})}^{2} (1 + {(\frac{g t}{c})}^{2})$

Rearranging further to get an equation for t ,

$t = \frac{u}{g \sqrt{1 - \frac{u^{2}}{c^{2}}}} = \frac{0.99 c}{9.8 \frac{m}{s^{2}} (\sqrt{1 - {(0.99)}^{2}})} = 2.143 \times 10^{8} s = 6.795 y r s$

Hence, the time the rocket really takes to reach $0.99 c$ is $6.8 y e a r s .$

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Short Answer

Step by step solution

Determine the time required to reach 0.99 cclassically

Determine the time required to reach 0.99 crelativistically

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