Q. 63 A rocket is fired from the earth... [FREE SOLUTION]

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Chapter 36: Q. 63 (page 1061)

A rocket is fired from the earth to the moon at a speed of $0.990 c$ . Let two events be 鈥渞ocket leaves earth鈥� and 鈥渞ocket hits moon.鈥� a. In the earth鈥檚 reference frame, calculate 鈭唜, 鈭唗, and the space time interval s for these events.
b. In the rocket鈥檚 reference frame, calculate 鈭唜鈥�, 鈭唗鈥�, and the space time interval s鈥� for these events.
c. Repeat your calculations of part a if the rocket is replaced with a laser beam.

Short Answer

Expert verified

a. $鈭� x$ =distance between earth and moon

$鈭� t = \frac{鈭� x}{0.99 c}$

$鈭� S^{2} = {(c 鈭� t)}^{2} - 鈭� x^{2}$

b. $鈭� x' = \frac{鈭� x - v 鈭� t}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$鈭� t' = \frac{鈭� t - \frac{v 鈭� x}{c^{2}}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$鈭� S'^{2} = {(c 鈭� t')}^{2} - 鈭� x'^{2}$

c. $鈭� x$ = distance between earth and moon

$鈭� t = \frac{鈭� x}{c}$

Step by step solution

Part (a) Step 1: Given information

We have given,

speed of rocket = $0.99 c$

We have to find the $鈭� x, 鈭� t$ in earth reference frame.

Simplify

We know that,

$鈭� x = \frac{鈭� x' + v 鈭� t}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ =distance between moon and earth

and,

$鈭� t = \frac{鈭� t + \frac{v 鈭� x}{c^{2}}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} = \frac{鈭� x}{0.99 c}$

then,

the space time interval of s is,

$鈭� S^{2} = {(c 鈭� t)}^{2} - 鈭� x^{2}$

Part (b) Step 1: Given information

We have given,

speed of rocket = $0.99 c$

We have to find the $鈭� x', 鈭� t'$ in rocket reference frame.

Simplify

We know that,

$鈭� x' = \frac{鈭� x - v 鈭� t}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

and,

$鈭� t' = \frac{鈭� t - \frac{v 鈭� x}{c^{2}}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

then,

the space time interval of s is,

$鈭� S'^{2} = {(c 鈭� t')}^{2} - 鈭� x'^{2}$

Part (c) Step 1: Given information

We have given,

Speed of rocket = $0.99 c$

We have to repeat the calculation a for laser light.

Simplify

We know that,

$鈭� x = \frac{鈭� x' + v 鈭� t}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

and

$鈭� t = \frac{鈭� t' + \frac{v 鈭� x}{c^{2}}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} = \frac{鈭� x}{c}$

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Short Answer

Step by step solution

Part (a) Step 1: Given information

Simplify

Part (b) Step 1: Given information

Simplify

Part (c) Step 1: Given information

Simplify

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