Question

An intergalactic rock star bangs his drum every 1.50 s. A person on earth measures that the time between beats is 2.70 s. How fast is the rock star moving relative to the earth?

Answer 1

Answer:

v = 83.1 % of speed of light

Explanation:

given,

T_e is the earth time = 2.7 s

T_s is the ship time = 1.5 s

we know,

$T_s = T_e \times \gamma$

where c is the speed of light

v is the speed of the rock star moving

$T_s = T_e\times \sqrt{1-\dfrac{v^2}{c^2}}$

$1.5= 2.7\times \sqrt{1-\dfrac{v^2}{c^2}}$

$\sqrt{1-\dfrac{v^2}{c^2}} =0.556$

squaring both side

$1-\dfrac{v^2}{c^2}=0.3086$

$v^2=0.6914c^2$

v = 0.831 c

v = 83.1 % of speed of light