Q. One day on a spacecraft corresponds to 2 days on the earth. The speed of the spacecraft relative to the earth is
A) $1.5\ \times 10^{8}\ ms^{- 1}$
B) $2.1\ \times 10^{8}\ ms^{- 1}$
C) $2.6\ \times 10^{8}\ ms^{- 1}$
D) $5.2\ \times 10^{8}\ ms^{- 1}$
Explanation
(c) $T = \frac{T_{0}}{\left\lbrack 1 - \left( v^{2}/c^{2} \right) \right\rbrack^{1/2}}$
By substituting $T_{0} = 1$ day and $T = 2$ days we get
$v = 2.6 \times 10^{8}\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.