One day on a spacecraft corresponds to 2 days on the earth.…

Physics Newton's Laws of Motion Mix Single Correct MCQ

Published by: CGP EDU Academic Team

Published on: September 12, 2026

One day on a spacecraft corresponds to 2 days on the earth. The speed of the spacecraft relative to the earth is

\(1.5 \times 10^{8} \, \mathrm{m\,s}^{-1}\)

\(2.1 \times 10^{8} \, \mathrm{m\,s^{-1}}\)

\(2.6 \times 10^{8} \, m s^{-1}\)

\(5.2 \times 10^{8} \, ms^{-1}\)

Text Solution

Verified by Experts

The correct answer is:

\(T = \frac{T_0}{\left[ 1 - \left( v^2 / c^2 \right) \right]^{1/2}}\)

By substituting \overline{T_0} = 1 day and \overline{T} = 2 days we get

\(\nu = 2.6 \times 10^{8} \, \mathrm{m\,s}^{-1}\)

Previous Question Next Question

Get practice questions, solutions, and test series in one place.

Share your experience with this question and solution.

Send your comment, doubt, correction, or feedback to admin.

What students say about this solution

No reviews yet. Be the first to write a review.