Question

An object lying at the equator of the earth will fly off the surface (will feel weightlessness), if the length of the day becomes

• A. $1.00h$
• B. $1.41h$
• C. $17.0h$
• D. $17.0min$

Answer 1

The correct option is A $1.00h$

Escape velocity at earth radius $R$ is given by $V=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}$
Now, time period $T=\frac{2\pi R}{\sqrt{2gR}}=\pi \sqrt{\frac{2R}{g}}=3.14\times \sqrt{\frac{2\times 6.4\times {10}^6}{9.8}}=1hr(approx.)$
A is the right answer.