Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Gravitation

Question:

A chord of length 64 m is used to connect a 100 kg astronaut to a spaceship whose mass is much larger than that of the astronaut. Then the value of the tension in the cord. (Assume that the spaceship is orbiting near earth surface. Also assume that the spaceship and the astronaut fall on a straight line from the earth centre. The radius of the earth is 6400 km).

Options:

2 × 10-2 N

3 × 10-2 N

4 × 10-2 N

5 × 10-2 N

Correct Answer:

3 × 10-2 N

Explanation:

The tension in the string T is given as, T + Fgr = mar

$\Rightarrow T+\frac{GMm}{(R+\ell)^2}=m(R+\ell) \omega^2$

$\Rightarrow T=m(R+\ell) \omega^2-\frac{GMm}{R^2}\left(\frac{R}{R+\ell}\right)^2$

$\Rightarrow T=m(R+\ell) \frac{g}{R}-mg\left[1+\frac{\ell}{R}\right]^{-2}$

$\Rightarrow \quad T=\frac{3 mgl}{R}=\frac{3 \times 100 \times 10 \times 64}{6400 \times 10^3}=3 \times 10^{-2} N$