The acceleration due to gravity at height R above the surface of the earth is g / 4. The periodic time of a simple pendulum in an artificial satellite at this height will be :

Question

The acceleration due to gravity at height R above the surface of the earth is g / 4. The periodic time of a simple pendulum in an artificial satellite at this height will be : (A) T =2 π √2 l / g (B) T =2 π √l / 2 g (C) Zero (D) Infinity

Tardigrade Admin · Accepted Answer

In satellite g texteff= zero ⇒ T=∞

Q. The acceleration due to gravity at height $R$ above the surface of the earth is $g /4$ . The periodic time of a simple pendulum in an artificial satellite at this height will be :

$T = 2 π 2 l / g$

$T = 2 π l /2 g$

Zero

Infinity

In satellite $g_{eff} =$ zero
$\Rightarrow T = \infty$

Q. The acceleration due to gravity at height R above the surface of the earth is g/4. The periodic time of a simple pendulum in an artificial satellite at this height will be :

T=2π2l/g​

T=2πl/2g​