Question

A small particle of mass m is attached at one end of a light string of
length and other end is connected to the ceiling of a satellite revolving
around earth in circular orbit. The satellite is at height h = 2Re above
earth surface. The period of oscillation of mass will be (Re = radius of
earth)
3R
6.
Zero
Infinity
​

Answer 1

Given:

A small particle of mass m is attached at one end of a light string of length and other end is connected to the ceiling of a satellite revolving around earth in circular orbit. The satellite is at height h = 2Re above earth surface.

To find:

Time period of oscillation ?

Calculation:

First of all, let's find out the net gravitational acceleration at that height :

$\rm\therefore \: g' = \dfrac{g}{ { \bigg(1 + \dfrac{h}{R_{e}} \bigg) }^{2} }$

$\rm\implies \: g' = \dfrac{g}{ { \bigg(1 + \dfrac{2 R_{e}}{R_{e}} \bigg) }^{2} }$

$\rm\implies \: g' = \dfrac{g}{ { \bigg(1 + 2\bigg) }^{2} }$

$\rm\implies \: g' = \dfrac{g}{ { \bigg(3\bigg) }^{2} }$

$\rm\implies \: g' = \dfrac{g}{9}$

Now, time period will be :

$\rm T = 2\pi \sqrt{ \dfrac{l}{g'} }$

$\implies \rm T = 2\pi \sqrt{ \dfrac{l}{( \frac{g}{9} )} }$

$\implies \rm T = 2\pi \sqrt{ \dfrac{9l}{g} }$

$\implies \rm T =3 \times 2\pi \sqrt{ \dfrac{l}{g} }$

$\implies \rm T =6\pi \sqrt{ \dfrac{l}{g} }$

So, time period is :

$\boxed{ \bf T =6\pi \sqrt{ \dfrac{l}{g} } }$

Answer 2

OPTION 2) IS THE RIGHT ANSWER , AIATS QUESTION IT IS!!❤