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
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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 :
[tex] \rm\therefore \: g' = \dfrac{g}{ { \bigg(1 + \dfrac{h}{R_{e}} \bigg) }^{2} } [/tex]
[tex] \rm\implies \: g' = \dfrac{g}{ { \bigg(1 + \dfrac{2 R_{e}}{R_{e}} \bigg) }^{2} } [/tex]
[tex] \rm\implies \: g' = \dfrac{g}{ { \bigg(1 + 2\bigg) }^{2} } [/tex]
[tex] \rm\implies \: g' = \dfrac{g}{ { \bigg(3\bigg) }^{2} } [/tex]
[tex] \rm\implies \: g' = \dfrac{g}{9} [/tex]
Now, time period will be :
[tex] \rm T = 2\pi \sqrt{ \dfrac{l}{g'} } [/tex]
[tex] \implies \rm T = 2\pi \sqrt{ \dfrac{l}{( \frac{g}{9} )} } [/tex]
[tex] \implies \rm T = 2\pi \sqrt{ \dfrac{9l}{g} } [/tex]
[tex] \implies \rm T =3 \times 2\pi \sqrt{ \dfrac{l}{g} } [/tex]
[tex] \implies \rm T =6\pi \sqrt{ \dfrac{l}{g} } [/tex]
So, time period is :
[tex] \boxed{ \bf T =6\pi \sqrt{ \dfrac{l}{g} } }[/tex]
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