The time period of a satellite of earth is 5 hour. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become

  • Option 1)

    10\: hour

  • Option 2)

    80\: hour

  • Option 3)

    40\: hour

  • Option 4)

    20\: hour

 

Answers (1)

As we learnt in 

 

Kepler's 3rd law -

T^{2}\: \alpha\: a^{3}

From fig.

AB=AF+FB

2a=r_{1}+r_{2}

\therefore\; a=\frac{r_{1}+r_{1}}{2}

a= semi major Axis

r_{1}= Perigee

- wherein

r_{2}= apogee

T^{2}\: \alpha \: \left ( \frac{r_{1}+r_{2}}{2} \right )^{3}

{r_{1}+r_{2}= 2a

 

 T^2\propto r^3

\left ( \tfrac{T_{1}}{T_{2}} \right )^2=\left ( \frac{_{r_{1}}}{_{r_{2}}} \right )^3 =\left ( \frac{_{1}}{4} \right )^3

= \frac{1}{64}

\frac{T_{1}}{T_{2}} = \frac{1}{8}

\Rightarrow T_{2} = 8T_{1} = 40 hrs

 


Option 1)

10\: hour

Incorrect

Option 2)

80\: hour

Incorrect

Option 3)

40\: hour

Correct

Option 4)

20\: hour

Incorrect

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