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  • Stuck here, help me understand: - Gravitation - JEE Main-2

The relative uncertainty in the period of a satellite orbiting around the earth is 10^{-2}.
If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is

 

  • Option 1)

    10^{-2}

  • Option 2)

    2 \times 10^{-2}

  • Option 3)

    3 \times 10^{-2}

  • Option 4)

    6 \times 10^{-2}

 

Answers (1)

best_answer

As we have learned

Time period of satellite -

T=2\pi\sqrt{\frac{r^{3}}{GM}} 

r= radius of orbit

T\rightarrow Time period

M\rightarrow Mass of planet

 

- wherein

T=\frac{Circumference\: of\: orbit}{orbital\: velocity}

 

v=\sqrt{\left ( \frac{QM}{R} \right )}\Rightarrow T= \frac{2\pi R}{v}= \frac{2\pi }{\sqrt (4m)}\cdot R^{3/2}

M=\frac{4\pi ^{2}}{4}\cdot \frac{R^{3}}{T^{2}}

\frac{\Delta M}{M} = 3\cdot \frac{\Delta R}{R}+ 2\frac{\Delta T}{T}=3.0 +2\times 10^{-2}

\frac{\Delta M}{M} =2\times 10^{-2}

 

 

 

 

 

 

 

 


Option 1)

10^{-2}

This is incorrect

Option 2)

2 \times 10^{-2}

This is correct

Option 3)

3 \times 10^{-2}

This is incorrect

Option 4)

6 \times 10^{-2}

This is incorrect

Posted by

Plabita

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