The distance of two planets from sun are nearly 1014 and 1012 meters. Assuming that they move in circular orbits, their periodic times will be in the ratio

  • Option 1)

    100

  • Option 2)

    10

  • Option 3)

    1000

  • Option 4)

    100\sqrt{10}

 

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

Known as law of periods

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}\Rightarrow \left ( \frac{T_{2}}{T_{1}} \right )^{2}=\left ( \frac{R_{2}}{R_{1}} \right )^{3}=\left ( \frac{10^{12}}{10^{14}} \right )^{3}=10^{-6}

\therefore T_{2}=10^{-3}T_{1}    or     \frac{T_{1}}{T_{2}}=1000


Option 1)

100

This is incorrect option

Option 2)

10

This is incorrect option

Option 3)

1000

This is correct option

Option 4)

100\sqrt{10}

This is incorrect option

Preparation Products

Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Exams
Articles
Questions