Question

35.
A satellite in a circular orbit of radius R has a period of 4 hours. Another satellite with
orbital radius 3 R around the same planet will have a period (in hours)
1)
16
2) 4
3) 4√27
4). 4√8



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Answer 1

Explanation:

According to the Kepler's third law T2∝R3 T1T2 = (R1R23/2) or R2 = (R1) (T2T1)3/2 = (R1) (162)2/3 = 4R1 = 4R (given R1 = R ) ....(i)Orbital velocity, ...

Answer 2

Kepler's Law of Period states that, if R is orbital radius and T is time period of a satellite,

$\longrightarrow\sf{T^2\propto R^3}$

Therefore,

$\longrightarrow\sf{\left(\dfrac{T_2}{T_1}\right)^2=\left(\dfrac{R_2}{R_1}\right)^3}$

$\longrightarrow\sf{\dfrac{T_2}{T_1}=\left(\dfrac{R_2}{R_1}\right)^{\frac{3}{2}}}$

$\longrightarrow\sf{T_2=T_1\left(\dfrac{R_2}{R_1}\right)^{\frac{3}{2}}}$

Here,

• $\sf{R_1=R}$

• $\sf{T_1=4\ hr}$

• $\sf{R_2=3R}$

So time period of second satellite will be,

$\longrightarrow\sf{T_2=4\left(\dfrac{3R}{R}\right)^{\frac{3}{2}}}$

$\longrightarrow\sf{T_2=4\times3^{\frac{3}{2}}\ hr}$

$\longrightarrow\underline{\underline{\sf{T_2=4\sqrt{27}\ hr}}}$

Hence (3) is the answer.