Question

The distance of two planet from the sun are 10^13 and 10^12 respectively the ratio of time period and the orbital speed of 2 planets.

Answer 1

Using Kepler's 3'd Law     T^2 = K R^3

T1^2 / T2^2 = R1^3 / R2^3

T1 ^2 / T2^2 = 10^39 / 10^36 = 1000

T1 = 32 T2

2 pi R = v T    using mean distance and time for 1 revolution        

v1 / v2 = R1 T2 / (R2 T1)  = (R1 / R2) * (T2 / T1) = 1000 /32 = 32

Answer 2

Answer:

$(\dfrac{T_1}{T_2})=10\sqrt{10}$

Explanation:

It is given that,

Distance of planet 1 from the sun, $R_1=10^{13}$

Distance of planet 2 from the sun, $R_2=10^{12}$

Let T₁ and T₂ are the time period of two planets. Using Kepler's third law of motion as :

$T^2\propto R^3$

$(\dfrac{T_1}{T_2})^2=(\dfrac{R_1}{R_2})^3$

$(\dfrac{T_1}{T_2})^2=(\dfrac{10^{13}}{10^{12}})^3$

$(\dfrac{T_1}{T_2})^2=1000$

$(\dfrac{T_1}{T_2})=10\sqrt{10}$

Let v₁ and v₂ are speeds of two planets. The speed of a planet is given by :

$v=\dfrac{2\pi R}{T}$

So,

$\dfrac{v_1}{v_2}=\dfrac{T_2R_1}{T_1R_2}$

$\dfrac{v_1}{v_2}=\dfrac{1}{10\sqrt{10}}\times \dfrac{10^{13}}{10^{12}}$

$\dfrac{v_1}{v_2}=0.316$

Hence, this is the required solution.