Question

time period of a planet around the sun is 64 times that of the earth the ratio of radius of planet of Orbital to the radius of the Earth orbit is

Answer 1

Is double................

Answer 2

Answer:

The ratio of the radius of the planet of Orbital to the radius of the Earth orbit is

$\mathrm{R}_{0}=16 \mathrm{R}$

Explanation:

Time period of earth $=\mathrm{T}$

distance of earth from $=\mathrm{R} .$

Time period of planet $=64 \mathrm{~T}$

let the distance from sun be $\mathrm{R}_{0}$

So, from Kepler's law's

The square of the time for a planet to complete a revolution about the sun is proportional to the cube root of its distance from sun.

So,$\mathrm{T}^{2} \alpha \mathrm{R}^{3} \cdots \cdots-(1)$

$(64 \mathrm{~T})^{2} \alpha \mathrm{R}_{0}^{3}---(2)$

dividing equation (1) and (2)

$\frac{\mathrm{T}^{2}}{(64)^{2} \mathrm{~T}^{2}}=\frac{\mathrm{R}^{3}}{\mathrm{R}_{0}^{3}}=\mathrm{R}_{0}^{3}=(64)^{2} \mathrm{R}^{3} \mathrm{~S}$

dividing equation (1) and (2)

$\frac{\mathrm{T}^{2}}{(64)^{2} \mathrm{~T}^{2}}=\frac{\mathrm{R}^{3}}{\mathrm{R}_{0}^{3}}=\mathrm{R}_{0}^{3}=(64)^{2} \mathrm{R}^{3} \mathrm{~S}$

$\mathrm{R}_{0}^{3}=\left(4^{3}\right)^{2} \mathrm{R}^{3}$

$\mathrm{R}_{0}=16 \mathrm{R}$

Hence,

We get, $\mathrm{R}_{0}=16 \mathrm{R}$