Question

The orbital radius of the Earth (from Earth to Sun) is 1.496 x 10^11 m.

Answer 1

Explanation:

The orbital radius of the Earth is $r_1=1.496\times 10^{11}\ m$

The orbital radius of the Mercury is $r_2=5.79 \times 10^{10}\ m$

The orbital radius of the Pluto is $r_3=5.91 \times 10^{12}\ m$

We need to find the time required for light to travel from the Sun to each of the  three planets.

(a) For Sun -Earth,

Kepler's third law :

$T_1^2=\dfrac{4\pi ^2}{GM}r_1^3$

M is mass of sun, $M=1.989\times 10^{30}\ kg$

So,

$T_1^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 1.496\times 10^{11}\\\\T_1=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times1.496\times10^{11}}\\\\T_1=2\times 10^{-4}\ s$

(b) For Sun -Mercury,

$T_2^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.79 \times 10^{10}\ m\\\\T_2=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.79 \times 10^{10}}\ m\\\\T_2=1.31\times 10^{-4}\ s$

(c) For Sun-Pluto,

$T_3^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.91 \times 10^{12}\\\\T_3=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.91 \times 10^{12}}\\\\T_3=1.32\times 10^{-3}\ s$