Question

The distances of two planets from the sun are 10¹³ m and 10¹² m respectively. Find the ratio of time periods and orbital speeds of the two planets.
(Ans : 31.62:1, 0.3162:1)

Answer 1

Answer10*10

10*12

Explanation:

Answer 2

Answer:

Ratio of time periods= 31.62:1

Orbital speeds of the two planets.=0.3162:1

Explanation:

Since we know the kepler law of planetary motion which sates $T^{2} \alpha R^{3}$ where R is the distance.

We have the distances of two planets from the sun are 10¹³ m and 10¹² m respectively.

Apply kepler law for first planet we get

$T^{2} \alpha 10^{13}$........................(1)

Apply kepler law for second   planet we get

$T^{2} \alpha 10^{12}$...............................(2)

from first and second equation

$\frac{T^{2} _{1} }{T^{2} _{2} } =\frac{10^{13} }{10^{12} }$

solving further we get

$\frac{T_{1} }{T_{2} } =\frac{R_{1} }{R_{2} }\times\sqrt\frac{R_{1} }{R_{2} }$

$\frac{T_{1} }{T_{2} } =\frac{10^{13} }{10^{12} } }\times\sqrt\frac{10^{13} }{10_{12} }$

after solving we get

$\frac{T_{1} }{T_{2} } =31.62$

So ratio become $T_{1} :T_{2} :31.62:1$

For orbital speed we know force of attraction is equals to the centripetal force.

$\frac{GMm}{R^{2} } =\frac{mv^{2} }{R}$

After rearranging we get

$v\alpha \frac{1}{\sqrt{R} }$..................(3)

apply for first planet we get

$v_{1} \alpha \frac{1}{\sqrt[]{10^{13} } }$

apply for second planet we get

$v_{2} \alpha \frac{1}{\sqrt[]{10^{12} } }$.....................(4)

from .................(3) and (4)

$\frac{v_{1} }{v_{2} } =\sqrt[]{\frac{10^{12} }{10^{13} } }$

after solving we get

$\frac{v_{1} }{v_{2} } =.3162$

so ratio become

$v_{1} :v_{2}=0.3162:1$