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Standard XII

Physics

Kepler's Law

Mercury orbit...

Question

Mercury orbits the sun in about one-fifth of the earth year. If $1 A U$ is defined as the distance from the earth to the sum, what is the approximate distance between mercury and the sun?

1 / 25^{1 / 3} A U

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1 / 9^{1 / 3} A U

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1 / 5^{1 / 3} A U

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1 / 3^{1 / 3} A U

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Solution

The correct option is A $1 / 25^{1 / 3} A U$
Using Kepler's third law,
$T^{2} \propto R^{3}$
where:
$T :$ Time period of revolution
$R :$ Radius of revolution

$\frac{T_{e a r t h}^{2}}{T_{m e r c u r y}^{2}} = \frac{R_{e a r t h}^{3}}{R_{m e r c u r y}^{3}}$
$R_{m e r c u r y} = R_{e a r t h} \frac{T_{m e r c u r y}^{2 / 3}}{T_{e a r t h}^{2 / 3}}$
$= 1 \times \frac{(1 / 5)^{2 / 3}}{1^{2 / 3}}$
$= (1 / 25)^{1 / 3} A U$

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Mercury orbits the sun in about one-fifth of the earth year. If 1 AU is defined as the distance from the earth to the sum, what is the approximate distance between mercury and the sun?

The correct option is A 1/251/3 AUUsing Kepler's third law, T2∝R3where:T: Time period of revolutionR: Radius of revolutionT2earthT2mercury=R3earthR3mercuryRmercury=RearthT2/3mercuryT2/3earth =1×(1/5)2/312/3 =(1/25)1/3 AU

Mercury orbits the sun in about one-fifth of the earth year. If $1 A U$ is defined as the distance from the earth to the sum, what is the approximate distance between mercury and the sun?