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(C) According to Kepler's third law of planetary motion,the square of the time period of revolution $(T)$ is directly proportional to the cube of the

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$3$

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(C) According to Kepler's third law of planetary motion,the square of the time period of revolution $(T)$ is directly proportional to the cube of the semi-major axis $(R)$ of the orbit: $T^2 \propto R^3$.Given:For Earth: $T_1 = 1 \, 	ext{year}$,$R_1 = 1.5 	imes 10^8 \, 	ext{km}$.For the imaginary planet: $T_2 = 2.83 \, 	ext{years}$,$R_2 = ?$.Using the ratio formula:$\left(\frac{T_2}{T_1}ight)^2 = \left(\frac{R_2}{R_1}ight)^3$Substituting the values:$\left(\frac{2.83}{1}ight)^2 = \left(\frac{R_2}{1.5 	imes 10^8}ight)^3$Since $2.83 \approx \sqrt{8}$,then $(2.83)^2 \approx 8$:$8 = \left(\frac{R_2}{1.5 	imes 10^8}ight)^3$Taking the cube root on both sides:$2 = \frac{R_2}{1.5 	imes 10^8}$$R_2 = 2 	imes 1.5 	imes 10^8 = 3 	imes 10^8 \, 	ext{km}$.Thus,the distance is $3 	imes 10^8 \, 	ext{km}$.

Column-$I$	Column-$II$
$(1)$ Kepler's first law	$(a)$ Law of periods
$(2)$ Kepler's second law	$(b)$ Law of orbits
$(3)$ Kepler's third law	$(c)$ Law of areas

If the distance of the earth from the Sun is $1.5 \times 10^8 \, km$. Then the distance of an imaginary planet from the Sun,if its period of revolution is $2.83$ years,is $............. \times 10^8 \, km$.

Similar Questions

If $J$ is the angular momentum of a planet revolving around the Sun,what is the areal velocity of the planet?

Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Kepler's first law $(a)$ Law of periods

$(2)$ Kepler's second law $(b)$ Law of orbits

$(3)$ Kepler's third law $(c)$ Law of areas

Choose the incorrect statement from the following.

The period of revolution of planet $A$ around the sun is $8$ times that of planet $B$. How many times is the distance of planet $A$ from the sun greater than that of planet $B$ from the sun (in $\times$)?

State Kepler's law of periods (Kepler's third law) for planetary motion.

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