The time period of a satellite in a circular | vedclass

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Question

(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$.Let the time period of the first satellite be $T_1 = T$ with radius $R_1 = R$.Let the time period of the second satellite be $T_2$ with radius $R_2 = 9R$.Using the ratio formula:$\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3$Substituting the given values:$\left(\frac{T_2}{T}\right)^2 = \left(\frac{9R}{R}\right)^3$$\left(\frac{T_2}{T}\right)^2 = (9)^3 = 729$Taking the square root on both sides:$\frac{T_2}{T} = \sqrt{729} = 27$Therefore,$T_2 = 27T$.

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

The time period of a satellite in a circular orbit of radius $R$ is $T$. The period of another satellite in a circular orbit of radius $9R$ is............ $T$.

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

Column-$I$ Column-$II$

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

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