Question 134: Gravitation

Question

If the distance of the earth from Sun is 1.5 $$	imes$$ 10$$^6$$ km. Then the distance of an imaginary planet from Sun, if its period of revolution is 2.83 years is :

Accepted Answer

We can use Kepler's third law to solve this problem. Kepler's third law states that the square of the period of revolution of a planet around the Sun is proportional to the cube of its average distance from the Sun. Let $T$ be the period of revolution of the imaginary planet in years, and let $d$ be its average distance from the Sun in kilometers. We can use the following equation to solve for $d$:
$$\frac{T_1^2}{T_2^2} = \frac{d_1^3}{d_2^3}$$
where $T_1$ and $d_1$ are the period of revolution and an average distance of the Earth from the Sun, respectively, and $T_2$ is the period of revolution of the imaginary planet.
Substituting the given values, we get:
$$
\begin{aligned}
& \frac{T_{1}}{T_{2}}=\left(\frac{d_{1}}{d_{2}}ight)^{\frac{3}{2}} \\
& \Rightarrow\left(\frac{1 	ext { year }}{2.83 	ext { year }}ight)^{\frac{2}{3}}=\left(\frac{1.5 	imes 10^{6} \mathrm{~km}}{d_{2}}ight) \\
& \Rightarrow \frac{1}{2}=\frac{1.5 	imes 10^{6} \mathrm{~km}}{d_{2}} \\
& d_{2}=3 	imes 10^{6} \mathrm{~km}
\end{aligned}
$$

Answer

$$6	imes10^6$$ km

Answer

$$3	imes10^7$$ km

Answer

$$6	imes10^7$$ km

Answer

$$3	imes10^6$$ km

Gravitation

If the distance of the earth from Sun is 1.5 $\times 10^6$ km. Then the distance of an imaginary planet from Sun, if its period of revolution is 2.83 years is :

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