Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

Asked by Pragya Singh | 1 year ago |  82

Solution :-

Time taken by the earth for one complete revolution, T= 1 Year

Radius of Earth’s orbit, R= 1 AU

Thus, the time taken by the planet to complete one complete revolution:

T$$\dfrac{1}{2}T_E = \dfrac{1}{2}year$$

Let, the orbital radius of this planet = RP

Now, according to the Kepler’s third law of planetary motion:

$$\left ( \dfrac{R_{P}}{R_{E}} \right )^{3}=\left ( \dfrac{T_{P}}{T_{E}} \right )^{2}\\\dfrac{R_{P}}{R_{E}}=\left ( \dfrac{T_{P}}{T_{E}} \right )^{\dfrac{2}{3}}$$

$$=\left ( \dfrac{\frac{1}{2}\;T_{E}}{T_{E}} \right )^{\dfrac{2}{3}}=\left ( \dfrac{1}{2} \right )^{\dfrac{2}{3}}=0.63$$

$$\dfrac{R_{P}}{R_{E}}=\left ( \dfrac{T_{P}}{T_{E}} \right )^{\dfrac{2}{3}}$$

Therefore, radius of orbit of this planet is 0.63 times smaller than the radius of orbit of the Earth.

Answered by Abhisek | 1 year ago

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