Gravitation

Question : 3 of 25

Marks: +1, -0

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?

Solution:

Using kepler’s third law,

We have,

r_P = r_E \left( \frac{T_P}{T_E} \right)^{\frac{2}{3}}

\because T_E = 1

year,

T_P = \frac{T_E}{2} = \frac{1}{2}

year,

r_E = 1 \text{AU}

\therefore r_P = 0.63 \text{AU}

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