Question

Two planets of equal mass orbit a much more massive star (figure). Planet $m_1$ moves in a circular orbit of radius $1\times {10}^8$ km with period 2 year. Planet $m_2$ moves in an elliptical orbit with closest distance $r_1=1\times {10}^8$ km and farthest distance $r_{2}1.8\times {10}^8$km, as shown. Using the fact that mean radius of an elliptical orbit is the length of the semi-major axis, find the period of $m_2^{{}^{\prime }}$ s orbit.

• A. 2.1 year
• B. 3.31 year
• C. 4.12 year
• D. 5.24 year

Answer 1

The correct option is B

3.31 year

Mean radius of planet's orbit of revolution,

$=\frac{r_1+r_2}{2}=1.4\times {10}^{8}kmNow,T\propto r^{\frac{3}{2}}\therefore Timeperiodofrevolutionof{m}_{2}is,T_2=T_1{\left(\frac{1.4\times {10}^8}{{10}^8}\right)}^{\frac{3}{2}}=3.31year$