Question

A comet of mass $10^{8}kg$ travels around the Sun in an elliptical orbit. When it is closest to the Sun it is $2.5\times 10^{11}m$ away and its speed is $2\times 10^{4}m{s}^{-1}$ . Find the change in kinetic energy when it is farthest from the Sun and is $5\times 10^{10}m$ away from the Sun

• A. $38\times 10^{8}J$
• B. $48\times 10^{8}J$
• C. $58\times 10^{8}J$
• D. $56\times 10^{8}J$

Answer 1

Answer: (B)

From the conservation of angular momentum,
$v_1{r}_1=v_2{r}_2$
$\Rightarrow v_2=\frac{2\times 10^4\times 2.5\times 10^{11}}{5\times 10^{10}}=10^{5}m{s}^{-1}$
Change in kinetic energy is,
$\Delta KE=\frac{1}{2}\times 10^8\left[{\left(10^5\right)}^2-4\times 10^8\right]=48\times 10^{8}J$