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 $2\times 10^{4}ms^{- 1}$ . 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,
$v_{1}r_{1}=v_{2}r_{2}$
$\Rightarrow \, \, \, v_{2}=\frac{2 \times 1 0^{4} \times 2.5 \times 1 0^{11}}{5 \times 1 0^{10}}=10^{5}ms^{- 1}$
Change in kinetic energy is,
$\Delta KE=\frac{1}{2}\times 10^{8}\left[\left(1 0^{5}\right)^{2} - 4 \times 1 0^{8}\right]=48\times 10^{8} \, J$