Question

A comet travels around the sun in elliptical orbit and its mass is ${10}^8$ kg. When comet is $2.5\times {10}^{11}m$ away, its speed is $2\times {10}^{4}m{s}^{-1}$. Find the change in $KE$, when it has reached $5\times {10}^{11}m$ 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

The correct option is B $48\times {10}^{8}J$

$v_1{r}_1=v_2{r}_2$ or
$v_2=\frac{2\times {10}^4\times 2.5\times {10}^{11}}{5\times {10}^{10}}={10}^5$
$\Delta KE=\frac{1}{2}\times {10}^8\left[\right({10}^5{)}^2-4\times {10}^8]$
$=48\times {10}^{8}J$