Question

The maximum and minimum distances of a comet from the sun are $8\times {10}^{12}m$ and $1.6\times {10}^{12}m$. If its velocity when nearest to the sun is $60m/s$, what will be its velocity in $m/s$ when it is farthest?

Answer 1

Given, $r_{\text{max}}=8\times {10}^{12}m$

$r_{\text{min}}=1.6\times {10}^{12}m$

$V_{\text{max}}=60m/s$

As we know from Kepler's law of planter motion, angular momentum of the planet about sun remains constant.
$mvr=$ constant

$V_{\text{min}}\times r_{\text{max}}=V_{\text{max}}\times r_{\text{min}}$

$V_{\text{min}}=\frac{60\times 1.6\times {10}^{12}}{8\times {10}^{12}}$

$=\frac{60}{5}=12m/s$