Question

The maximum and minimum distances of a comet from the Sun are 1.6 × 10$^{12}$ m and 8.0 × 10$^{10}$ m respectively. If the speed of the comet at the nearest point is 6 × 10$^4$ ms$^{-1}$, the speed at the farthest point is:

Answer 1

3000 m/s

Answer 2

Using conservation of angular momentum,

$$v_{\min} \, r_{\min} = v_{\max} \, r_{\max}$$

Thus,

$$v_{\max} = \frac{v_{\min} \, r_{\min}}{r_{\max}} = \frac{6 \times 10^4 \times 8.0 \times 10^{10}}{1.6 \times 10^{12}}$$

Simplify:

$$\frac{8.0 \times 10^{10}}{1.6 \times 10^{12}} = 5 \times 10^{-2}$$

Then,

$$v_{\max} = 6 \times 10^4 \times 5 \times 10^{-2} = 3000 \, \text{m/s}$$

Explanation (minimal):
Conservation of angular momentum gives $v_{\min}r_{\min} = v_{\max}r_{\max}$. Substitute values and simplify to find $v_{\max} = 3000 \, \text{m/s}$.