Question

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.8 1010 m (inside the orbit of Mercury), at which point its speed is 9.1 104 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6 1012 m from the Sun

Answer 1

Answer:

Explanation:

From the given information:

Distance $d_i = 4.8 \times 10^{10} \ m$

Speed of the comet $V_i = 9.1 \times 10^{4} \ m/s$

At distance $d_2 = 6 \times 10^{12} \ m$

where;

mass of the sun = $1.98 \times 10^{30}$

$G = 6.67 \times 10^{-11}$

To find the speed $V_f$:

Using the formula:

$E_f = E_i + W \\ \\ where; \ \ W = 0 \ \ \text{since work done by surrounding is zero (0)}$

$E_f = E_i + 0 \\ \\ K_f + U_f = K_i + U_i \\ \\ = \dfrac{1}{2}mV_f^2 + \dfrac{-GMm}{d^2} = \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [ \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}$

$V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [ \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}$

$\mathbf{V_f =53.125 \times 10^4 \ m/s}$