Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.1*104 m/s when at a distance of 2.5*1011 m from the center of the sun, what is its speed when at a distance of 4.9*1010 m. Express your answer using two significant figures.

v_f = 6.92 x 10^ (4) m/s

Explanation:

From conservation of energy,

E = (1/2) mv² - GmM/r

Where M is mass of sun

Thus,

E_i = E_f will give;

(1/2) mv_i² - GmM / (r_i) = (1/2) mv_f² - GmM / (r_f)

m will cancel out to give;

(1/2) v_i² - GM / (r_i) = (1/2) v_f² - GM / (r_f)

Let's make v_f the subject;

v_f = √[ (v_i) ² + 2MG ((1/r_f) - (1/r_i)) ]

G is Gravitational constant and has a value of 6.67 x 10^ (-11) N. m²/kg²

Mass of sun is 1.9891 x 10^ (30) kg

v_i = 2.1*10⁴ m/s

r_i = 2.5 * 10^ (11) m

r_f = 4.9 * 10^ (10) m

Plugging in all these values, we have;

v_f = √[ (2.1*10⁴) ² + 2 (1.9891 x 10^ (31)) (6.67 x 10^ (-11)) ((1 / (4.9 * 10^ (10))) - (1 / (2.5 * 10^ (11))) ] 20.408 e12

v_f = √[ (441000000) + 2 (1.9891 x 10^ (30)) (6.67 x 10^ (-11)) ((16.408 x 10^ (-12)) ]

v_f = √[ (441000000) + (435.38 x 10^ (7))

v_f = 6.92 x 10^ (4) m/s