Mercury orbits the Sun at an average distance of 5.79 times 10^7 km with an orbi

Question

Mercury orbits the Sun at an average distance of 5.79 times 10^7 km with an orbital period of 0.241 yr. The Moon, which is one of the satellites of the Earth orbits its parent at an average distance of 3.84 times 10^5 km with an orbital period of 0.07481 yr. (a) Use the above information to find the orbital speeds of Mercury around the Sun and of tie Moot around the Earth.^v Mercury = m/s^v Moon = m/s (b) What is the expression for the mass M at the parent in terms at the orbital speed v of the satellite, the orbital radius R of the satellite and the gravitational constant G? (Do not substitute numerical values: use variables only.) M = (c) Now use your answers from parts (a) and (b) to find the ratio of the mass of the Earth to that of the Sun. M_E/M_S =

Explanation / Answer

(A) For mercury:

w = 2pi / T = 2pi / (0.241 x 365 x 24 x 3600s )

w = 8.267 x 10^-7 rad/s

V = w r

V_mercury = (8.267 x 10^-7) (5.79 x 10^10 m ) = 47866.84 m/s ......Ans

For moon:

w = 2pi / (0.07481 x 365 x 24 x 3600) = 2.663 x 10^-6 rad/s

w = v r

v_moon = (2.663 x 10^-6) (3.84 x 10^8) = 1022.7 m/s ..........Ans

(b) for circular path,

G M m / R^2 = m v^2 / R

M = v^2 R / G

(C) mE / mS = (1022.7^2 x 3.84 x 10^8) / (47866.84^2 x 5.79 x 10^10)

= 3.03 x 10^-6

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