Question

G = 6.67x10^-11 m^3/kgs^2

Neptune has an orbital radius of 4.5 x 10^9 km and orbital period of 164.8 years.

Jupiter has a mass of 1.89 x 10^27 kg and an orbital period of 11.9 years.

a) Based on the table and Kepler's laws and/or Newton's Gravity, what would Jupiter's orbital radius be?

b) Use the values to find the mass of the sun.

Answer 1

Answer:

(b) Using Kepler's third law, T² = [4²/GM_s] R³

where T is orbital period of the planet, R is orbital radius, G is gravitational constant and M_s is mass of Sun

Using the Neptune's data we will calculate the mass of Sun

From the above expression, M_s = 4²R³/GT²

Then, M_s = 4²(4.5 x 10⁹ x 10³ m)³/(6.67 x 10^-11m³/kg.s²) (164.8 x 365 x 24 x 60 x 60 s)²

M_s = (3.593 x 10³⁹) / (180.09 x 10⁷) kg = 1.99 x 10³⁰ kg

(a) Using the above result and the Kepler's third law we will able to calculate the orbital radius of Jupiter.

T² = [4²/GM_s] R³

or R = [GM_sT²/4²]^1/3 = [(6.67 x 10^-11m³/kg.s²) (1.99 x 10³⁰ kg) (11.9 x 365 x 24 x 60 x 60 s)²/4²]^1/3

R = [(1.867 x 10³⁷)/4²]^1/3 = 7.779 x 10¹¹ m or 7.779 x 10⁸ km.

Hence, this is the orbital radius of Jupiter.