Answer:
T_b / T_a = sqrt ( 8 )
Explanation:
Given:
- The mass of planet A = m _a
- The mass of planet B = 2*m _a
- The semi-major axis of plant A = a
- The semi-major axis of plant B = 2*a
Find:
- What is the ratio of the orbital period of planet B to that of planet A?
Solution:
- Kepler's Third Law for planetary motion of a satellite in an elliptical orbit is gives us the relation of time period as follows:
T = 2*pi*sqrt ( a^3 / G*M_s )
Where,
a : Semi major axis of the period.
M_s : It is the mass of the sun.
G: Gravitational constant
- So for planet A, we can develop an expression for time period:
T_a = 2*pi*sqrt ( a^3 / G*M_s )
- And for planet B, we can develop an expression for time period:
T_b = 2*pi*sqrt ( (2a)^3 / G*M_s )
- Now take the ratio of T_b to T_a:
T_b / T_a = 2*pi*sqrt ( (2a)^3 / G*M_s ) / 2*pi*sqrt ( a^3 / G*M_s )
Simplify,
T_b / T_a = sqrt ( (2a)^3 / a^3 ) = sqrt ( 8a^3 / a^3 )
T_b / T_a = sqrt ( 8 )