Question 171: Gravitation

Question

Consider a binary star system of star A and star B with masses mA and mB revolving in a circular orbit of radii rA an rB, respectively. If TA and TB are the time period of star A and star B, respectively,Then :

Accepted Answer

In a binary star system, the two stars orbit around a common center of mass. When considering periods of revolution, Kepler's Third Law comes into play. This law states that the square of the period of revolution (T) is proportional to the cube of the semi-major axis (r) of the orbit. It's often written in the following form for a single object orbiting another:
TÃÂ² Ã¢ÂÂ rÃÂ³
For a binary star system, this would still hold true. The periods of revolution for both stars A and B will be the same because they are both orbiting the same common center of mass, regardless of their individual masses or individual orbital radii. In other words, star A and star B complete one orbit in the same amount of time.
So, Option B: T
A
= T
B
is correct.
The other options (Option A, C, and D) would not be correct. Kepler's Third Law is not a ratio between the periods and radii of two different bodies, and the periods do not depend on the masses of the individual stars or their individual distances from the center of mass.

Answer

$${{{T_A}} \over {{T_B}}} = {\left( {{{{r_A}} \over {{r_B}}}} \right)^{{3 \over 2}}}$$

Answer

$${T_A} = {T_B}$$

Answer

$${T_A} > {T_B}$$ (if $${m_A} > {m_B}$$)

Answer

$${T_A} > {T_B}$$ (if $${r_A} > {r_B}$$)

Gravitation

Consider a binary star system of star A and star B with masses mA and mB revolving in a circular orbit of radii rA an rB, respectively. If TA and TB are the time period of star A and star B, respectively,Then :

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