Question:
A binary star system consists of two stars of masses $M_1\,and\,M_2$ revolving in circular orbits of radii $R_1\,and\,R_2$ respectively. If the respective time periods are $T_1\,and\,T_2$
A binary star system consists of two stars of masses $M_1\,and\,M_2$ revolving in circular orbits of radii $R_1\,and\,R_2$ respectively. If the respective time periods are $T_1\,and\,T_2$
Updated On: Jul 5, 2022
- $T_1>T_2\,if\,R_1>R_2$
- $T_1>T_2\,if\,M_1>M_2$
- $T_1=T_2$
- $\frac{T_1}{T_2}=\left(\frac{R_1}{R_2}\right)^{3/2}$
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The Correct Option is C
Solution and Explanation
In a binary star system, there are two stars moving under their mutual gravitational force of attraction. Their angular velocities are equal and hence their periodic times are also equal.
$\left[\because \omega=\frac{2 \pi}{T}\right]$
$\therefore T_{1}=T_{2}$
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].