Mass of each star, \(M\) = \(2\times 10^{30}\) \(kg\)
Radius of each star, \(R\) = \(10^4\) \(km\) = \(10^7 m\)
Distance between the stars,\(r\) = \(10^9\; km\) = \(10^{12}\;m\)
For negligible speeds, v = 0 total energy of two stars separated at distance r
= \(\frac{-GMM}{r}+\frac{1}{2}mv^2\)
= \(\frac{-GMM}{r}+0\) ...(i)
Now, consider the case when the stars are about to collide:
Velocity of the stars =\(v\)
Distance between the centers of the stars = \(2R\)
Total kinetic energy of both stars = \(\frac{1}{2}Mv^2+\frac{1}{2}Mv^2\) = \(Mv^2\)
Total energy of both stars =\(Mv^2-\frac{GMM}{2R}\) ...(ii)
Total energy of the two stars = Using the law of conservation of energy, we can write:
\(MV^2-\frac{GMM}{2R}\) = \(\frac{-GMM}{r}\)
\(v^2 \)= \(\frac{-GM}{r}+\frac{GM}{2R}= GM\) \(\bigg(-\frac{1}{r}+\frac{1}{2R}\bigg)\)
= \(6.67 \times 10^ {-11} \times2 \times10^{30} \bigg[-\frac{1}{ 10^{12}} + \frac{1}{2\times 10^7}\bigg ]\)
= \(13.34 \times 10^{19} [-10^{-12} + 5 \times 10^{-8}] ∼ 13.34 \times 10^{19} \times 5 \times10^{-8} ∼ 6.67 \times 10^{12}\)
\(v\) = \(\sqrt{6.67 \times 10^{12}}\) = \(2.58 \times 10^6 m/s\)
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
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.
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
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].