Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released, they interact only through gravitational force and collide after a time T=4$T=4$ seconds. If the sides of the triangle are increased to length 2a and also the masses of the spheres are made 2m, then they will collide after ________ seconds.

### Related Formula By Dimensional Analysis or Scaling of Kepler's Third Law: T propto m^x G^y a^z$$T \propto m^x G^y a^z$$ ### Core Logic Let's perform a dimensional matching to express the collision time T$T$ in terms of physical scaling variables m$m$, G$G$, and a$a$: [T] = [M]^x [M^-1L^3T^-2]^y [L]^z$$[T] = [M]^x [M^{-1}L^3T^{-2}]^y [L]^z$$ Equating dimensions on both sides: - Mass (M$M$): x - y = 0 implies x = y$x - y = 0 \implies x = y$ - Length (L$L$): 3y + z = 0 implies z = -3y$3y + z = 0 \implies z = -3y$ - Time (T$T$): -2y = 1 implies y = -1/2$-2y = 1 \implies y = -1/2$ Solving these equations: x = -1/2, quad y = -1/2, quad z = 3/2$$x = -1/2, \quad y = -1/2, \quad z = 3/2$$ ### Step 1: Scaling Formula of Time Therefore, the scaling relationship for time T$T$ is: T propto m^-1/2 G^-1/2 a^3/2 implies T propto sqrtfraca^3m$$T \propto m^{-1/2} G^{-1/2} a^{3/2} \implies T \propto \sqrt{\frac{a^3}{m}}$$ Let's write the ratio for two cases: fracT_2T_1 = sqrtleft(fraca_2a_1right)^3 cdot left(fracm_1m_2right)$$\frac{T_2}{T_1} = \sqrt{\left(\frac{a_2}{a_1}\right)^3 \cdot \left(\frac{m_1}{m_2}\right)}$$ ### Step 2: Calculating Final Time Given values: - a_1 = a$a_1 = a$, a_2 = 2a$a_2 = 2a$ - m_1 = m$m_1 = m$, m_2 = 2m$m_2 = 2m$ - T_1 = 4mathrm~s$T_1 = 4\mathrm{~s}$ fracT_24 = sqrtleft(frac2aaright)^3 cdot left(fracm2mright) = sqrt2^3 cdot frac12 = sqrt4 = 2$$\frac{T_2}{4} = \sqrt{\left(\frac{2a}{a}\right)^3 \cdot \left(\frac{m}{2m}\right)} = \sqrt{2^3 \cdot \frac{1}{2}} = \sqrt{4} = 2$$ T_2 = 4 times 2 = 8mathrm~seconds$$T_2 = 4 \times 2 = 8\mathrm{~seconds}$$ ### Pattern Recognition Kepler's Third Law / free-fall collapse scaling: whenever a orbit or a direct gravitational collapse scale is involved, the time scales as T propto sqrtfracR^3GM$T \propto \sqrt{\frac{R^3}{GM}}$. Thus doubling R$R$ multiplies time by sqrt8$\sqrt{8}$ and doubling M$M$ divides time by sqrt2$\sqrt{2}$. The combination results in a clean doubling of time: sqrt8/sqrt2 = 2$\sqrt{8}/\sqrt{2} = 2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation Class 11 Physics: Units and Measurements: Dimensional Analysis