Kevin S. Huang

2013: Problem 16

Topic: Other
Concepts: Dimensional analysis, Gravitational force

Solution:

We can use dimensional analysis to find how the time $T$ of collapse scales with $\rho_0$ , $r_0$ , and $G$ :

T \propto \rho_0^{\alpha}r_0^{\beta}G^{\delta}

We first find the dimensions of each quantity:

[T]=T

[\rho_0]=\frac{M}{L^3}

[r_0]=L

To find the dimensions of $G$ , we use

[F]=\frac{[G][m_1][m_2]}{[r]^2}

\frac{ML}{T^2}=\frac{[G]M^2}{L^2}

[G]=\frac{L^3}{MT^2}

We take the dimensions of the first equation,

[T]=[\rho_0]^{\alpha}[r_0]^{\beta}[G]^{\delta}

T=\left(\frac{M}{L^3}\right)^{\alpha}L^{\beta}\left(\frac{L^3}{MT^2}\right)^{\delta}

Counting powers of time $T$ ,

1=-2\delta

\delta=-\frac{1}{2}

Counting powers of mass $M$ ,

0=\alpha-\delta

\alpha=\delta=-\frac{1}{2}

Counting powers of length $L$ ,

0=-3\alpha+\beta+3\delta

\beta=3\alpha-3\delta=0

Thus,

T \propto \frac{1}{\sqrt{G\rho_0}}

so the answer is D.