Question

What is the dimensional formula of escape velocity?

Answer 1

Escape velocity:

The lowest velocity required for a moving body is just to leave the gravitational field of such a celestial body and go outward towards space.

Formula:

${\mathrm{V}}_{\mathrm{e}}=\sqrt{\frac{2\mathrm{MG}}{\mathrm{r}}}$

Here,

V_e is the escape velocity.

M is the mass of the earth (kg).

r is the distance to the center of the earth (meters).

G is a universal gravitational constant.

Dimensional formula:

We know,

Dimensional formula of Universal gravitational constant: $\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]$

Dimensional formula of distance to the center of the earth: $\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]$

Dimensional formula of mass of the earth: $\left[{\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^0\right]$

By substituting the formula in the above equation,

$$\begin{gathered} V_e=\sqrt{\frac{\left[{\mathrm{M}}^{-1+1}{\mathrm{L}}^{3+0}{\mathrm{T}}^{-2+0}\right]}{\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]}} \\ =\sqrt{\frac{\left[{\mathrm{M}}^0{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]}} \end{gathered}$$

By using the law of algorithm $\frac{{\mathrm{a}}^{\mathrm{x}}}{{\mathrm{a}}^{\mathrm{y}}}={\mathrm{a}}^{\mathrm{x}-\mathrm{y}}$,

$$\begin{gathered} V_e=\sqrt{\left[{\mathrm{M}}^{0-0}{\mathrm{L}}^{3-1}{\mathrm{T}}^{-2-0}\right]} \\ =\sqrt{\left[{\mathrm{M}}^0{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]} \end{gathered}$$

By taking the square root, we get

$V_e=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-1}\right]$

Therefore, the dimensional formula of escape velocity will be $\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-1}\right]$.