Question 6cea0

Jul 1, 2015

I think it's $\left[{\text{M"^0"L T}}^{- 1}\right]$.

Explanation:

Dimensional formulas use mass, length, and time to express physical quantities as a propduct of powers of fundamental quantities.

Since velocity is defined as distance divided by time, you can say that

${\text{velocity" = "distance"/"time" = "meters"/"seconds" = "m"/"s" = "m s}}^{- 1}$

This means that the dimensional formula for velocity will have

$\left[{\text{M"^(0) "L T"^(-1)] <=> ["L T}}^{- 1}\right]$

This tells you that

• you have no units of mass in equation, since $\text{M}$ is raised to the power of zero;
• you have units of length, meters, raised to the power of 1;
• you have units of time, seconds, raised to the power of negative 1;

So, take the equation for escape velocity and test to see if these are the units that you're left with for the dimensional formula.

${v}_{\text{esc}} = \sqrt{\frac{2 \cdot G \cdot M}{r}}$, where

$G$ - the universal gravitational constant, equal to $6.67 \cdot {10}^{- 11} {\text{m"^(3)"kg"^(-1)"s}}^{- 2}$
$M$ - the mass of the body, expressed in $\text{kg}$;
$r$ - the distance from the center of the body's mass to a point in space.

If you use units, the equation will look like this

v_"esc" = sqrt( ((2 * "m"^(3)cancel("kg"^(-1))"s"^(-2) * cancel("kg"))/"m")

v_"esc" = sqrt( (2 * "m"^(2) * "s"^(-2))#

As far as units are concerned, this will give you

${v}_{\text{esc" = "m" * "s}}^{- 1}$

As you can see, this confirms the initial dimensional formula, since you have meters multiplied by seconds to the power of negative 1. Therefore, the dimensional formula for escape velocity is

$\left[{\text{M"^(0)"L T"^(-1)] <=> ["L T}}^{- 1}\right]$